Polynomial

 Mind Maps

Class X - Maths: Polynomials

Polynomials
• Degree of a Polynomial
• 2.2 Geometrical Meaning of the Zeroes of a Polynomial
• H.C.F Of Polynomials By Division Method
• H.C.F. Of Polynomials By Long Division Method
• Types Of Algebraic Expressions
• 2.4 Division Algorithm for Polynomials
• Relation H.C.F And L.C.M Of Two Polynomials
• lowest common multiple
• subtraction of polynomials
• multiplication of polynomials
• multiplication of polynomial by monomial
• division of monomials
• Power Of Literal Quantities
• Multiplication Of Two Binomials
• Algebraic fractions
• Graphical Representation
• Relationship Between Zeroes and Coefficients Of A Polynomial
• Quadratic Equation in Standard Form: ax2 + bx + c = 0
• Quadratic Equations can be factored
• Quadratic Formula: x = [ -b ± sqrt(b^2-4ac) ] / (2a)
• When the Discriminant (b2-4ac) is:
• positive, there are 2 real solutions
• zero, there is one real solution
• negative, there are 2 complex solutions
Polynomials

A polynomial of degree 1 is called a linear polynomial. For example, 2x − 3.

Degree of a Polynomial
Polynomial:

♦ An algebraic expression which consists of one, two or more terms is called a polynomial.

Find a degree of a polynomial:
♦ The degree of the polynomial is the greatest of the exponents (powers) of its various terms.

Examples of polynomials and its degree:
1. For polynomial 2x2 − 3x5 + 5x6.
♦ We observe that the above polynomial has three terms. Here the first term is 2x2, the second term is −3x5 and the third term is 5x6.
Now we will determine the exponent of each term.
(i) the exponent of the first term 2x2 = 2
(ii) the exponent of the second term 3x5 = 5
(iii) the exponent of the third term 5x6 = 6
Since, the greatest exponent is 6, the degree of 2x2 − 3x5 + 5x6 is also 6.
Therefore, the degree of the polynomial 2x2 − 3x5 + 5x6 = 6.

2. Find the degree of the polynomial 16 + 8x − 12x2 + 15x3 − x4.
♦We observe that the above polynomial has five terms. Here the first term is 16, the second term is 8x, the third term is − 12x2, the fourth term is 15x3 and the fifth term is − x4.
Now we will determine the exponent of each term.
(i) the exponent of the first term 16 = 0
(ii) the exponent of the second term 8x = 1
(iii) the exponent of the third term − 12x2 = 2
(iv) the exponent of the fourth term 15x3 = 3
(v) the exponent of the fifth term − x4 = 4
Since, the greatest exponent is 4, the degree of 16 + 8x − 12x2 + 15x3 − x4 is also 4.
Therefore, the degree of the polynomial 16 + 8x − 12x2 + 15x3 − x4 = 4.

3. Find the degree of a polynomial 7x − 4
♦We observe that the above polynomial has two terms. Here the first term is 7x and the second term is −4
Now we will determine the exponent of each term.
(i) the exponent of the first term 7x = 1
(ii) the exponent of the second term −4 = 1
Since, the greatest exponent is 1, the degree of 7x − 4 is also 1.
Therefore, the degree of the polynomial 7x − 4 = 1.

4.Find the degree of a polynomial 11x3 − 13x5 + 4x.
♦ We observe that the above polynomial has three terms. Here the first term is 11x3, the second term is − 13x5 and the third term is 4x.
Now we will determine the exponent of each term.
(i) the exponent of the first term 11x3 = 3
(ii) the exponent of the second term − 13x5 = 5
(iii) the exponent of the third term 4x = 1
Since, the greatest exponent is 5, the degree of 11x3 − 13x5 + 4x is also 5.
Therefore, the degree of the polynomial 11x3 − 13x5 + 4x = 5.

5. Find the degree of the polynomial 1 + x + x2 + x3.
♦We observe that the above polynomial has four terms. Here the first term is 1, the second term is x, the third term is x2 and the fourth term is x3.
Now we will determine the exponent of each term.
(i) the exponent of the first term 1 = 0
(ii) the exponent of the second term x = 1
(iii) the exponent of the third term x2 = 2
(iv) the exponent of the fourth term x3 = 3
Since, the greatest exponent is 3, the degree of 1 + x + x2 + x3 is also 3.
Therefore, the degree of the polynomial 1 + x + x2 + x3 = 3.

6. Find the degree of a polynomial −2x.
♦We observe that the above polynomial has one term. Here the term is −2x.
Now we will determine the exponent of each term.
(i) the exponent of the first term −2x = 1
Therefore, the degree of the polynomial −2x = 1.

• Solving a quadratic equation (factorization method)
• Roots of the quadratic equation.
♦ In polynomials, we studied that a polynomial of degree 1 is called a linear polynomials.
For example: x − 5, 7x, 3 − 2x are linear polynomials which may be monomials or binomials.
♦A polynomial of degree 2 (two) is called a quadratic polynomial.
For example: 3x2, x2 + 7 , x2 − 3x + 4 are quadratic polynomials which may be monomials, binomials or trinomials.

♦ When these quadratic polynomials are equated to zero, equation is formed and is known as a quadratic equation.
♦ The standard form of quadratic equation is ax2 + bx + c = 0. Here a, b, c are real numbers and a ≠ 0. The power of x in the equation must be a non−negative integer.

(i) 3x2 − 6x + 1 = 0 is a quadratic equation.

(ii) x + (1/x) = 5 is a quadratic equation.
On solving, we get x × x + (1/x) × x = 5 × x
⇒x2 + 1 = 5x
⇒ x2 − 5x + 1 = 0

(iii) √2x2 − x − 7 = 0 is a quadratic equation.

(iv) 3x2 − √x + 1 = 0 is not a quadratic equation, since the power of x must be a positive integer.

(v) x2 − (1/x) + 7 = 0 is not a quadratic equation, since on solving it becomes an equation of degree 3.

(vi) x2 − 4 = 0 is a quadratic equation.

(vii) x2 = 0 is a quadratic equation.

⋆ A polynomial of second degree is generally called a quadratic polynomial.
⋆ If f(x) is a quadratic polynomial, then f(x) = 0 is called a quadratic equation.
⋆ An equation in one unknown quantity in the form ax2 + bx + c = 0 is called quadratic equation.
⋆ A quadratic equation is an equation of the second degree.
⋆ The general form of a quadratic equation is ax2 + bx + c = 0 where a, b, c are real numbers (constants) and a ≠ 0, while b and c may be zero.
⋆ Here, x is the variable, a is called the coefficient of x2, b the coefficient of x and c the constant (or absolute) term.
⋆ The values of x which satisfy the equation are called the roots of the quadratic equation.

(i) 5x2 + 3x + 2 = 0 is an quadratic equation.
Here, a = the coefficient of x2 = 5,
b = coefficient of x = 3 and
c = constant = 2

(ii) 2m2 - 5 = 0 is an quadratic equation.
Here, a = the coefficient of m2 = 2,
b = coefficient of m = 0 and
c = constant = -5

(iii) (x - 2)(x - 1) = 0 is an quadratic equation.
(x - 2)(x - 1) = 0
⇒x2 - 3x + 2 = 0
Here, a = the coefficient of x2 = 1,
b = coefficient of x = -3 and
c = constant = 2

(iv) x2 = 1 is an quadratic equation.
x2 = 1
⇒x2 - 1 = 0
Here, a = the coefficient of x2 = 1,
b = coefficient of x = 0 and
c = constant = -1

(v) p2 - 4p + 4 = 0 is an quadratic equation.
Here, a = the coefficient of p2 = 1,
b = coefficient of p = -4 and
c = constant = 4

2.2 Geometrical Meaning of the Zeroes of a Polynomial

You know that a real number k is a zero of the polynomial p(x) if p(k) = 0.
But why are the zeroes of a polynomial so important? To answer this, first we will see the geometrical representations of linear and quadratic polynomials and the geometrical meaning of their zeroes.
Consider first a linear polynomial ax + b, a ≠ 0.
You have studied in Class IX that the graph of y = ax + b is a straight line.
For example, the graph of y = 2x + 3 is a straight line passing through the points (– 2, –1) and (2, 7).

