How do you factor completely $45 x^{4} + 43 x^{2} y^{2} - 28 y^{4}$ $45x^4 + 43x^2 y^2 - 28y^4$ ?

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How do you factor completely $45 x^{4} + 43 x^{2} y^{2} - 28 y^{4}$ $45x^4 + 43x^2 y^2 - 28y^4$ ?

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Answer 1 · 2016-04-06T18:55:05

$(3 x + 2 y) (3 x - 2 y) (5 x^{2} + 7 y^{2})$ $(3x+2y)(3x-2y)(5x^2+7y^2)$

Explanation:

Given
$XXX 45 x^{4} + 43 x^{2} y^{2} - 28 y^{4}$ $color(white)("XXX")45x^4+43x^2y^2-28y^4$

Replacing $p = x^{2}$ $p=x^2$ and $q = y^{2}$ $q=y^2$
$XXX = 45 p^{2} + 43 p q - 28 q^{2}$ $color(white)("XXX")=45p^2+43pq-28q^2$

Factoring by examining factors of $45$ $45$ and $28$ $28$
or by using the quadratic formula:
$XXX = (9 p - 4 q) (5 p + 7 q)$ $color(white)("XXX")=(9p-4q)(5p+7q)$

Replacing $p$ $p$ and $q$ $q$ with $x^{2}$ $x^2$ and $y^{2}$ $y^2$ respectively
and
Noting that the first of these factors is the difference of squares
$XXX = (3 x + 2 y) (3 x + 2 y) (5 x^{2} + 7 y^{2})$ $color(white)("XXX")=(3x+2y)(3x+2y)(5x^2+7y^2)$

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How do you factor completely $45 x^{4} + 43 x^{2} y^{2} - 28 y^{4}$ $45x^4 + 43x^2 y^2 - 28y^4$ ?

1 Answer

Explanation:

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How do you factor completely 45x4+43x2y2−28y445x^4 + 43x^2 y^2 - 28y^4?

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How do you factor completely $45 x^{4} + 43 x^{2} y^{2} - 28 y^{4}$ $45x^4 + 43x^2 y^2 - 28y^4$ ?