Question

Find the HCF of the following triples:
$a^2(2x^2+5x+2),a^{2}b(3x^2+8x+4)$ and $a{b}^2(2x^2+3x-2)$

Answer 1

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

Factorise $a^2(2x^2+5x+2)$ as follows:

$a^2(2x^2+5x+2)=a^2(2x^2+4x+x+2)=a^2\left[2x\right(x+2)+1(x+2\left)\right]=a^2(2x+1)(x+2)$

Now, factorise $a^{2}b(3x^2+8x+4)$ as follows:

$a^{2}b(3x^2+8x+4)=a^{2}b(3x^2+6x+2x+4)=a^{2}b\left[3x\right(x+2)+2(x+2\left)\right]=a^{2}b(3x+2)(x+2)$

Finally, factorise $a{b}^2(2x^2+3x-2)$ as follows:

$a{b}^2(2x^2+3x-2)=a{b}^2(2x^2+4x-x-2)=a{b}^2\left[2x\right(x+2)-1(x+2\left)\right]=a{b}^2(x+2)(2x-1)$

Since the common factor between the polynomials $a^2(2x^2+5x+2)$,$a^{2}b(3x^2+8x+4)$and $8(x^2+3x-4)$ is $a(x+2)$.

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