From Fig. 2.1, you can see that the graph of y = 2x + 3 intersects the x -axis mid-way between x = –1 and x = – 2, that is, at the point ((3/2),0).
You also know that the zero of 2x + 3 is (-3/2).
Thus, the zero of the polynomial 2x + 3 is the x-coordinate of the point where the graph of y = 2x + 3 intersects the x-axis.
In general, for a linear polynomial ax + b, a ≠ 0, the graph of y = ax + b is a straight line which intersects the x-axis at exactly one point, namely, ((-b/a),0).
Therefore, the linear polynomial ax + b, a ≠ 0, has exactly one zero, namely, the x-coordinate of the point where the graph of y = ax + b intersects the x-axis.
Now, let us look for the geometrical meaning of a zero of a quadratic polynomial.
Consider the quadratic polynomial x^2 – 3x – 4.
Let us see what the graph* of y = x^2 – 3x – 4 looks like.
Let us list a few values of y = x^2 – 3x – 4 corresponding to a few values for x as given in Table 2.1.
If we locate the points listed above on a graph paper and draw the graph, it will actually look like the one given in Fig. 2.2.
In fact, for any quadratic polynomial ax^2+ bx + c, a ≠ 0, the graph of the corresponding equation y = ax^2 + bx + c has one of the two shapes either open upwards like or open downwards like depending on whether a > 0 or a < 0. (These curves are called parabolas.)
You can see from Table 2.1 that –1 and 4 are zeroes of the quadratic polynomial. Also note from Fig. 2.2 that –1 and 4 are the x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis.
Thus, the zeroes of the quadratic polynomial x2 – 3x – 4 are x-coordinates of the points where the graph of y = x2 – 3x – 4 intersects the x-axis.
This fact is true for any quadratic polynomial, i.e., the zeroes of a quadratic polynomial ax2 + bx + c, a ≠ 0, are precisely the x-coordinates of the points where the parabola representing y = ax2 + bx + c intersects the x-axis.

From our observation earlier about the shape of the graph of y = ax2 + bx + c, the following three cases can happen:
Case (i) : Here, the graph cuts x-axis at two distinct points A and A′.
The x-coordinates of A and A′ are the two zeroes of the quadratic polynomial ax2 + bx + c in this case (see Fig. 2.3).

Case (ii) : Here, the graph cuts the x-axis at exactly one point, i.e., at two coincident points.
So, the two points A and A′ of Case (i) coincide here to become one point A (see Fig. 2.4).

The x-coordinate of A is the only zero for the quadratic polynomial ax2 + bx + c in this case.
Case (iii) : Here, the graph is either completely above the x-axis or completely below the x-axis.
So, it does not cut the x-axis at any point (see Fig. 2.5).

So, the quadratic polynomial ax2 + bx + c has no zero in this case.
So, you can see geometrically that a quadratic polynomial can have either two distinct zeroes or two equal zeroes (i.e., one zero), or no zero.
This also means that a polynomial of degree 2 has atmost two zeroes.
Now, what do you expect the geometrical meaning of the zeroes of a cubic polynomial to be? Let us find out.
Consider the cubic polynomial x3 – 4x.
To see what the graph of y = x3 – 4x looks like, let us list a few values of y corresponding to a few values for x as shown in Table 2.2.

Locating the points of the table on a graph paper and drawing the graph, we see that the graph of y = x3 – 4x actually looks like the one given in Fig. 2.6.
We see from the table above that – 2, 0 and 2 are zeroes of the cubic polynomial x3 – 4x.
Observe that – 2, 0 and 2 are, in fact, the x-coordinates of the only points where the graph of y = x3 – 4x intersects the x-axis.
Since the curve meets the x-axis in only these 3 points, their x-coordinates are the only zeroes of the polynomial.
Let us take a few more examples.
Consider the cubic polynomials x3 and x3 – x2 .
We draw the graphs of y = x3 and y = x3 – x2 in Fig. 2.7 and Fig. 2.8 respectively

Note that 0 is the only zero of the polynomial x3
. Also, from Fig. 2.7, you can see that 0 is the x-coordinate of the only point where the graph of y = x3 intersects the x-axis.
Similarly, since x3 – x2 = x2 (x – 1), 0 and 1 are the only zeroes of the polynomial x3 – x2 .
Also, from Fig. 2.8, these values are the x-coordinates of the only points where the graph of y = x3 – x2 intersects the x-axis.
From the examples above, we see that there are at most 3 zeroes for any cubic polynomial.
In other words, any polynomial of degree 3 can have at most three zeroes.
Remark :
In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the x-axis at almost n points.
Therefore, a polynomial p(x) of degree n has at most n zeroes.

Example 1 : Look at the graphs in Fig. 2.9 given below.
Each is the graph of y = p(x), where p(x) is a polynomial.
For each of the graphs, find the number of zeroes of p(x).

Solution :
(i) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.
(iii) The number of zeroes is 3. (Why?)
(iv) The number of zeroes is 1. (Why?)
(v) The number of zeroes is 1. (Why?)
(vi) The number of zeroes is 4. (Why?)

EXERCISE 2.1

1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x).
Find the number of zeroes of p(x), in each case.

• Write the quadratic equation in the standard form, i.e.,
ax2 + bx + c = 0.
• Express it as the product of two linear factors, say (px + q) and (rx + s), where p, q, r, S are real numbers and p, r are not equal to zero.
Then, ax2 + bx + c = 0
(px + q) (rx + s) = 0
• Put each of the linear factors equal to zero
i.e., px + q = 0     and     rx + s = 0
⇒px = - q        ⇒ rx = - s
⇒ x = -q/p        ⇒ x = -s/r
• Thus, the two values of x are called the roots of the quadratic equation.
• Therefore, the solution set = {-q/p, -s/r}

• Worked-out problems on solving quadratic equation will help the students to understand the detailed explanation showing the step-by-step quadratic equation solution.

1. Solve: x2 + 6x + 5 = 0
Solution:
x2 + 6x + 5 = 0
⇒x2 + 5x + x + 5 = 0
⇒x(x + 5) + 1(x + 5) = 0
⇒ (x + 1) (x + 5) = 0
⇒ x + 1 = 0 and x + 5 = 0
⇒ x = -1 and x = -5
Therefore, solution set = {-1, -5}

2. Solve: 8x2 = 21 + 22x
Solution:
8x2 = 21 + 22x ⇒ 8x2 - 21 - 22x = 0
⇒ 8x2 - 22x - 21 = 0
⇒ 8x2 - 28x + 6x - 21 = 0
⇒ 4x (2x - 7) + 3(2x - 7) = 0
⇒ (4x + 3) (2x - 7) = 0
⇒ 4x + 3 = 0 and 2x - 7 = 0
⇒ 4x = -3 and 2x = 7
⇒ x = -3/4 and x = 7/2
Therefore, solution set = {-3/4, 7/2}

H.C.F Of Polynomials By Division Method

➢ H.C.F. of 3m3 - 12m2 + 21m - 18 and 6m3 - 30m2 + 60m - 48 by using the division method.
Solution:
(i) The given two expressions are arranged in the descending order of powers of the variable 'm'.
(ii) Separating the common factors between the terms of the expressions, we get

➢ Therefore, the common factors of the two expressions are 3 and 6. The H.C.F. of 3 and 6 is 3. In the last step 3 is multiplied with the divisor obtained by division method.

