How do you factor the trinomial $b^2-4b-32$ ?

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Answer 1 · 2017-06-18T20:25:06

See a solution process below:

We can use the quadratic formula to factor this expression as:

The quadratic formula states:

For $ax^2 + bx + c = 0$ , the values of $x$ which are the solutions to the equation are given by:

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

Substituting $1$ for $a$ ; $-4$ for $b$ and $-32$ for $c$ gives:

$b = (-(-4) +- sqrt((-4)^2 - (4 * 1 * -32)))/(2 * 1)$

$b = (4 +- sqrt(16 - (-128)))/(2)$

$b = (4 +- sqrt(16 + 128))/(2)$

$b = (4 +- sqrt(144))/(2)$

$b = (4 +- 12)/(2)$

$b = (4 + 12)/(2)$ or $b = (4 - 12)/(2)$

$b = 16/2 or$ b = -8/2#

$b = 8$ or $b = -4$

$b - 8 = 0$ or $b + 4 = 0$

$b^2 − 4b − 32 => (b - 8)(b + 4)$

How do you factor the trinomial b^2-4b-32b^2-4b-32?