Question

Factorise $x^2–5x+6$ by using the Factor Theorem.

• A. $(x–2)(x–10)$
• B. $(x–7)(x–3)$
• C. $(x–2)(x–3)$
• D. $(x–2)(x–4)$

Answer 1

The correct option is C

$(x–2)(x–3)$

Let $p\left(x\right)=x^2–5x+6$.

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 = 6.

The factors of 6 are 1, 2 and 3.

Now, $p\left(2\right)=2^2–(5\times 2)+6=4–10+6=0$.

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

Also, $p\left(3\right)=3^2–(5\times 3)+6=9–15+6=0$

So, $(x–3)$ is also a factor of $x^2–5x+6$

Therefore, $x^2–5x+6=(x–2)(x–3)$