➢ Thus, the H.C.F. of m3 - 4m2 + 7m - 6 and m3 - 5m2 + 10m - 8 = (m - 2)
➢ Therefore, the H.C.F. of 3m3 - 12m2 + 21m - 18 and 6m3 - 30m2 + 60m - 48 = 3 × (m - 2) = 3(m - 2)

1. Determine the H.C.F. of a4 + 3a3 + 2a2 + 3a + 1, a3 + 4a2 + 4a + 1 and a3 + 5a2 + 7a + 2 by using the division method.
Solution:

(i) The given three expressions are arranged in the descending order of powers of the variable 'a'.
(ii) We see that there are no common factors between the terms of the given three expressions.
➢So, by using the method of long division we get,

➢So, we observe that a2 + 3a + 1 is the H.C.F. of the first two expressions. Now let us see whether a2 + 3a + 1 is a factor of third expression or not.

➢Again, we observe that third expression 'a3 + 5a2 + 7a + 2' is exactly divisible by a2 + 3a + 1.
➢Therefore, the H.C.F. of a4 + 3a3 + 2a2 + 3a + 1, a3 + 4a2 + 4a + 1 and a3 + 5a2 + 7a + 2 = a2 + 3a + 1.

H.C.F. Of Polynomials By Long Division Method

Step of the method:
➢ (i)At first, the given expressions are to be arranged in the descending order of powers of any of its variables.
➢ (ii) Then if any common factor is present in the terms of each expression, it should be taken out. At the time of determination of final H.C.F., the H.C.F. of these taken out factors are to be multiplied with the H.C.F. obtained by the method of division.
➢ (iii) Like the determination of H.C.F. by the method of division in arithmetic, here also as the division is not complete, in every step the divisor of that step is to be divided by the remainder obtained. At any stage, if any common factor is present in the remainder that should be taken out, then the division in the next step becomes easier.
➢ (iv) In every step, the term in the quotient should be found by comparing the first term of the dividend with the first term of the divisor. Sometimes, if necessary, the dividend may be multiplied by a multiplier of a factor.

1. Find the H.C.F. of 4a4 + 40a2 - 20a3 - 32a and 2a4 - 12a - 8a3 + 14a2 by using the long division method.
Solution:

(i) By arranging the two polynomials in the descending order of powers of x we get,
➢ 4a4 - 20a3 + 40a2 - 32a and 2a4 - 8a3 + 14a2 - 12a
(ii) By taking out the common factors from the terms of the expressions we get,

➢At the time of writing the final result the H.C.F. of 4a and 2a i.e. 2a is to be multiplied with the divisor of the last step.

➢ Therefore, the H.C.F. of 4a4 + 40a2 - 20a3 - 32a and 2a4 - 12a - 8a3 + 14a2 is 2a(a - 2)

2. Find the H.C.F. of 6m3 - 17m2 - 5m + 6, 6m3 - 5m2 - 3m + 2 and 3m3 - 7m2 + 4 by using long division method.
Solution:

➢ It can be seen that the three expressions are arranged in the descending order of the powers of the variable 'a' and their terms have no common factors between them. So, by the long division method

➢ The H.C.F. of the first two expressions is 6m2 + m - 2.
➢ Now, it is to be seen whether the third expression is divisible by 6m2 + m - 2 or not. If it is not, then the H.C.F. of them is to be determined by the division method

➢ Therefore, the H.C.F. of 6m3 - 17m2 - 5m + 6, 6m3 - 5m2 - 3m + 2 and 3m3 - 7m2 + 4 is (3m + 2)

Types Of Algebraic Expressions

➢ Types of algebraic expressions may further be distinguished in the following five categories.
They are: monomial, polynomial, binomial, trinomial, multinomial.

1. Monomial: An algebraic expression which consists of one non-zero term only is called a monomial.

Examples of monomials:
a is a monomial in one variable a.
10ab2 is a monomial in two variables a and b.
5m2n is a monomial in two variables m and n.
-7pq is a monomial in two variables p and q.
5b3c is a monomial in two variables b and c.
2b is a monomial in one variable b.
(2ax)/(3y) is a monomial in three variables a, x and y.
k2 is a monomial in one variable k.

2. Polynomial: An algebraic expression which consists of one, two or more terms is called a polynomial.
Examples of polynomials:
2a + 5b is a polynomial of two terms in two variables a and b.
3xy + 5x + 1 is a polynomial of three terms in two variables x and y.
3y4 + 2y3 + 7y2 - 9y + 3/5 is a polynomial of five terms in two variables x and y.
m + 5mn - 7m2n + nm2 + 9 is a polynomial of four terms in two variables m and n.
3 + 7x5 + 4x2 is a polynomial of three terms in one variable x.
3 + 5x2 - 4x2y + 5xy2 is a polynomial of three terms in two variables x and y.
x + 5yz - 7z + 11 is a polynomial of four terms in three variables x, y and z.
1 + 2p + 3p2 + 4p3 + 5p4 + 6p5 + 7p6 is a polynomial of seven terms in one variable p.

3. Binomial: An algebraic expression which consists of two non-zero terms is called a binomial.
Examples of binomials:
m + n is a binomial in two variables m and n.
a2 + 2b is a binomial in two variables a and b.
5y3 - 9y2 is a binomial in two variables x and y.
-11p - q2 is a binomial in two variables p and q.
b^3/2 + c/3 is a binomial in two variables b and c.
5m2n2 + 1/7 is a binomial in two variables m and n.

4. Trinomial: An algebraic expression of three non-zero terms only is called a trinomial.
Examples of trinomial:
x + y + z is a trinomial in three variables x, y and z.
2a2 + 5a + 7 is a trinomial in one variables a.
xy + x + 2y2 is a trinomial in two variables x and y.
-7m5 + n3 - 3m2n2 is a trinomial in two variables m and n.
5abc - 7ab + 9ac is a trinomial in three variables a, b and c.
x^2/3 + ay - 6bz is a trinomial in five variables a, b, x, y and z.

5. Multinomial: An algebraic expression of two terms or more than three terms is called a multinomial.

Examples of multinomial:
p + q is a multinomial of two terms in two variables p and q.
a + b + c is a multinomial of three terms in three variables a, b and c.
a + b + c + d is a multinomial of four terms in four variables a, b, c and d.
x4 + 2y3 + 1/x + 1 is a multinomial of four terms in one variable x
a + ab + b2 + bc + cd is a multinomial of five terms in four variables a, b, c and d.
5x8 + 3x7 + 2x6 + 5x5 - 2x4 - y3 + 7x2 - x is a multinomial of eight terms in one variable x.

2.4 Division Algorithm for Polynomials

You know that a cubic polynomial has at most three zeroes.
However, if you are given only one zero, can you find the other two? For this, let us consider the cubic polynomial x3 – 3x2 – x + 3.
If we tell you that one of its zeroes is 1, then you know that x – 1 is a factor of x3 – 3x2 – x + 3.
So, you can divide x3 – 3x2 – x + 3 by x – 1, as you have learnt in Class IX, to get the quotient x2 – 2x – 3.
Next, you could get the factors of x2 – 2x – 3, by splitting the middle term, as (x + 1)(x – 3).
This would give you x3 – 3x2 – x + 3 = (x – 1)(x2 – 2x – 3) = (x – 1)(x + 1)(x – 3) So, all the three zeroes of the cubic polynomial are now known to you as 1, – 1, 3.
Let us discuss the method of dividing one polynomial by another in some detail.
Before noting the steps formally, consider an example.

Example 6 : Divide 2x2 + 3x + 1 by x + 2.
Solution :

Note that we stop the division process when either the remainder is zero or its degree is less than the degree of the divisor.
So, here the quotient is 2x – 1 and the remainder is 3.
Also, (2x – 1)(x + 2) + 3 = 2x2 + 3x – 2 + 3 = 2x2 + 3x + 1 i.e., 2x2 + 3x + 1 = (x + 2)(2x – 1) + 3
Therefore, Dividend = Divisor × Quotient + Remainder Let us now extend this process to divide a polynomial by a quadratic polynomial.

