How do you factor the trinomial ${(a + b)}^{2} - 9 (a + b) - 36$ $(a + b)^2 - 9(a + b) - 36$ ?

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How do you factor the trinomial ${(a + b)}^{2} - 9 (a + b) - 36$ $(a + b)^2 - 9(a + b) - 36$ ?

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Answer 1 · 2015-11-29T02:45:34

${(a + b)}^{2} - 9 (a + b) - 36 = (a + b + 3) (a + b - 12)$ $(a+b)^2-9(a+b)-36=(a+b+3)(a+b-12)$

Explanation:

${(a + b)}^{2} - 9 (a + b) - 36$ $(a+b)^2-9(a+b)-36$ is a quadratic equation in the form $a x^{2} + b x + c$ $ax^2+bx+c$ , where $a = 1, b = - 9, and c = - 36$ $a=1, b=-9, and c=-36$ .

For the time being, let $(a + b) = x$ $(a+b)=x$ . After factoring with $x$ $x$ , $(a + b)$ $(a+b)$ will be substituted into the factors.

$x^{2} - 9 x - 36$ $x^2-9x-36$ . Find two numbers that when added equal $- 9$ $-9$ and when multiplied equal $- 36$ $-36$ . The numbers $3$ $3$ and $- 12$ $-12$ meet the criteria.

Rewrite the expression.

$(x + 3) (x - 12)$ $(x+3)(x-12)$

Now substitute $(a + b)$ $(a+b)$ back into the expression in place of $x$ $x$ .

${(a + b)}^{2} - 9 (a + b) - 36 = ((a + b) + 3) ((a + b) - 12) = (a + b + 3) (a + b - 12)$ $(a+b)^2-9(a+b)-36=((a+b)+3)((a+b)-12)=(a+b+3)(a+b-12)$

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How do you factor the trinomial ${(a + b)}^{2} - 9 (a + b) - 36$ $(a + b)^2 - 9(a + b) - 36$ ?

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Explanation:

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How do you factor the trinomial (a+b)2−9(a+b)−36 (a + b)^2 - 9(a + b) - 36?

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How do you factor the trinomial ${(a + b)}^{2} - 9 (a + b) - 36$ $(a + b)^2 - 9(a + b) - 36$ ?