How do you factor completely: $b^{2} - 7 b - 44$ $b^2 -7b -44$ ?

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How do you factor completely: $b^{2} - 7 b - 44$ $b^2 -7b -44$ ?

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Answer 1 · 2015-07-18T03:00:56

I found:
$(b - 11) (b + 4)$ $(b-11)(b+4)$

Explanation:

I would first try to solve the equation:
$b^{2} - 7 b - 44 = 0$ $b^2-7b-44=0$ using the Quadratic Formula:
$b_{1, 2} = \frac{7 \pm \sqrt{49 + 176}}{2} = \frac{7 \pm 15}{2} =$ $b_(1,2)=(7+-sqrt(49+176))/2=(7+-15)/2=$
$b_{1} = \frac{7 + 15}{2} = 11$ $b_1=(7+15)/2=11$
$b_{2} = \frac{7 - 15}{2} = - 4$ $b_2=(7-15)/2=-4$

So we can use these two results (with opposite signs) and get:
$(b - 11) (b + 4) = b^{2} - 7 b - 44$ $(b-11)(b+4)=b^2-7b-44$

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How do you factor completely: $b^{2} - 7 b - 44$ $b^2 -7b -44$ ?

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How do you factor completely: b2−7b−44b^2 -7b -44?

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How do you factor completely: $b^{2} - 7 b - 44$ $b^2 -7b -44$ ?