How do you factor $(x^{2} - 9) \times (x^{3} - 8)$ $(x^2-9)×(x^3-8)$ ?

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How do you factor $(x^{2} - 9) \times (x^{3} - 8)$ $(x^2-9)×(x^3-8)$ ?

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Answer 1 · 2015-04-09T01:07:41

$(x^{2} - 9) = (x - 2) (x + 2)$ $(x^2-9) = (x-2)(x+2)$

$(x^{3} - 8) = (x - 2) (x^{2} + 2 x + 4)$ $(x^3-8) = (x-2)(x^2+2x+4)$
(note that $(x^{2} + 2 x + 4)$ $(x^2+2x+4)$ has no Real factors; this may have been an intentional trick)

So
$(x^{2} - 9) \times (x^{3} - 8)$ $(x^2-9)xx(x^3-8)$
can be factored as
${(x - 2)}^{2} (x + 2) (x^{2} + 2 x + 4)$ $(x-2)^2(x+2)(x^2+2x+4)$

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How do you factor $(x^{2} - 9) \times (x^{3} - 8)$ $(x^2-9)×(x^3-8)$ ?

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