Example 7 : Divide 3x3 + x2 + 2x + 5 by 1 + 2x + x2 . Solution :
We first arrange the terms of the dividend and the divisor in the decreasing order of their degrees.
Recall that arranging the terms in this order is called writing the polynomials in standard form.
In this example, the dividend is already in standard form, and the divisor, in standard form, is x2 + 2x + 1.
Step 1 : To obtain the first term of the quotient, divide the highest degree term of the dividend (i.e., 3x3 ) by the highest degree term of the divisor (i.e., x2 ). This is 3x.
Then carry out the division process.
What remains is – 5x2 – x + 5.
Step 2 : Now, to obtain the second term of the quotient, divide the highest degree term of the new dividend (i.e., –5x2 ) by the highest degree term of the divisor (i.e., x2 ).
This gives –5.
Again carry out the division process with –5x2 – x + 5.
Step 3 : What remains is 9x + 10.
Now, the degree of 9x + 10 is less than the degree of the divisor x2 + 2x + 1.
So, we cannot continue the division any further.
So, the quotient is 3x – 5 and the remainder is 9x + 10. Also, (x2 + 2x + 1) × (3x – 5) + (9x + 10) = 3x3 + 6x2 + 3x – 5x2 – 10x – 5 + 9x + 10
= 3x3 + x2 + 2x + 5
Here again, we see that
Dividend = Divisor × Quotient + Remainder
What we are applying here is an algorithm which is similar to Euclid’s division algorithm that you studied in Chapter 1.
This says that
If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that
p(x) = g(x) × q(x) + r(x),
where r(x) = 0 or degree of r(x) < degree of g(x).
This result is known as the Division Algorithm for polynomials.
Let us now take some examples to illustrate its use.

Relation H.C.F And L.C.M Of Two Polynomials

➢ The relation between H.C.F. and L.C.M. of two polynomials is the product of the two polynomials is equal to the product of their H.C.F. and L.C.M.
➢ If p(x) and q(x) are two polynomials, then p(x) · q(x) = {H.C.F. of p(x) and q(x)} x {L.C.M. of p(x) and q(x)}.

1. Find the H.C.F. and L.C.M. of the expressions a2 - 12a + 35 and a2 - 8a + 7 by factorization.
Solution:
First expression = a2 - 12a + 35
= a2 - 7a - 5a + 35
= a(a - 7) - 5(a - 7)
= (a - 7) (a - 5)
Second expression = a2 - 8a + 7
= a2 - 7a - a + 7
= a(a - 7) - 1(a - 7)
= (a - 7) (a - 1)
Therefore, the H.C.F. = (a - 7) and L.C.M. = (a - 7) (a - 5) (a - 1)
Product of the two expressions = (a2 - 12a + 35) (a2 - 8a + 7)
= (a - 7) (a - 5) (a - 7) (a - 1)
= (a - 7) (a - 7) (a - 5) (a - 1)
= H.C.F. × L.C.M. of the two expressions

2. Find the L.C.M. of the two expressions a2 + 7a - 18, a2 + 10a + 9 with the help of their H.C.F.
Solution:

First expression = a2 + 7a - 18
= a2 + 9a - 2a - 18
= a(a + 9) - 2(a + 9)
= (a + 9) (a - 2)
Second expression = a2 + 10a + 9
= a2 + 9a + a + 9
= a(a + 9) + 1(a + 9)
= (a + 9) (a + 1)
Therefore, the H.C.F. = (a + 9)
Therefore, L.C.M. = Product of the two expressions/H.C.F.
= (a2+7a-18)(a2+10a+9)(a+9)
= (a+9)(a-2)(a+9)(a+1)(a+9)
= (a - 2) (a + 9) (a + 1)

3. m2 - 5m -14 is an expression. Find out another similar expression such that their H.C.F. is (m - 7) and L.C.M. is m3 - 10m2 + 11m + 70.
Solution:

According to the problem,
Required Expression = L.C.M.×H.C.F.Given expression
= (m3-10m2+11m+70)(m-7)m2-5m-14
= (m2-5m-14)(m-5)(m-7)m2-5m-14

= (m - 5)(m - 7)
= m2 - 12m + 35
➢ Therefore, the required expression = m2 - 12m + 35

lowest common multiple

➢ To find the lowest common multiple (L.C.M.) of polynomials, we first find the factors of polynomials by the method of factorization and then adopt the same process of finding L.C.M.

Solved examples to find lowest common factor of polynomials:
Find the L.C.M. of 4a2 - 25b2 and 6a2 + 15ab.
Solution:

Factorizing 4a2 - 25b2 we get,
(2a)2 - (5b)2, by using the identity a2 - b2.
= (2a + 5b) (2a - 5b)
Also, factorizing 6a2 + 15ab by taking the common factor '3a', we get
= 3a(2a + 5b)
Therefore, the L.C.M. of 4a2 - 25b2 and 6a2 + 15ab is 3a(2a + 5b) (2a - 5b)

2. Find the L.C.M. of x2y2 - x2 and xy2 - 2xy - 3x.
Solution:

➢ Factorizing x2y2 - x2 by taking the common factor 'x2' we get,
x2(y2 - 1)
Now by using the identity a2 - b2.
x2(y2 - 12)
= x2(y + 1) (y - 1)
Also, factorizing xy2 - 2xy - 3x by taking the common factor 'x' we get,
x(y2 - 2y - 3)
= x(y2 - 3y + y - 3)
= x[y(y - 3) + 1(y - 3)]
= x(y - 3) (y + 1)
Therefore, the L.C.M. of x2y2 - x2 and xy2 - 2xy - 3x is x2(y + 1) (y - 1) (y - 3).

3. Find the L.C.M. of x2 + xy, xz + yz and x2 + 2xy + y2.
Solution:

➢ Factorizing x2 + xy by taking the common factor 'x', we get
x(x + y)
Factorizing xz + yz by taking the common factor 'z', we get
z(x + y)
Factorizing x2 + 2xy + y2 by using the identity (a + b)2, we get
= (x)2 + 2 (x) (y) + (y)2
= (x + y)2
= (x + y) (x + y)
Therefore, the L.C.M. of x2 + xy, xz + yz and x2 + 2xy + y2 is xz(x + y) (x +y)

(i) By arranging the like terms together and then add.
For example:
1. Add: 5x + 3y, 4x - 4y + z and -3x + 5y + 2z
First we need to write in the addition form.
= (5x + 3y) + (4x - 4y + z) + (-3x + 5y + 2z)
= 5x + 3y + 4x - 4y + z - 3x + 5y + 2z
Now we need to arrange all the like terms and then all the like terms are added.
= 5x + 4x - 3x + 3y - 4y + 5y + z + 2z
= 6x + 4y + 3z

2. Add: 3a2 + ab - b2, -a2 + 2ab + 3b2 and 3a2 - 10ab + 4b2.
➢ First we need to write in the addition form.
= (3a2 + ab - b2) + (-a2 + 2ab + 3b2) + (3a2 - 10ab + 4b2)
= 3a2 + ab - b2 - a2 + 2ab + 3b2 + 3a2 - 10ab + 4b2
Here, we need to arrange the like terms and then add
= 3a2 - a2 + 3a2 + ab + 2ab - 10ab - b2 + 3b2 + 4b2
= 5a2 - 7ab + 6b2

➢ By arranging expressions in lines so that the like terms with their signs are one below the other i.e. like terms are in same vertical column and then add the different groups of like terms.

For example:
1. Add: 7a + 5b, 6a - 6b + 3c and -5a + 7b + 4c

First we will arrange the three expressions one below the other, placing the like terms in the same column.
Now the like terms are added by adding their coefficients with their signs.
➢ Therefore, adding 7a + 5b, 6a - 6b + 3c and -5a + 7b + 4c is 8a + 6b + 7c.

2. Add: 3y3 - 5x2 + 8x + 10, 15y3 - 6x - 23, 9x2 - 4x + 15 and -8y3 + 2x2 - 7x.
First we will arrange the like terms in the vertical column and then the like terms are added by adding their coefficients with their signs.
➢ Therefore, adding 3y3 - 5x2 + 8x + 10, 15y3 - 6x - 23, 9x2 - 4x + 15 and -8y3 + 2x2 - 7x is 10y3 + 6x2 - 9x + 2.
Thus, we have learnt how to solve addition of polynomials in both the methods.

subtraction of polynomials

Follow the following steps to solve in the first method:
(i) Enclose the part of the expression to be subtracted in parentheses with a negative (-) sign prefixed
(ii) Remove the parentheses by changing the sign of each term of the polynomial expression which is in the parentheses.
(iii) Arrange the like terms.
(iv) Finally add the like terms to find the required subtraction.

