Question

Factorise $x^2+3x+2$ by using the Factor Theorem.

• A. $(x+4)(x+2)$
• B. $(x+5)(x+2)$
• C. $(x+1)(x+6)$
• D. $(x+1)(x+2)$

Answer 1

The correct option is D

$(x+1)(x+2)$

Let $p\left(x\right)=x^2+3x+2$.

Let factors be $(x-a)and(x-b)$

So, $p\left(x\right)=(x–a)(x–b)$

$\Rightarrow p\left(x\right)=x^2–bx–ax+ab$

On comparing the constants, we get ab = 2.

The factors of 2 are +1,-1,-2 and 2.

Now, $p(-2)=(-2{)}^2+(3\times (-2))+2=4-6+2=0$.

So, (x + 2) is a factor of p(x).

Also, $p(-1)=(-1{)}^2+(3\times (-1))+2=1-3+2=0$

So, (x +1) is also a factor of p(x).

Therefore, $x^2+3x+2=(x+1)(x+2)$