How do you factor $9 t^{2} + 42 t + 49$ $9t^2+42t+49$ ?

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How do you factor $9 t^{2} + 42 t + 49$ $9t^2+42t+49$ ?

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Answer 1 · 2016-05-01T10:44:40

$9 t^{2} + 42 t + 49 = {(3 t + 7)}^{2}$ $9t^2+42t+49 = (3t+7)^2$

Explanation:

This is a perfect square trinomial.

Notice that $9 t^{2} = {(3 t)}^{2}$ $9t^2 = (3t)^2$ and $49 = 7^{2}$ $49 = 7^2$ are both perfect squares, so it remains to check whether the middle term is correct:

${(3 t + 7)}^{2} = {(3 t)}^{2} + 2 (3 t) (7) + 7^{2} = 9 t^{2} + 42 t + 49$ $(3t+7)^2 = (3t)^2+2(3t)(7)+7^2 = 9t^2+42t+49$

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How do you factor $9 t^{2} + 42 t + 49$ $9t^2+42t+49$ ?

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