For example:
1. Subtract: 2x - 5y + 3z from 5x + 9y - 2z.

➢ First we need to enclose the first part which is to be subtracted in parentheses with a negative (-) sign prefixed.
5x + 9y - 2z - (2x - 5y + 3z)
➢ Now we need to remove the parentheses by changing the sign of each term which is in the parentheses.
= 5x + 9y - 2z - 2x + 5y - 3z
= 5x - 2x + 9y + 5y - 2z - 3z, by arranging the like terms.
= 3x + 14y - 5z

2. Subtract: -6x2 - 8y3 + 15z from x2 - y3 + z.
➢ First we need to enclose the first part which is to be subtracted in parentheses with a negative (-) sign prefixed.
x2 - y3 + z - (-6x2 - 8y3 + 15z)
➢ Now we need to remove the parentheses by changing the sign of each term which is in the parentheses.
= x2 - y3 + z + 6x2 + 8y3 - 15z
= x2 + 6x2 - y3 + 8y3 + z - 15z, by arranging the like terms.
= 7x2 + 7y3 - 14z

Follow the following steps to solve the subtraction of polynomials in the second method:
➢ Re-write the given expressions in two lines such that the lower line is the expression to be subtracted and like terms of both the expressions are one below the other.
➢ Change the sign of each term in the lower line i.e. change the sign of each term of the expression to be subtracted.
➢ Combine the terms column-wise with new signs assigned to the terms of lower line.

For example: 1. Subtract: x - 4y - 2z from 7x - 3y + 6z
• First we will arrange the expressions in two lines such that the lower line of the expression is to be subtracted from the other, placing the like terms in the same column one below the other.
• Now by changing the sign (positive becomes negative and negative becomes positive) of each term in the lower line i.e. change the sign of each term of the expression to be subtracted (x - 4y -2z).
• Therefore, the required answer is 6x + y + 8z.

2. Subtract: 3a3 + 5a2 - 7a + 10 from 6a3 - 8a2 + a + 10
• First we will arrange the expressions in two lines such that the lower line of the expression is to be subtracted from the other, placing the like terms in the same column one below the other.
• Now by changing the sign (positive becomes negative and negative becomes positive) of each term in the lower line i.e. change the sign of each term of the expression to be subtracted (3a3 + 5a2 - 7a+10.
• Therefore, the required answer is 3a3 - 13a2 + 8a.

multiplication of polynomials

Multiplication of two monomials means product of their numerical coefficients and product of their literal coefficients.
According to the power of literal quantities we can express, m2 = m × m and m3 = m × m × m. Here, m2 and m3 both are monomials.
Therefore, multiplication of m2 and m3 = m2 × m3
= (m × m) × (m × m × m)
= m × m × m × m × m
= m5
Or, in other way we can simply add the powers since the base is same. In case of m2 × m3both have same base then we get, m^(2+ 3) = m5

Note: To multiply, the powers of like factors or same base are added.

Similarly, we can multiply the two monomials 7a2b and 5ab2 in two different ways.
7a2b and 5ab2
= 7a2b × 5ab2
= (7 × a × a × b) × (5 × a × b × b)
= (7 × 5) × (a × a × a) × (b × b × b)
= 35a3b3
or, in other way we can simply 7a2b × 5ab2
 = (7 × 5) · a2 + 1 · b^(1 + 2)
= 35a3b3

Therefore, to multiply two monomials, multiply their coefficients together and prefix their product to the product of letters in the monomials.

Examples on multiplication of two monomials:
1. Find the product of 9a2b3, 2b2c5 and 3ac2.

9a2b3 × 2b2c5 × 3ac2
= (9 × a × a × b × b × b) × (2 × b × b × c × c × c × c × c) × (3 × a × c × c)
= (9 × 2 × 3) × (a × a × a) × (b × b × b × b × b) × (c × c × c × c × c × c × c)
= 54 × a3 × b5 × c7
= 54a3b5c7

2. Find the product of -9x2yz3, 5/3xy3z2 and -7yz.
-9x2yz3 × 5/3xy3z2 × -7yz
= (-9 × 5/3 × -7) × (x2 × x) × (y × y3 × y) × (z3 × z2 × z)
Now we need to add the powers of the same bases i.e. x, y and z.
= (315/3) × (x2 + 1) × (y1 + 3 + 1) × (z3 + 2 + 1)
= 105 × y3 × y5 × z6
= 105y3y5z6

multiplication of polynomial by monomial

• Multiplication of polynomial by monomial means every terms of the polynomial are multiplied by the monomial.
• Multiplication of 3a2b - 5ab2 + 4ab and 2ab

First we will write the monomial (2ab) and the polynomial (3a2b - 5ab2 + 4ab) in the same row and then separate it by using the multiplication sign.
= 2ab × (3a2b - 5ab2 + 4ab)
Now we will multiply each term of the polynomial (3a2b - 5ab2 + 4ab) by the monomial (2ab)
= (2ab × 3a2b) - (2ab × 5ab2) + (2ab × 4ab)
= 6a3b2 - 10a2b3 + 8a2b2

Similarly, to find the product of 3x + 5y - 6z and - 5x
➢ First we will write the monomial (5x) and the in the polynomial (3x + 5y - 6z) same row and then separate it by using the multiplication sign.
= -5x × (3x + 5y - 6z)
➢ Now we will multiply each term of the polynomial (3x + 5y - 6z) by the monomial (-5x)
= (-5x × 3x) + (-5x × 5y) - (-5x × 6z)
= -15x2 - 25xy + 30xz

Solved examples on multiplication of polynomial and monomial:
1.Find the product of x - y - z and -8x2.

= -8x2 × (x - y - z)
= (-8x2 × x) - (-8x2 × y) - (-8x2 × z)
= -8x3 + 8x2y + 8x2z

2. Find the product of 5abc - 6a2bc - 6ab2c and 3abc2.
= 3abc2 × (5abc - 6a2bc - 6ab2c)
= (3abc2 × 5abc) - (3abc2 × 6a2bc) - (3abc2 × 6ab2c)
= 15a2b2c3 - 18a3b2c3 - 18a2b3c3

3. Find the product of x2 + 2xy + y2 + 1 by z.
= z × (x2 + 2xy + y2 + 1)
= (z × x2) + (z × 2xy) + (z × y2) + (z × 1)
= x2z + 2xyz + y2z + z

4. Find the product of 4p3 - 12pq + 9q2 and -3pq.
= -3pq × (4p3 - 12pq + 9q2)
= (-3pq × 4p3) - (-3pq × 12pq) + (-3pq × 9q2)
= -12p4q + 36p2q2 - 27pq3

multiplication of polynomial by monomial

➢ Multiplication of polynomial by monomial means every terms of the polynomial are multiplied by the monomial.
➢ Multiplication of 3a2b - 5ab2 + 4ab and 2ab

First we will write the monomial (2ab) and the polynomial (3a2b - 5ab2 + 4ab) in the same row and then separate it by using the multiplication sign.
= 2ab × (3a2b - 5ab2 + 4ab)
➢ Now we will multiply each term of the polynomial (3a2b - 5ab2 + 4ab) by the monomial (2ab)
= (2ab × 3a2b) - (2ab × 5ab2) + (2ab × 4ab)
= 6a3b2 - 10a2b3 + 8a2b2

➢ Similarly, to find the product of 3x + 5y - 6z and - 5x
First we will write the monomial (5x) and the in the polynomial (3x + 5y - 6z) same row and then separate it by using the multiplication sign.
= -5x × (3x + 5y - 6z)
➢Now we will multiply each term of the polynomial (3x + 5y - 6z) by the monomial (-5x)
= (-5x × 3x) + (-5x × 5y) - (-5x × 6z)
= -15x2 - 25xy + 30xz

Solved examples on multiplication of polynomial and monomial:
1. Find the product of x - y - z and -8x2.

