Question

Use the Factor Theorem to determine whether $\text{g(x)}$ is a factor of $\text{p(x)}$ in the following case: ${\text{p(x) = x}}^3-4x^2+x+6,g\left(x\right)=x–3$

Answer 1

Solve using Factor Theorem

A polynomial is an algebraic expression in which the exponent on any variable is a whole number. Polynomial’s degree is the highest or the greatest power of a variable in a polynomial equation.

Factor Theorem

A polynomial $\text{P(x)}$ divided by $\text{Q(x)}$ results in $\text{R(x)}$ with zero remainders if and only if $\text{Q(x)}$ is a factor of $\text{P(x)}$.

Given

• Dividend is ${\text{p(x) = x}}^3–4x^2+x+6$
• Divisor is $\text{g(x) = x – 3}$

Let us put divisor is equal to zero.

$\begin{array}{rcl}& & \text{g(x) = x – 3 = 0}\\ & & \text{x = 3}\end{array}$

∴ Zero of $\text{g(x)}$ is $3$.

Let us consider

Let ${\text{p(x) = x}}^3–4x^2+x+6$

put $x=3$ in the above equation

$\begin{array}{rcl}& & \text{p(3) = (3)3– 4(3)2 + (3) + 6}\\ & & \text{p(3) = 27- 36 + 3 + 6}\\ & & \text{p(3) = 0}\end{array}$

Since reminder is zero, $\text{x – 3}$is a factor of ${\text{x}}^{\text{3}}{\text{– 4x}}^{\text{2}}\text{+ x + 6}$

∴ By factor theorem, $\text{g(x)}$ is a factor of $\text{p(x)}$.