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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 $x^2+y^2$.
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