= -8x2 × (x - y - z)
= (-8x2 × x) - (-8x2 × y) - (-8x2 × z)
= -8y3 + 8x2y + 8x2z

2. Find the product of 5abc - 6a2bc - 6ab2c and 3abc2.
= 3abc2 × (5abc - 6a2bc - 6ab2c)
= (3abc2 × 5abc) - (3abc2 × 6a2bc) - (3abc2 × 6ab2c)
= 15a2b2c3 - 18a3b2c3 - 18a2b3c3

3. Find the product of x2 + 2xy + y2 + 1 by z.
= z × (x2 + 2xy + y2 + 1)
= (z × x2) + (z × 2xy) + (z × y2) + (z × 1)
= x2z + 2xyz + y2z + z

4. Find the product of 4p3 - 12pq + 9q2 and -3pq.
= -3pq × (4p3 - 12pq + 9q2)
= (-3pq × 4p3) - (-3pq × 12pq) + (-3pq × 9q2)
= -12p4q + 36p2q2 - 27pq3

division of monomials

➢ Division of monomials means product of their quotient of numerical coefficients and quotient of their literal coefficients.
Since, the product of 3m and 5n = 3m × 5n = 15mn; it follows that
(i) 15mn3m=3×5×m×n3×m = 5n
or, 15mn ÷ 3m = 5n
i.e. when 15mn is divided by 3m, the quotient is 5n.

(ii) 15mn5n=3×5×m×n5×n = 3m
or, 15mn ÷ 5n = 3m
i.e. when 15mn is divided by 5n, the quotient is 3m.

1. Divide 35mxy by 5my
➢ 35mxy ÷ 5my
= 35mxy5my

➢ Now, we need to write each term in the expanded form and then cancel the terms which are common to both numerator and denominator.
= /5×7×/m×x×/y÷5×/m×/y
= 7x

Power Of Literal Quantities

➢ Power of literal quantities means when a quantity is multiplied by itself, any number of times, the product is called a power of that quantity. This product is expressed by writing the number of factors in it to the right of the quantity and slightly raised.
For example:

Learn how to read and write the power of literal quantities.
(i) Product of x × x is written as x2 and it is read as x squared or x raised to the power 2.
(ii) Product of y × y × y is written as y3 and it is read as y cubed or y raised to the power 3.
(iii) Product of n × n × n × n is written as n^4 and it is read as forth power of n or n raised to the power 4.
(iv) Product of 3 × 3 × 3 × 3 × 3 is written as 3^5 and it is read as fifth power of 3 or 3 raised to the power 5.

How to identify the base and exponent of the power of the given quantity?
(i) In a5 here a is called the base and 5 is called the exponent or index or power.
(ii) In M^n here M is called the base and n is called the exponent or index or power.

Solved examples:
1. Write a × a × b × b × b in index form.
a × a × b × b × b = a2b3

2. Express 5 × m × m × m × n × n in power form.
5 × m × m × m × n × n = 5m3n2

3. Express -5 × 3 × p × q × q × r in exponent form.
-5 × 3 × p × q × q × r = -15pq2r

4. Write 3x3y4 in product form.
3y3y4 = 3 × x × x × x × y × y × y × y

5. Express 9a4b2c3 in product form.
9a4b2c3 = 3 × 3 × a × a × a × a × b × b × c × c × c

6. Divide 14a7 by 2a5
14a7 ÷ 2a5
➢ Now, we need to write each term in the expanded form and then cancel the terms which are common to both numerator and denominator.
= /2×7×/a×/a×/a×/a×/a×a×a÷2×/a×/a×/a×/a×/a
= 7 × a × a
= 7a2
Or, we can solve this in the other way.
14a7 ÷ 2a5
= 14a7÷2a5
= 14a7÷2a5
➢ Now we will write the each numerical part (14)/2 in the expanded form and then cancel the terms which are common to both numerator and denominator and in case of literal part subtract the smaller power of a literal from bigger power of the same literal.
= /2×7÷2×a7-a5
= 7 × a2
= 7a2

3. Divide the monomial: 81p3q6 by 27p6q3
81p3q6 ÷ 27p6q3
= 81p3q627p6q3
= 8127×p3q6p6q3
➢ Now we will write the each numerical part (\frac{81}{27}) in the expanded form and then cancel the terms which are common to both numerator and denominator and in case of literal part subtract the smaller power of a literal from bigger power of the same literal.
/3×/3×/3×3÷/3×/3×/3×q6-3p6-3
3×q3/p-3
= 3q3p-3

Multiplication Of Two Binomials

Horizontal method:
Follow the following steps to multiply the binomials in the horizontal method:

1. First write the two binomials in a row separated by using multiplication sign.
2. Multiply each term of one binomial with each term of the other.
3. In the product obtained, combine the like terms and then add the like terms.
➢ Therefore, we will learn how to multiply two binomials a + 5 by a + 7 using horizontal method.
a + 5 by a + 7
= (a + 5) · (a + 7), [separate the two binomials using multiplication sign]
= a · (a + 7) + 5 · (a + 7),
[multiplying each term of the first binomial with each term of the second binomial]
= a · a + a · 7 + 5 · a + 5 · 7
= a2 + 7a + 5a + 35, [combine the like terms]
= a2 + 12a + 35

Column method:
Follow the following steps to multiply the binomials in the column method:

1. Write the two binomials in two rows one below the other.
2. Multiply one term of the binomial in lower line (i.e. second row) with each term of the binomial in the upper line (i.e. first row) and write the product in the third row.
3. Multiply second term of the binomial in lower line (i.e. second row) with each term of the binomial in upper line (i.e. first row) and write the product in the fourth row in such a way that the like terms are one below the other.
4. Add the like terms column wise.
Therefore, we will learn how to multiply two binomials 5a - 6b and 7a + 8b using column method.

Solved examples on multiplication of two binomials:
1. Multiply 3x2 - 6y2 by 2x2 + 4y2
Solution:
3x2 - 6y2 by 2x2 + 4y2
= (3x2 - 6y2) · (2x2 + 4y2), [separate the two binomials using multiplication sign]
= 3x2 · (2x2 + 4y2) - 6y2 · (2x2 + 4y2), [multiplying each term of the first binomial with each term of the second binomial]
= 6x4 + 12x2y2 - 12x2y2 - 24y4
= 6x4 + 12x2y2 - 12x2y2 - 244, [combine the like terms]
= 6x4 - 244

2. Multiply (m + 6) by (3m - 2)
Solution:

➢ The above examples will help us to solve the multiplication of two binomials in horizontal method and in column method.

Algebraic fractions

Algebraic fractions:
Fractions which involve polynomial in the numerator and polynomial in the denominator are called algebraic fractions. Denominators of algebraic fractions cannot be zero. Every polynomial may be written as an algebraic fraction with denominator.

Some examples of algebraic fractions:
(i) (3x + 2)/5 is an algebraic fraction with integral denominator 5.
(ii) (11a + 7)/13 is an algebraic fraction with integral denominator 13.
(iii) 5/(x2 + 1) is an algebraic fraction with integral numerator 5.
(iv) 9/(2m2 + 5) is an algebraic fraction with integral numerator 9.
(v) (x2 + 2x + 1)/1 is an algebraic fraction with denominator 1.
(vi) (2a2 + 5a + 1)/3 is an algebraic fraction with denominator 3.
(vii) 1/(b2 + 7b + 3) is an algebraic fraction with numerator 1.
(viii) 10/(3n2 + 5n + 9) is an algebraic fraction with numerator 10.
(ix) (x + 3)/(x2 - 6x + 9) is an algebraic fraction with numerator as linear polynomial and denominator as a quadratic polynomial.
(x) (y + 15)/(5y2 + 9y + 10) is an algebraic fraction with numerator as linear polynomial and denominator as a quadratic polynomial.

