## Topic outline

• ### Zeroes of Polynomials / Roots of Polynomials

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• Zeros (Roots) Of a Polynomial

General form of a polynomial in ‘x’ is anxn + an−1xn−1 +….. + a1x + a0, where an, an−1, ….. , a1, a0 are constants, an ≠ 0 and n is a whole number.

Value of a Polynomial

Let’s consider the polynomial f(x) = x2 -3x + 2

If we replace ‘x’ by 2 in the polynomial x2 -3x + 2 everywhere, we get

f(2) = (2)2 - 3 x 2 + 2 = 4 – 6 + 2 = 0

So, we can say that the value of f(x) at x = 2 is 0.

Therefore, the value of a polynomial f(x) at x = α is obtained by replacing x = α in the given polynomial and is denoted by f(α).

The values of the quadratic polynomial f(x) = 4x2 -2x + 3 at x = -1 and x = 2 are given by

f(-1) = 4 x (-1)2 – 2 x (-1) + 3 = 4 + 2 + 3 = 9

and, f(2) = 4 x (2)2 – 2 x 2 + 3 = 16 – 4 + 3 = 15

Zero or Root of a Polynomial

Let’s consider the polynomial f(x) = x3 – 6x2 + 11x – 6

The value of this polynomial at x = 2 is

f(2) = 23 – 6 x 22 + 11 x 2 - 6 = 8 - 24 + 22 – 6 = 0

So, we say that 2 is a zero or a root of the polynomial f(x).

Therefore, a real number α is a root or zero of a polynomial

f(x) = anxn + an-1 xn-1 + an-2 xn-2 +  …. +a1x + a0, if (α) = 0

i.e. anxn + an-1 xn-1 + an-2 xn-2 +  …. +a1x + a0 = 0

If we draw the graph of f(x) =0, the values where the curve cuts the X-axis are called Zeros of the polynomial

Some Important Points

If f(x) is a polynomial with integral coefficients and the leading coefficient is equal to 1, then any integer root of f(x) is a factor of the constant term.

Example: Thus, if f(x) = x3 + 2x2 – 11x – 12 has integer root, then it is one of the factors of 12 which are ±1, ±2, ±3, ±4, ±6, ±12.

If b/c is a root of the polynomial f(x) = anxn + an-1 xn-1 + an-2 xn-2 +  …. +a1x + a0, an ≠ 0 with integral coefficients. Then, b is a factor of constant term a0 and c is a factor of the leading coefficient an.

Example: If b/c is a rational root of the polynomial f(x) = 6x3 + 5x2 – 3x -2,then the values of b are limited to the factors of -2 which are ±1, ±2; and the value of c are limited to the factors of 6, which are ±1, ±2, ±3, ±6. Hence, the possible rational roots of f(x) are ±1, ±2, ±1/2, ±1/3, ±1/6, ±2/3.

An nth degree polynomial can have at most n real roots. So, a polynomial of degree 3 will have 3 roots. A polynomial of degree 4 will have 4 roots. And so on.

Finding a zero or root of a polynomial f(x) means solving the polynomial equation f(x) = 0.

Example: If f(x) = ax + b, a ≠ 0 is a linear polynomial, then it has only one root given by f(x) = 0

i.e. ax + b= 0

ax = -b

x = -b/a

Thus, x = -b/a is the only root of f(x) = ax + b.

Few More Examples

Example1: If f(x) = 3x3 – 7x2 + 5x + 2, find f(2).

Solution: Here, f(x) = 3x3 – 7x2 + 5x + 2

f(2) = 3 x (2)3 – 7 x (2)2 + 5 x 2 + 2

= 3 x 8 – 7 x4 + 10 + 2 = 24 - 28 + 10 +2 = 8

Example 2: Show that x= 1 is a root of the polynomial 3x3 – 4x2 + 7x – 6

Solution: Let f(x) = 3x3 – 4x2 + 7x – 6. Then,

f(1) = 3 x 13 – 4 x 12 + 7 x 1 – 6 = 3 - 4 + 7 – 6 = 0

Hence, x = 1 is a root of polynomial f(x).

