How do you factor $10w² + 11w - 8$ ?

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Answer 1 · 2015-05-17T11:59:35

$10w^2+11w-8$ is of the form $aw^2+bw+c$ with $a=10$ , $b=11$ and $c=-8$ .

The discriminant of this quadratic is given by the formula

$Delta = b^2-4ac = 11^2-(4xx10xx-8) = 121+320 = 441 = 21^2$

Being a positive square integer the equation $10w^2+11w-8 = 0$ has two distinct rational roots, given by the formula:

$w = (-b+-sqrt(Delta))/(2a) = (-11+-21)/20$

That is

$w = (-11+21)/20 = 10/20 = 1/2$

and

$w = (-11-21)/20 = -32/20 = -8/5$

These give us factors $(2w-1)$ and $(5w+8)$

So $10w^2+11w-8 = (2w-1)(5w+8)$

How do you factor 10w² + 11w - 810w² + 11w - 8?