Question

Find the LCM of the following: $x^2-5x+6,x^2+4x-12$ whose GCD is $x-2$

• A. $(x-3)(x-2)(x+6)$
• B. $(x+3)(x-2)(x+6)$
• C. $(x-3)(x+2)(x+6)$
• D. None of these

Answer 1

The correct option is A $(x-3)(x-2)(x+6)$

We first factorize the given polynomials $x^2-5x+6$ and $x^2+4x-12$ as shown below:

$x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)$

$x^2+4x-12=x^2+6x-2x-12=x(x+6)-2(x+6)=(x-2)(x+6)$

We know that if $p\left(x\right)$ and $q\left(x\right)$ are two polynomials, then $p\left(x\right)\times q\left(x\right)=$ {GCD of $p\left(x\right)$ and $q\left(x\right)$}$\times$ {LCM of $p\left(x\right)$ and $q\left(x\right)$}.

Now, it is given that the GCD of the polynomials is $x-2$, therefore, we have:

$(x^2-5x+6)(x^2+4x-12)=(x-2)\times LCM$

$\Rightarrow (x-2)(x-3)\times (x-2)(x+6)=(x-2)\times LCM$

$\Rightarrow (x-2{)}^2(x-3\left)\right(x+6)=(x-2)\times LCM$

$\Rightarrow LCM=\frac{(x-2{)}^2(x-3\left)\right(x+6)}{(x-2)}$

$\Rightarrow LCM=(x-2)(x-3)(x+6)$

Hence, the LCM of $x^2-5x+6$ and $x^2+4x-12$ is $(x-2)(x-3)(x+6)$.