Question

$x^3-5x^2-2x+24$

• A. $(x+2)(x-3)(x-4)$
• B. $(x-2)(x-3)(x-4)$
• C. $(x+2)(x+3)(x-4)$
• D. None of these

Answer 1

The correct option is A $(x+2)(x-3)(x-4)$

Let the given polynomial be $p\left(x\right)=x^3-5x^2-2x+24$.

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

$Forx=0p\left(0\right)=(0{)}^3-5(0{)}^2-(2\times 0)+24=0-0-0+24=24\ne 0\therefore p\left(0\right)\ne 0$

$Forx=1p\left(1\right)=(1{)}^3-5(1{)}^2-(2\times 1)+24=1-5-2+24=25-7=18\ne 0\therefore p\left(1\right)\ne 0$

$Forx=-2p(-2)=(-2{)}^3-5(-2{)}^2-(2\times -2)+24=-8-20+4+24=28-28=0\therefore p(-2)=0$

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 $g\left(x\right)=x^2-7x+12$ and now we factorize it as follows:

$x^2-7x+12=x^2-4x-3x+12=x(x-4)-3(x-4)=(x-3)(x-4)$

From equation 1, we get $p\left(x\right)=(x+2)(x-3)(x-4)$.

Hence, $x^3-5x^2-2x+24=(x+2)(x-3)(x-4)$.