How do you factor $3t^3+2t^2-48t-32$ ?

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Answer 1 · 2017-05-19T04:26:13

Use group factoring and difference of squares: $(3t+2)(t-4)(t+4)$

Given: $3t^3 + 2t^2 - 48t - 32$

Use group factoring by factoring the GCF from the two groups:

$3t^3 + 2t^2 - 48t - 32 = (3t^3 + 2t^2) - (48t + 32)$

$= t^2(3t + 2) - 16(3t + 2)$

Notice that both groups have the factor $(3t+2)$ . Factor this from each group:

$=(3t + 2) (t^2 - 16)$

Realize that $t^2 - 16 = t^2 - 4^2$ is the difference of squares.

Difference of squares $a^2 - b^2 = (a + b) (a - b)$

$3t^3 + 2t^2 - 48t - 32 = (3t + 2) (t - 4)(t + 4)$

How do you factor 3t^3+2t^2-48t-323t^3+2t^2-48t-32?