How do you factor $16 + 20y - 6y^2$ ?

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Answer 1 · 2015-05-16T13:50:00

$16+20y-6y^2 = -2(3y^2-10y-8)$

$3y^2-10y-8$ is of the form $ay^2+by+c$ , with $a=3$ , $b=-10$ and $c=-8$ .

The general quadratic solution tells us that $ay^2+by+c = 0$ when

$y=(-b+-sqrt(b^2-4ac))/(2a)$

So substituting our values of $a$ , $b$ and $c$ , that gives

$y=(10+-sqrt(10^2-4*3*(-8)))/(2*3)$

$=(10+-sqrt(100+96))/6$

$=(10+-sqrt(196))/6$

$=(10+-14)/6$

That is for $y=4$ and $y=-2/3$

So $(y-4)$ and $(3y+2)$ are both factors of $3y^2-10y-8$

In summary $16+20y-6y^2 = -2(y-4)(3y+2) = 2(4-y)(3y+2)$

How do you factor 16 + 20y - 6y^216 + 20y - 6y^2?