Example 3: If x = 2 and x = 0 are roots of the polynomial f(x) = 2x3 – 5x2 + cx + d. Find the values of c and d.

Solution: We have, f(x) = 2x3 – 5x2 + cx + d

Therefore, f(2) = 2 x 23 – 5 x 22 + c x 2 + d = 16 – 20 + 2c + d = 2c + d -4

And, f(0) = 2 x 0 - 5 x 0 + c x 0 + d = d

Since x = 2 and x = 0 are root of the polynomial f(x)

Therefore f(2) = 0 and f(0) = 0

2c + d – 4 = 0 and d = 0

2c – 4 = 0 and d = 0

c = 2 and d = 0

Example 4: Find the integral roots of the polynomial x3 – 6x2 + 11x – 6.

Solution: Let f(x) = x3 – 6x2 + 11x – 6

Clearly, f(x) is a polynomial with integer coefficients and the coefficient of the highest degree term i.e. the leading coefficients is 1. Therefore, integer roots of f(x) are limited to the integer factors of 6, which are ±1, ±2, ±3, ±6.

Therefore;

f(1) = 1 - 6 + 11 – 6 = 0

f(2) = 8 – 24 + 22 – 6 = 0

f(3) = 27 – 54 + 33 – 6 = 0

Hence, integral roots of f(x) are 1, 2, 3.

Example 5: Find the rational roots of the polynomial 2x3 + 3x2 - 11x – 6.

Solution: Let f(x) = 2x3 + 3x2 - 11x – 6

Clearly, f(x) is a cubic polynomial with integer coefficients. If b/c is a rational root in lowest terms, then the values of b are limited to the factors of 6 which are ±1, ±2, ±3, ±6; and the values of c are limited to the factor of 2 which are ±1, ±2. Hence, the possible rational roots of f(x) are ±1, ±2, ±3, ±6, ±1/2, ±3/2

We see that

f(2) = 2 x 8 + 3 x 4 – 11 x 2 – 6 = 16 + 12 – 22 – 6 = 0

f(-3) = 2 x -27 + 3 x 9 – 11 x -3 – 6 = -54 + 27 + 33 – 6 = 0

and f(-1/2) = 2 x -1/8 + 3 x ¼ - 11 x -1/2 – 6 = -1/4 + 3/4 + 11/2 – 6 = 0

Hence, 2, -3 and -1/2 are rational roots of f(x)

Example 6: Find the zero (root) of the polynomial in each of the following cases: (i) f(x) = x – 7   (ii) g(x) = 3x + 4   (iii) p(x) = 3x   (iv) f(x) = cx + d, c ≠ 0    (v) p(x) = bx, b ≠ 0

Solution: We know that the roots (zeros) of a polynomial f(x) are given by the solving the polynomial equation f(x) = 0. Therefore,

(i)                  A root of f(x) = x – 7 is given by

f(x) = 0

x – 7 = 0

x = 7

Thus, x = 7 is a root of f(x) = x – 7

(ii)                 A root of g(x) = 3x + 4 is given by

g(x) = 0

3x + 4 = 0

3x = -4

X = -4/3

Thus, x = -4/3 is a root of g(x) = 3x + 4

(iii)                A root of p(x) = 3x is given by

p(x) = 0

3x = 0

x = 0

Thus, x = 0 is a root of p(x) = 3x.

(iv)               Roots of f(x) = cx + d are given by

f(x) = 0

cx + d = 0

cx = -d

x = -d/c

Thus, x = -d/c is the root of f(x)

(v)                Roots of p(x) = bx are given by

p(x) = 0

bx = 0

x = 0

Hence, x = 0 is the roots of p(x)

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