### Zeroes of Polynomials / Roots of Polynomials

**Tags:**Find the zero of the polynomial for Grade 9, Verify whether the following are zeroes of the polynomial for Grade IX, Zero of a polynomial worksheet PDF for Class 9, Find the roots of polynomials practice page class IX, Zeros of polynomial exercise for ninth class, Roots of polynomial examples for 9^{th}classZeros (Roots) Of a Polynomial

General form of a polynomial in ‘x’ is a

_{n}x^{n}+ a_{n−1}x^{n−1}+….. + a_{1}x + a0, where a_{n}, a_{n−1}, ….. , a1, a0 are constants, a_{n}≠ 0 and n is a whole number.**Value of a Polynomial**Let’s consider the polynomial f(x) = x

^{2 }-3x + 2If we replace ‘x’ by 2 in the polynomial x

^{2 }-3x + 2 everywhere, we getf(2) = (2)

^{2 }- 3 x 2^{ }+ 2 = 4 – 6 + 2 = 0So, 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) = 4x

^{2}-2x + 3 at x = -1 and x = 2 are given byf(-1) = 4 x (-1)

^{2}– 2 x (-1) + 3 = 4 + 2 + 3 = 9and, 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) = x

^{3}– 6x^{2}+ 11x – 6The value of this polynomial at x = 2 is

f(2) = 2

^{3}– 6 x 2^{2}+ 11 x 2 - 6 = 8 - 24 + 22 – 6 = 0So, 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 polynomialf(x) = a

_{n}x^{n}+ a_{n-1}x^{n-1}+ a_{n-2}x^{n-2}+ …. +a_{1}x + a_{0}, if (α) = 0i.e. a

_{n}x^{n}+ a_{n-1}x^{n-1}+ a_{n-2}x^{n-2}+ …. +a_{1}x + a_{0}= 0If 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) = x^{3}+ 2x^{2}– 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) = a

_{n}x^{n}+ a_{n-1}x^{n-1}+ a_{n-2}x^{n-2}+ …. +a_{1}x + a_{0}, a_{n}≠ 0 with integral coefficients. Then, b is a factor of constant term a_{0}and c is a factor of the leading coefficient a_{n}.**Example:**If b/c is a rational root of the polynomial f(x) = 6x^{3}+ 5x^{2 }– 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) = 0i.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) = 3x^{3}– 7x^{2}+ 5x + 2, find f(2).**Solution:**Here, f(x) = 3x^{3}– 7x^{2}+ 5x + 2f(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 3x^{3}– 4x^{2}+ 7x – 6**Solution:**Let f(x) = 3x^{3}– 4x^{2}+ 7x – 6. Then,f(1) = 3 x 1

^{3}– 4 x 1^{2}+ 7 x 1 – 6 = 3 - 4 + 7 – 6 = 0Hence, x = 1 is a root of polynomial f(x).

**Example 3:**If x = 2 and x = 0 are roots of the polynomial f(x) = 2x^{3}– 5x^{2}+ cx + d. Find the values of c and d.**Solution:**We have, f(x) = 2x^{3}– 5x^{2}+ cx + dTherefore, f(2) = 2 x 2

^{3}– 5 x 2^{2}+ c x 2 + d = 16 – 20 + 2c + d = 2c + d -4And, 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 x^{3}– 6x^{2}+ 11x – 6.**Solution:**Let f(x) = x^{3}– 6x^{2}+ 11x – 6Clearly, 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 2x^{3}+ 3x^{2}- 11x – 6.**Solution:**Let f(x) = 2x^{3}+ 3x^{2}- 11x – 6Clearly, 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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