Question

80X⁴-405y⁴ divide by 20X²+45y²​

Answer 1

1. $4 {x }^{2} - 9 {y}^{2}$

Answer 2

Answer:

$\frac{80X^{4}-405Y^{4}}{20X^{2}+45Y^{2}} = 4X^{2}-9Y^{2}$

Step-by-step explanation:

Given $\frac{80X^{4}-405Y^{4}}{20X^{2}+45Y^{2}}$

Taking $5$ as common in both the numerator and the denominator,

$\frac{5(16X^{4}-81Y^{4})}{5(4X^{2}+9Y^{2})}$

Cancelling out the $5$,

$\frac{16X^{4}-81Y^{4}}{4X^{2}+9Y^{2}}=\frac{(4X^{2})^{2}-(9Y^{2})^{2}}{4X^{2}+9Y^{2}}$

Now the numerator is in the form $a^{2}-b^{2}$, which can be written as $(a-b)(a+b)$,

$=\frac{(4X^{2}-9Y^{2})(4X^{2}+9Y^{2})}{4X^{2}+9Y^{2}}$

Cancel out the same terms in the numerator and the denominator,

$=4X^{2}-9Y^{2}$.