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### Theory:

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