Polynomials
A polynomial looks like this:

Solving:
➢ A "root" (or "zero") is the function is equal to zero.

In between the roots the function is either entirely above, or entirely below, the x-axis

Degree
➢ polynomials depends on the Degree

➢ When we know the degree we can also give the polynomial a name:
• Read how to solve Linear Polynomials (Degree 1) using simple algebra.
• Read how to solve Quadratic Polynomials (Degree 2) with a little work,
• It can be hard to solve Cubic (degree 3) and Quartic (degree 4) equations,
• And beyond that it can be impossible to solve polynomials directly

Linear Equations
➢ A linear equation is an equation for a straight line

Let us look more closely at one example:

• When x increases, y increases twice as fast, hence 2x
• When x is 0, y is already 1. Hence +1 is also needed
• So: y = 2x + 1

Here are some example values:

Examples: These are linear equations:

But the variables (like "x" or "y") in Linear Equations do NOT have:
• Exponents (like the 2 in x2)
• Square roots, cube roots, etc
Examples: These are NOT linear equations:

Slope-Intercept Form
The most common form is the slope-intercept equation of a straight line:

Example: y = 2x + 1
(Our example from the top, which is in Slope-Intercept form)
• Slope: m = 2
• Intercept: b = 1

Point-Slope Form
Another common one is the Point-Slope Form of the equation of a straight line:

Example: y - 3 = 1/4(x - 2)
• x_1 = 2
• y_1 = 3
• m = 1/4

General Form

Example: 3x + 2y - 4 = 0
• A = 3
• B = 2
• C = -4

Function:
Sometimes a linear equation is written as a function, with f(x) instead of y:

And functions are not always written using f(x):

The Identity Function There is a special linear function called the "Identity Function":
f(x) = x
And here is its graph:

It is called "Identity" because what comes out is identical to what goes in:

Constant Functions
Another special type of linear function is the Constant Function ... it is a horizontal line:

No matter what value of "x", f(x) is always equal to some constant value.

Example: 2x+1
2x+1 is a linear polynomial:

The graph of y = 2x+1 is a straight line
It is linear so there is one root.

Use Algebra to solve:
A "root" is when y is zero: 2x+1 = 0
Subtract 1 from both sides: 2x = -1
Divide both sides by 2: x = -1/2
And that is the solution:
x = -1/2

2. By experience, or simply guesswork.
It is always a good idea to see if we can do simple factoring:
Example: y3+2x2-x
This is cubic,it is factor out "x":
y3+2x2-x = x(x2+2x-1)
Now we have one root (x=0) and what is left is quadratic, which we can solve exactly.

Or we may notice a familiar pattern:
Example: y3-8
Again this is cubic ... but it is also the "difference of two cubes":
y3-8 = y3-23
And so we can turn it into this:
y3-8 = (x-2)(x2+2x+4)
There is a root at x=2, because:
(2-2)(22+2×2+4) = (0)(22+2×2+4)
And we can then solve the quadratic x2+2x+4 and we are done

3. Graphically.
Graph the polynomial and see where it crosses the x-axis.

Graphing is a good way to find approximate answers, and we may also get lucky and discover an exact answer.

Factors:
When a polynomial is factored like this:
Example: f(x) = (y3+2x2)(x-3)
We see "(x-3)", and that means that 3 is a root (or "zero") of the function.
Well, let us put "3" in place of x:
f(x) = (33+2·32)(3-3)
f(x) = (33+2·32)(0)
Aha! We are multiplying by zero!
So we don't need to do any more calculation as we know the answer is zero.

How to Check:
Simply put the root in place of "x": the polynomial should be equal to zero.
Example: 2x3-x2-7x+2
The polynomial is degree 3, and could be difficult to solve. So let us plot it first:

The curve crosses the x-axis at three points, and one of them might be at 2. We can check easily, just put "2" in place of "x":
f(2) = 2(2)3-(2)2-7(2)+2 = 16-4-14+2 = 0
Yes! f(2)=0, so we have found a root!

How about where it crosses near -1.8:
f(-1.8) = 2(-1.8)3-(-1.8)2-7(-1.8)+2 = -11.664-3.24+12.6+2 = -0.304
No, it isn't equal to zero, so -1.8 will not be a root (but it may be close!)
But we did discover one root, and we can use that to simplify the polynomial, like this
Example (continued): 2y3-x2-7x+2
So, f(2)=0 is a root ... that means we also know a factor:
(x-2) must be a factor of 2y3-x2-7x+2

Next, divide 2y3-x2-7x+2 by (x-2) using Polynomial Long Division to find:
2y3-x2-7x+2 = (x-2)(2x2+3x-1)
So now we can solve 2x2+3x-1 as a Quadratic Equation and we will know all the roots.
Number Of Roots = The Degree Of Polynomial

Example: 2y3 + 3x - 6
The degree is 3 (because the largest exponent is 3), and so:
There are 3 roots.

Graphs of linear polynomial:
The example will help us to understand the graph of multiples of different numbers.

1. Draw the graph of the function y = 3x.
2. From the graph, find the value of y, when
(a) x = 4
(b) x = 5
Solution:
1. The given function is y = 3x.
For some different values of x, the corresponding values of y are given below.

On a graph paper plot the points O (0, 0), A (1, 3), B (2, 6) and C (3, 9) and join them successively to obtain the required graph.

2. Reading off from the graph of simple function:
(a) On the x-axis, take the point L at x = 4.
Draw LP ⊥ x-axis, meeting the graph at P.
Clearly, PL = 12 units.
Therefore, x = 4 ⇒ y = 12.
(b) On the x-axis, take the point M at x = 5.
Draw MQ ⊥ x-axis, meeting the graph at Q.
Clearly MQ = 15 units
Therefore, x = 5 ⇒ y = 15

Linear polynomial:
Properties for graphing linear equation:

1. Linear equations have infinitely many solutions.
2. Every point (h, k) on the line AB gives the solution x = h and y = k.
3. Every point which lies on AB satisfies the equation of AB.
4. To draw an exact line on the graph paper you can plot as many points you like, but it is necessary to plot minimum three points.

Method to draw the graph of linear equation in two variables:
1. Convert the given equation in the form of y = mx + b (slope intercept form).
2. Apply trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation.
3. Plot these points on the graph paper.
4. Join the points marked on the graph paper to get a straight line which represent the given equation graphically.

Note:
1. Linear equation in two variables has infinitely many solutions.
2. A graph of linear equation is always a straight line.
3. Every point on the straight line is the solution of the linear equation.
4. Equation of y-axis is x = 0. The standard form of this equation is x + 0.y = 0.
5. Equation of x-axis is y = 0. The standard form of this equation is 0.x + y = 0.
6. x = a is a graph of straight line parallel to y-axis and standard form of this equation is x + 0.y = a
7. y = b is a graph of straight line parallel to x-axis and standard form of this equation is 0.x + y = b.
8. The equation y = mx is always passing through the origin (0, 0).

Learn the steps for graphing linear equation in two variables:
1. Draw the graph of the linear equation y = 2x.
Solution:

The given linear equation y = 2x is already in the form of y = mx + b [here b = 0].
Now we will apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation y = 2x.
When the value of x = 0, then y = 2 × 0 = 0
When the value of x = 1, then y = 2 × 1 = 2
When the value of x = 3, then y = 2 × 3 = 6
When the value of x = -1, then y = 2 × -1 = -2
When the value of x = -2, then y = 2 × -2 = -4
Arrange the values of the linear equation y = 2x in the table.
Now, plot the points P (0, 0), Q (1, 2), R (2, 4), S (3, 6), T (-1, -2), U (-2, -4) on the graph paper.

Join the points of P, Q, R, S, T and U.
We get a straight line passing through origin. This straight line is the graph of the equation y = 2x.

