How do you factor $x^{3} + 2 x^{2} + 4 x + 8$ $x^3+2x^2+4x+8$ by grouping?

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Answer 1 · 2017-03-23T01:03:50

See the entire solution process below:

First, rewrite the expression by grouping as:

$(x^{3} + 2 x^{2}) + (4 x + 8)$ $(x^3 + 2x^2) + (4x + 8)$

Now factor each of these as:

$((x^{2} \cdot x) + (x^{2} \cdot 2)) + ((4 \cdot x) + (4 \cdot 2)) \to$ $((x^2 * x) + (x^2 * 2)) + ((4 * x) + (4 * 2)) ->$

#(x^2(x + 2)) + (4(x + 2))

We can now factor out the $(x + 2)$ $(x + 2)$ term leaving:

$(x^{2} + 4) (x + 2)$ $(x^2 + 4)(x + 2)$

How do you factor x3+2x2+4x+8x^3+2x^2+4x+8 by grouping?