Divide:(i) $ax^2-ay^2$ by $ax+ay$(ii) $x^4-y^4$ by $x^2-y^2$

Given:

The given expressions are:

(i) $ax^2-ay^2$ by $ax+ay$

(ii) $x^4-y^4$ by $x^2-y^2$

To do:

We have to divide the given expressions.

Solution:

We have to divide the given polynomials by simplifying them using algebraic formulas.

Polynomials:

Polynomials are expressions in which each term is a constant multiplied by a variable raised to a whole number power.

Therefore,

(i) The given expression is $ax^2-ay^2$ by $ax+ay$.

$ax^2-ay^2$ can be written as,

$ax^2-ay^2=a(x^2-y^2)$            (Taking $a$ common)

$ax^2-ay^2=a(x+y)(x-y)$.........(I)               [Since $a^2-b^2=(a+b)(a-b)$]

Therefore,

$ax^2-ay^2 \div (ax+ay)=\frac{ax^2-ay^2}{ax+ay}$

$ax^2-ay^2 \div (ax+ay)=\frac{a(x+y)(x-y)}{a(x+y)}$             [Using (I) and taking $a$ common in $ax+ay$]

$ax^2-ay^2 \div (ax+ay)=(x-y)$

Hence, $ax^2-ay^2$ divided by $ax+ay$ is $x-y$.

(ii) The given expression is $x^4-y^4$ by $x^2-y^2$.

$x^4-y^4$ can be written as,

$x^4-y^4=(x^2)^2-(y^2)^2$            [Since $x^4=(x^2)^2$ and $y^4=(y^2)^2$)

$x^4-y^4=(x^2+y^2)(x^2-y^2)$...........(I)               [Since $a^2-b^2=(a+b)(a-b)$]

Therefore,

$x^4-y^4 \div x^2-y^2=\frac{x^4-y^4}{x^2-y^2}$

$x^4-y^4 \div x^2-y^2=\frac{(x^2+y^2)(x^2-y^2)}{x^2-y^2}$             [Using (I)]

$x^4-y^4 \div x^2-y^2=x^2+y^2$

Hence, $x^4-y^4$ divided by $x^2-y^2$ is $x^2+y^2$.

Updated on: 13-Apr-2023

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