Question

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

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

Answer 1

The correct option is C

$(x-1)(x-2)$

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

So, $(x-2)$ is a factor of $p\left(x\right)$.

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

So, $(x-1)$ is also a factor of $p\left(x\right)$.

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