Question

factorize z4-81
find factors of y2-y-72
​

Answer 1

$\star\:\:\bf\large\underline\blue{Question\:1:-}$

• Factorise ${z}^{4} - 81$

$\star\:\:\bf\underline\blue{Solution:-}$

${z}^{4} - 81$

$= ({z}^{2})^{2} - {(9)}^{2}$

$= ( {z}^{2} + 9)( {z}^{2} - 9)$

$= ( {z}^{2} + 9)( {(z)}^{2} - {(3)}^{2} )$

$= ( {z}^{2} + 9)(z + 3)(z - 3)$

$\star\:\:\bf\underline\blue{Answer:-}$

• $( {z}^{2} + 9)(z + 3)(z - 3)$

$\star\:\:\bf\large\underline\blue{Question\:2:-}$

• Find the factors of ${y}^{2} -y - 72$

$\star\:\:\bf\underline\blue{Solution:-}$

${y}^{2} -y - 72$

$= {y}^{2} - 9y + 8y - 72$

$= y(y - 9) + 8(y - 9)$

$= (y + 8)(y - 9)$

$\star\:\:\bf\underline\blue{Answer:-}$

• $(y + 8)(y - 9)$

Answer 2

$\star\:\:\:\bf\large{Question\:1\::-}$

• $Factorise : \: z ^{4} - 81$

$\star\:\:\:\bf\large{Solution\::-}$

${z}^{4} - 81$

$= {(z {}^{2} )}^{2} - {(9)}^{2}$

$= {(z}^{2} + 9)( {z}^{2} - 9)$

$= ( {z}^{2} + 9)(( {z)}^{2} - ( {3})^{2} )$

$= ( {z}^{2} + 9)(z + 3)(z - 3)$

$\star\:\:\:\bf\large{Answer\::-}$

• $= ( {z}^{2} + 9)(z + 3)(z - 3)$

Formulae used :

• ${a}^{2} - {b}^{2} = (a + b)(a - b)$

$\star\:\:\:\bf\large{Question\:2\::-}$

• $Find \: the \: factors \: of \: {y}^{2} - y - 72$

$\star\:\:\:\bf\large{Solution\::-}$

${y}^{2} - y - 72$

$= {y}^{2} - 9y + 8y - 72$

$= y(y - 9) + 8(y - 9)$

$= (y + 8)(y - 9)$

$\star\:\:\:\bf\large{Answer\::-}$

• $= (y + 8)(y - 9)$

__________________________________

Hope it helps you ❤️