Question

Factorize the following polynomial.

$4x^3-7x+3$

• A. $(x-1)(2x+3)(2x-1)$
• B. $(x+1)(2x+3)(2x-1)$
• C. $(x-1)(2x+3)(2x+1)$
• D. None of these

Answer 1

The correct option is A $(x-1)(2x+3)(2x-1)$

Let the given polynomial be $p\left(x\right)=4x^3-7x+3$.

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

$Forx=0p\left(0\right)=4(0{)}^3-(7\times 0)+3=0-0+3=3\ne 0\therefore p(0)\ne 0$

$Forx=1p\left(1\right)=4(1{)}^3-(7\times 1)+3=4-7+3=7-7=0\therefore p(1)=0$

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 $g\left(x\right)=4x^2+4x-3$ and now we factorize it as follows:

$4x^2+4x-3=4x^2-2x+6x-3=2x(2x-1)+3(2x-1)=(2x+3)(2x-1)$

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

Hence, $4x^3-7x+3=(x-1)(2x+3)(2x-1)$.