Factor theorem — lesson. Mathematics State Board, Class 9.

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Theorem

If

$p(x)$ is a polynomial of degree

$n \geq 1$ and ‘

$a$ ’ is any real number then
(i)

$p(a) = 0$ implies

$(x - a)$ is a factor of

$p(x)$ .
(ii)

$(x-a)$ is a factor of

$p(x)$ implies

$p(a) = 0$ .

Proof:

Let

$p(x)$ be the dividend and

$(x-a)$ be the divisor.

Division algorithm

Given any two integers

$a$ and

$b$ with

$a > 0$ , there exists a unique interger

$q$ and

$r$ such that

$b = qa + r$ with

$0 \leq r < a$ .

By division algorithm we have,

$p(x) = q(x)(x-a) + p(a)$ where

$q(x)$ is the quotient and

$p(a)$ is the remainder.

(i) Assume

$p(a) = 0$ .

To prove:

$(x - a)$ is a factor of

$p(x)$ .

Substitute

$p(a) = 0$ in

$p(x) = q(x)(x-a) + p(a)$ .

Then,

$p(x) = q(x)(x-a)$

This implies,

$(x - a)$ is a factor of

$p(x)$ .

(ii) Assume

$(x-a)$ is a factor of

$p(x)$ .

To prove:

$p(a) = 0$ .

By assumption, we have

$p(x) = q(x)(x-a)$ , where

$q(x)$ is some polynomial.

Substitute

$(x = a)$ in

$p(x) = q(x)(x-a)$ .

Then,

$p(a) = q(a)(a-a)$

$= q(a)(0)$

$= 0$

Hence, the proof.

Example:

Check whether

$x - 5$ is the factor of

$x^{3} + x^{2} + 2x - 3$ .

Given:

The polynomial

$p(x) = x^{3} + x^{2} + 2x - 3$ .

To check:

If

$x - 5$ is the factor of

$p(x)$ .

Solution:

Step 1: Find the zero of the polynomial

$x = a$ .

Equate

$x - 5$ to zero and solve for

$x$ .

$x$

$-$

$5$

$=$

$0$

$x$

$=$

$5$

Step 2: Check if

$p(5) = 0$ .

By the factor theorem, for

$x - 5$ to be a factor of

$p(x)$ it must satisfy

$p(5) = 0$ .

Substitute

$x = 5$ in

$p(x) = x^{3} + x^{2} + 2x - 3$ .

$p(5) = 5^{3} + 5^{2} + 2(5) - 3$

$= 125 + 25 + 10 - 3$

$= 157$

Here,

$p(5) \neq 0$ .

This implies that,

$x - 5$ is not the factor of

$x^{3} + x^{2} + 2x - 3$ .

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