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How do you divide $\frac{y^{4} + 9 y^{2} + 20}{y^{2} + 4}$ $(y^4+9y^2+20)/(y^2+4)$ ?

1 Answer

Jun 12, 2015

$y^{2} + 5$ $y^2+5$

Explanation:

We can simplify this by replacing $y^{2}$ $y^2$ with $x$ $x$

$\frac{y^{4} + 9 y^{2} + 20}{y^{2} + 4} = \frac{x^{2} + 9 x + 20}{x + 4}$ $(y^4+9y^2+20)/(y^2+4) = (x^2+9x+20)/(x+4)$

which has a (perhaps) more apparent factoring of the numerator:
$XXXX$ $color(white)("XXXX")$ $((x+5)cancel((x+4)))/(cancel((x+4))$

Then restoring $x$ with $y^2$
$rArr (y^2+5)$

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