Question

The LCM of $2x+2,3x^2-12$ and $4x^2+12x+8$ is

• A. $12(x+1)(x-3)(x+2)$
• B. $12(x+1)(x-2)(x+3)$
• C. $12(x+1)(x-3)(x+3)$
• D. $12(x+1)(x+2)(x-2)$

Answer 1

Answer: (D)

$12(x+1)(x+2)(x-2)$

We first factorize the given terms $2x+2,3x^2-12$ and $4x^2+12x+8$ as follows:

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

$3x^2-12=3(x^2-4)=3(x^2-2^2)=3(x+2)(x-2)(\because a^2-b^2=(a+b\left)\right(a-b\left)\right)$

$4x^2+12x+8=4(x^2+3x+2)=4(x^2+x+2x+2)=4\left[x\right(x+1)+2(x+1\left)\right]=4(x+2)(x+1)$

We know that the LCM is the smallest common multiple of two or more numbers, therefore, the LCM of $2x+2,3x^2-12$ and $4x^2+12x+8$ is:

$LCM=4\times 3\left[\right(x+1\left)\right(x+2\left)\right(x-2\left)\right]=12(x+1)(x+2)(x-2)$

Hence, the LCM of $2x+2,3x^2-12$ and $4x^2+12x+8$ is $12(x+1)(x+2)(x-2)$.