How do you factor $27 x^{3} + 54 x^{2} + 36 x + 8$ $27x^3 + 54 x^2 + 36x +8$ by grouping?

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Answer 1 · 2016-06-19T08:00:01

Note that the first and last terms are perfect cubes and find:

$27 x^{3} + 54 x^{2} + 36 x + 8 = {(3 x + 2)}^{3}$ $27x^3+54x^2+36x+8 = (3x+2)^3$

Notice that the ratio between the first and second coefficients is not the same as that between the third and fourth.

That is: $\frac{27}{54} \neq \frac{36}{8}$ $27/54 != 36/8$

So this does not factor by grouping.

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Note that the first and last terms are both perfect cubes:

$27 x^{3} = {(3 x)}^{3}$ $27x^3 = (3x)^3$

$8 = 2^{3}$ $8 = 2^3$

So we might try ${(3 x + 2)}^{3}$ $(3x+2)^3$ :

${(3 x + 2)}^{3} = 27 x^{3} + 54 x^{2} + 36 x + 8$ $(3x+2)^3 = 27x^3+54x^2+36x+8$

Yes - that's our factoring.

How do you factor 27x3+54x2+36x+8 27x^3 + 54 x^2 + 36x +8 by grouping?