Question

Factorise $P\left(x\right)=6x^3+25x^2+23x+6$ using synthetic division

• A. $(x+1)(x+2)(x+3)$
• B. $(x-1)(x-2)(x-3)$
• C. $(2x-1)(3x-2)(x-3)$
• D. $(2x+1)(3x+2)(x+3)$

Answer 1

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

Let the given polynomial be $p\left(x\right)=6x^3+25x^2+23x+6$.

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

$Forx=1p(-1)=6(-1{)}^3+25(-1{)}^2+(23\times -1)+6=-6+25-23+6=31-29=2\ne 0\therefore p(-1)\ne 0$

$Forx=-2p(-2)=6(-2{)}^3+25(-2{)}^2+(23\times -2)+6=(6\times -8)+(25\times 4)-46+6=-48+100-46+6=12\ne 0\therefore p(-2)\ne 0$

$p(-3)=6(-3{)}^3+25(-3{)}^2+(23\times -3)+6=(6\times -27)+(25\times 9)-69+6=-162+225-69+6=231-231=0\therefore p(-3)=0$

Thus, $(x+3)$ is a factor of $p\left(x\right)$.

Now,

$p\left(x\right)=(x+3)\cdot g\left(x\right).....\left(1\right)\Rightarrow g\left(x\right)=\frac{p\left(x\right)}{(x+3)}$

Therefore, $g\left(x\right)$ is obtained by after dividing $p\left(x\right)$ by $(x+3)$ as shown in the above image:

From the division, we get the quotient $g\left(x\right)=6x^2+7x+2$ and now we factorize it as follows:

$6x^2+7x+2=6x^2+3x+4x+2=3x(2x+1)+2(2x+1)=(2x+1)(3x+2)$

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

Hence, $6x^3+25x^2+23x+6=(x+3)(2x+1)(3x+2)$.