How do you factor by grouping $t^{3} + 2 t^{2} - 7 t - 14$ $t^3+2t^2-7t-14$ ?

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How do you factor by grouping $t^{3} + 2 t^{2} - 7 t - 14$ $t^3+2t^2-7t-14$ ?

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Answer 1 · 2015-05-10T11:50:01

$t^{3} + 2 t^{2} - 7 t - 14 = (t^{3} + 2 t^{2}) - (7 t + 14)$ $t^3 + 2t^2 - 7t - 14 = (t^3 + 2t^2) - (7t + 14)$
$= t^{2} (t + 2) - 7 (t + 2) = (t^{2} - 7) (t + 2)$ $= t^2(t + 2) - 7(t + 2) = (t^2 - 7)(t + 2)$

So grouping the first two terms together and the last two terms together allows us to notice the common factor $(t + 2)$ $(t + 2)$ .

If we allow irrational coefficients, $(t^{2} - 7)$ $(t^2 - 7)$ factors as $(t - \sqrt{7}) (t + \sqrt{7})$ $(t - sqrt(7))(t + sqrt(7))$ .

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How do you factor by grouping $t^{3} + 2 t^{2} - 7 t - 14$ $t^3+2t^2-7t-14$ ?

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How do you factor by grouping t^3+2t^2-7t-14t3+2t2−7t−14t^3+2t^2-7t-14?

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How do you factor by grouping $t^{3} + 2 t^{2} - 7 t - 14$ $t^3+2t^2-7t-14$ ?