Question

Factorise and then divide the given algebraic expressions.

$1.\left(12r^2+8r^2-4r^2\right)by\left(-4r^2\right)$

$2.\left(\frac{4}{9}{x}^2-49z^2\right)by\left(2x+21z\right)$
$3.\left(5a^2{b}^2+15a^2{b}^2-20a^{4}b\right)by25a{b}^2$

$4.\left(49a^2-56a\right)by\left(21a-24\right)$

Answer 1

$\left(a\right)\frac{12r^2+8r^2-4r^2}{-4r^2}$

$=\frac{16r^2}{-4r^2}=-4$

$\left(b\right)\frac{\frac{4}{9}{x}^2-49z^2}{2x+21z}$

$=\frac{(\frac{2}{3}x-7z)(\frac{2}{3}x+7z)}{2x+21z}$

$=\frac{(2x-21z)(2x+21z)}{9(2x+21z)}$

$=\frac{(2x-21z)}{9}$

$\left(c\right)\frac{5a^2{b}^2+15a^2{b}^2-20a^{4}b}{25a{b}^2}$

$=\frac{20a^2{b}^2-20a^{4}b}{25a{b}^2}$

$=\frac{20a^{2}b(b-a^2)}{25a{b}^2}$

$=\frac{4a(b-a^2)}{5b}$

$\left(d\right)\frac{49a^2-56a}{21a-24}$

$=\frac{7a(7a-8)}{3(7a-8)}$

$=\frac{7a}{3}$