Question

Find the HCF of the following triples:
$6(x^2+10x+24),4(x^2-x-20)$ and $8(x^2+3x-4)$

Answer 1

We know that $HCF$ is the highest common factor.

Factorise $6(x^2+10x+24)$ as follows:

$6(x^2+10x+24)=6(x^2+6x+4x+24)=6\left[x\right(x+6)+4(x+6\left)\right]=(2\times 3)(x+4)(x+6)$

Now, factorise $4(x^2-x-20)$ as follows:

$4(x^2-x-20)=4(x^2-5x+4x-20)=4\left[x\right(x-5)+4(x-5\left)\right]=(2\times 2)(x+4)(x-5)$

Finally, factorise $8(x^2+3x-4)$ as follows:

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

Since the common factor between the polynomials $6(x^2+10x+24)$,$4(x^2-x-20)$and $8(x^2+3x-4)$ is $2(x+4)$.

Hence, the $HCF$ is $2(x+4)$.