How do you factor $5 x^{2} + 41 x y - 36 y^{2}$ $5x^2 + 41xy - 36y^2$ ?

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How do you factor $5 x^{2} + 41 x y - 36 y^{2}$ $5x^2 + 41xy - 36y^2$ ?

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Answer 1 · 2015-05-20T14:50:17

The coefficients of $x^{2}$ $x^2$ and $y^{2}$ $y^2$ are, respectively, $5$ $5$ and $- 36$ $-36$ . The product $5 (- 36) = - 180$ $5(-36)=-180$ , which, can be factored as $(- 4) (45)$ $(-4)(45)$ .

So, we can now rewrite $5 x^{2} + 41 x y - 36 y^{2}$ $5x^2+41xy-36y^2$ as

$5 x^{2} - 4 x y + 45 x y - 36 y^{2}$ $5x^2-4xy+45xy-36y^2$

Now, we can factor this by two groups:

$x (5 x - 4 y)$ $x(5x-4y)$ and $9 y (5 x - 4 y)$ $9y(5x-4y)$

We can see that the parenthesis are shared by the two factors. It becomes clearer when we rewrite:

$x (5 x - 4 y) + 9 y (5 x - 4 y) = (x + 9 y) (5 x - 4 y)$ $x(5x-4y)+9y(5x-4y)=(x+9y)(5x-4y)$

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How do you factor $5 x^{2} + 41 x y - 36 y^{2}$ $5x^2 + 41xy - 36y^2$ ?

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