Question

Factorise

$x^3+3x^2+3x+2$.

Answer 1

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

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

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

$Forx=-1p(-1)=(-1{)}^3+3(-1{)}^2+3(-1)+2=-1+3-3+2=5-4=1\ne 0\therefore p(-1)\ne 0$

$Forx=-2p(-2)=(-2{)}^3+3(-2{)}^2+3(-2)+2=-8+12-6+2=14-14=0\therefore p(-2)=0$

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 $g\left(x\right)=x^2+x+1$.

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

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