How do you factor $12 a^{2} + 7 a b - 10 b^{2}$ $12a^2 + 7ab - 10b^2$ ?

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How do you factor $12 a^{2} + 7 a b - 10 b^{2}$ $12a^2 + 7ab - 10b^2$ ?

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Answer 1 · 2015-05-21T22:37:28

Since this quadratic is homogeneous (that is all of the terms have the same order, viz 2), this problem is equivalent to the problem of factoring $12 x^{2} + 7 x - 10$ $12x^2+7x-10$ .

$12 x^{2} + 7 x - 10$ $12x^2+7x-10$ is of the form $a x^{2} + b x + c$ $ax^2+bx+c$ , with $a = 12$ $a=12$ , $b = 7$ $b=7$ and $c = - 10$ $c=-10$ (not to be confused with the $a$ $a$ and $b$ $b$ in your original problem).

This has discriminant given by the formula:

$Delta = b^2-4ac = 7^2-(4xx12xx-10) = 49+480$

$= 529 = 23^2$

Since this is a perfect square, the polynomial equation $12x^2+7x-10 = 0$ has rational roots, given by:

$x = (-b+-sqrt(Delta))/(2a) = (-7+-23)/24$

that is $x=-30/24=-5/4$ or $x = 16/24 = 2/3$ .

This allows us to deduce that:

$12x^2+7x-10 = (4x+5)(3x-2)$

and hence that:

$12a^2+7ab-10b^2 = (4a+5b)(3a-2b)$

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How do you factor $12 a^{2} + 7 a b - 10 b^{2}$ $12a^2 + 7ab - 10b^2$ ?

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