2. Draw the graph of the equation 4x - y = 3.
Solution:

The given linear equation 4x - y = 3.
Now convert the given equation in the form of y = mx + b
4x - y = 3
⇒ 4x - 4x - y = - 4x + 3
⇒ - y = - 4x + 3
⇒ y = 4x - 3
Now we will apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation y = 4x - 3.
When the value of x = 0, then y = (4 × 0) - 3 = - 3
When the value of x = 1, then y = (4 × 1) - 3 = 1
When the value of x = 2, then y = (4 × 2) - 3 = 5
Arrange these value of the linear equation y = 4x - 3 in the table.

Now, plot the point P (0, -3), Q (1, 1), R(2, 5) on the graph.

Join the points of P, Q and R.
We get a straight line passing through origin. This straight line is the graph of linear equation 4x - y = 3.

➢ A quadratic polynomial is a polynomial of degree 2.
➢ A univariate quadratic polynomial has the form . An equation involving a quadratic polynomial is called a quadratic equation. A closed-form solution known as the quadratic formula exists for the solutions of an arbitrary quadratic equation
An example of a Quadratic Equation:

Quadratic Equations make nice curves, like this one:

Name
The name Quadratic comes from "quad" meaning square, because the variable gets squared (like x2).
It is also called an "Equation of Degree 2" (because of the "2" on the x)

Standard Form The Standard Form of a Quadratic Equation looks like this:
ax^2 + bx + c  = 0
a, b and c are known values. a can't be 0.
"x" is the variable or unknown (we don't know it yet).

Here are some more examples:

The "Standard Form" of a Quadratic Equation is ax2 + bx + c = 0

But sometimes a quadratic equation doesn't look like that! For example:

The "solutions" to the Quadratic Equation are where it is equal to zero.
There are usually 2 solutions (as shown in the graph above).
They are also called "roots", or sometimes "zeros"

There are 3 ways to find the solutions:
1. We can use the special Quadratic Formula:
(x = ((-b )+- (sqrt (b^2 - 4ac) ))/ (2a)) 
Just plug in the values of a, b and c, and do the calculations.

Plus/Minus

First of all what is that plus/minus thing that looks like ± ?

Here is why we can get two answers:
• when b2 - 4ac is positive, we get two Real solutions
• when it is zero we get just ONE real solution (both answers are the same)
• when it is negative we get two Complex solutions

Just put the values of a, b and c into the Quadratic Formula, and do the calculations.
Example: Solve 5x2 + 6x + 1 = 0

The Formula:
I don't know of an easy way to remember the Quadratic Formula, but a kind reader suggested singing it to "Pop Goes the Weasel":
x = ((-b +- sqrt (b^2 - 4ac) )/ (2a))
"All around the mulberry brush the monkey chased the weasel.
The monkey thought it was all in fun pop.goes the weasel"

Complex Solutions:
When the Discriminant (the value b2 - 4ac) is negative we get Complex solutions .
Example: Solve 5x2 + 2x + 1 = 0
The coefficients are a = 5, b = 2 , c = 1
Note that the discriminant is negative b^2-4ac= 2^2 - 4 × 5 × 1 = -16
Use the quadratic Formula (x = ((-2) +- v(-16))/(10))
The square root of -16 is 4i
so x = ((-2 +- 4i )/(10))
Answer : - 0.2 +- 4i

In some ways it is easier: we don't need more calculation, just leave it as -0.2 ± 0.4i.

Cubic polynomial:
It can be hard to solve Cubic(degree 3) and Quartic (degree 4) equations.
Ex: 2y3-5x2

Examples of cubic polynomial:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.

### Relationship Between Zeroes and Coefficients Of A Polynomial:

Let us first understand what polynomial is;
➢ Any algebraic expression like 4x+2, 2y2 -3y +4, 5y3 -4x2 + x -2 etc are called polynomials.
➢ 4x+2 is a polynomial in the variable x of degree1 and is called linear polynomial.
➢ 2y2 -3y +4 is a polynomial in the variable y of degree 2 and is called quadratic polynomial.
➢ 5y3 -4x2 +x -2 is a polynomial in the variable x of degree 3 and is called cubic polynomial and so on.
Now in general let us take a quadratic polynomial as ax2 + bx +c. Here coefficient of x2 and x is 'a' and 'b' respectively. 'c' is a constant term.

'zero of polynomial':
➢ If p(x) is a polynomial in x, and if k is any real number, then the value obtained by replacing x by k in p(x), is called the value of p(x) at x =k, and is denoted by p(k). A real number k is said to be a zero of a polynomial of p(x), if p(k) =0.
Linear the relationship between zeros and coefficient of a polynomial:
The general form of linear polynomial is p(x) =ax +b , its zero is -b/a or minus of constant term divided by coefficient of x.
➢ If 'Α' is the zero of the above linear polynomial then,
➢ Α = -b/a
➢ In the general form of quadratic polynomial ax2 + bx + c, there are two zeros say Α and Β, then;
Sum of the zeros = Α + Β = -b/a = -(coefficient of x) / (coefficient of x2), and
Product of zeros = Α.Β = c/a = (Constant Term) / (Coefficient of x2)

Example: Find the zeros of the quadratic polynomial x2 + 7x +10, and verify the relationship between the zeroes and the coefficients.
We have, x2 + 7x +10 = (x +2) (x+5)
So, the value of x2 + 7x + 10 is zero, when x+2 = 0 or x+5 = 0, i.e., when, x = -2 or x = -5.
Therefore, the zeroes of x2 + 7x + 10 are -2 and -5. Now, the relationship between zeros and coefficient of above polynomial can be shown as:-
➢ Sum of zeroes = -2 + (-5) = -7 = -(7)/1 = -(coefficient of x) / (coefficient of x2)
➢ Product of zeroes = (-2) ×` (-5) = 10 =10/1 = (constant Term) / (coefficient of x2)
➢ A real number k is a zero of the polynomial p(x) if p(k) = 0. But why
➢ Consider first a linear polynomial ax + b, a ≠ 0. Graph of y = ax + b is a straight line.

### Worksheet on Addition and Subtraction of Polynomials

1. Solve the following polynomials:
(i)Take -pq + qr - rp from qr - rp + pq
(ii) Take 5a + 6b - 3c from 3a + 5b - 4c
2. (i) How much is -3k less than 3k?
(ii) How much is 2a + b greater than a - 2b?
(iii) What should be added to 2p - q + r to make it p + q - 2r?
3. From the sum of 3x - 2y + 4z and 3y - 2z, subtract x - y - z.
4. Subtract a - 2b - c from the sum of 3a - b + c and a + b - 3c.
5. How much should a + 2b - 3c be increased to get 3a?
6. The sum of two expressions is 3x2 + 2xy - y2. If one of them is 2x2 + 3y2, find the other.
7. From the sum of 4b2 + 5bc, -2b2 - 2bc - 2z2 and 2bc + 4c2, subtract the sum of 5b2 - c2 and -3b2 + 2bc + c2.
8. What should be subtracted from 3x + 5y + 9 to get - 2x + 3y + 15?
9. What should be taken away from p2 - q2 + 2pq + 10 to obtain - 2p2 - 2q2 + 7pq + 10?
10. By how much must 5p2 - 3q2 be diminished to give 2p2 + q2?
11. Subtract 2m - 5q + 7 from 4p + 2q and add your result to m2 - 3m3 + m - 1.
12. Subtract 2a2b - 3b2 from 3a2 - 5ab2 + 2b2 and subtract your result from the sum of the two expressions 3ab2 + 5b2 - 2a2b and a2b + 5a2 - 3ab2.

Answers for the worksheet on addition and subtraction of polynomials are given below to check the exact answers of the above word problems.

1. (i) 2pq
(ii) -2a - b - c
2. (i) 6k
(ii) a + 3b
(iii) -p + 2q - 3r
3. 2x + 2y + 3z
4. 3a + 2b - c
5. 2a - 2b + 3c
6. x2 + 2xy - 4y2
7. 3bc + 2c2
8. 5x + 2y - 6
9. 3p2 + q2 - 5pq
10. 3p2 - 4q2
11. 4p + 7q - m + m2 - 3m3 - 8
12. 2a2 + a2b + 5ab2

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