Question

Factorize each of the following polynomials.
$x^3-5x+4$

• A. $(x-1)(x^2+x-4)$
• B. $(x+1)(x^2+x-4)$
• C. $(x-1)(x^2-x-4)$
• D. None of these

Answer 1

Answer: (A)

$(x-1)(x^2+x-4)$

Let the given polynomial be $p\left(x\right)=x^3-5x+4$.

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

$Forx=0p\left(0\right)=(0{)}^3-(5\times 0)+4=4\ne 0\therefore p(0)\ne 0$

$Forx=1p\left(1\right)=(1{)}^3-(5\times 1)+4=1-5+4=5-5=0\therefore p(1)=0$

Thus, $(x-1)$ is a factor of $p\left(x\right)$.

Now,

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

Therefore, $g\left(x\right)$ is obtained by after dividing $p\left(x\right)$ by $(x-1)$ as shown in the above image:

From the division, we get the quotient $g\left(x\right)=x^2+x-4$.

From equation 1, we get $p\left(x\right)=(x-1)(x^2+x-4)$.

Hence, $x^3-5x+4=(x-1)(x^2+x-4)$.