Question

Factorize each of the following polynomials.
$x^3+x^2+x-14$

• A. $(x-2)(x^2+3x+7)$
• B. $(x+2)(x^2+3x+7)$
• C. $(x-2)(x^2+3x-7)$
• D. None of these

Answer 1

Answer: (A)

$(x-2)(x^2+3x+7)$

Let the given polynomial be $p\left(x\right)=x^3+x^2+x-14$.

We will now substitute various values of $x$ until we get $p\left(x\right)=0$ as follows:

$Forx=0p\left(0\right)=(0{)}^3+(0{)}^2+0-14=0+0+0-14=-14\ne 0\therefore p\left(0\right)\ne 0$

$Forx=1p\left(1\right)=(1{)}^3+(1{)}^2+1-14=1+1+1-14=3-14=-11\ne 0\therefore p\left(1\right)\ne 0$

$Forx=2p\left(2\right)=(2{)}^3+(2{)}^2+2-14=8+4+2-14=14-14=0\therefore p\left(2\right)=0$

Thus, $(x-2)$ is a factor of $p\left(x\right)$.

Now,

$p\left(x\right)=(x-2)\cdot g\left(x\right).....\left(1\right)\Rightarrow g\left(x\right)=\frac{p\left(x\right)}{(x-2)}$

Therefore, $g\left(x\right)$ is obtained by after dividing $p\left(x\right)$ by $(x-2)$ as shown in the above image:

From the division, we get the quotient $g\left(x\right)=x^2+3x+7$.

From equation 1, we get $p\left(x\right)=(x-2)(x^2+3x+7)$.

Hence, $x^3+x^2+x-14=(x-2)(x^2+3x+7)$.