Q: What is the factored version of x3 - 8?

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x3 - 5x2 + 6x - 8 = x3 - 4x2 - x2 + 4x + 2x - 8 = x2(x - 4) - x(x - 4) + 2(x - 4) = (x - 4)(x2 - x + 2) which has no further real factors.

An expression. It can be factored out slightly too: 2x6 + 5x3 - 7x4 = x3(2x3 - 7x + 5)

x3 - 2x2 - 4x + 8 = (x2 - 4)(x - 2) = (x + 2)(x - 2)(x - 2)

If you mean: n2+16n+64 then it is (n+8)(n+8) when factored

11

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71

x3-x2

x3 - x2 + 2x = x*(x2 - x + 2) which cannot be factored further.

x3 - x2 + x - 2 has no rational factors.

273-8 factored = 265

(x-2)(x^2+3)

(x3 - 8) is factored thus: (x - 2)(x2 + 2x + 4) The easiest way to do this is to remember the formula: (a3 - b3) = (a - b)(a2 + ab + b2)

x3 - 5x2 + 6x - 8 = x3 - 4x2 - x2 + 4x + 2x - 8 = x2(x - 4) - x(x - 4) + 2(x - 4) = (x - 4)(x2 - x + 2) which has no further real factors.

y5(x3 - 8)

It is (x-1)(x-8) when factored

[ x3 + 3x2 + 2x ] is a trinomial. It's factors are [ x, (x + 1), (x + 2) ] .

An expression. It can be factored out slightly too: 2x6 + 5x3 - 7x4 = x3(2x3 - 7x + 5)