How do you factor by grouping $x^{3} + 6 x^{2} - x - 30$ $x^3+6x^2-x-30$ ?

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How do you factor by grouping $x^{3} + 6 x^{2} - x - 30$ $x^3+6x^2-x-30$ ?

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Answer 1 · 2015-05-10T14:31:18

$x^{3} + 6 x^{2} - x - 30$ $x^3+6x^2-x-30$
$= (x^{3} + 8 x^{2} + 15 x) - (2 x^{2} + 16 x + 30)$ $= (x^3+8x^2+15x) - (2x^2+16x+30)$
$= x (x^{2} + 8 x + 15) - 2 (x^{2} + 8 x + 15)$ $= x(x^2+8x+15)-2(x^2+8x+15)$
$= (x - 2) (x^{2} + 8 x + 15)$ $= (x-2)(x^2+8x+15)$
$= (x - 2) (x^{2} + (3 + 5) x + 3 \cdot 5)$ $= (x-2)(x^2+(3+5)x+3*5)$
$= (x - 2) (x + 3) (x + 5)$ $= (x-2)(x+3)(x+5)$

I actually noticed that $x = 2$ $x = 2$ is a solution of $x^{3} + 6 x^{2} - x - 30 = 0$ $x^3+6x^2-x-30 = 0$ , so $(x - 2)$ $(x-2)$ is a factor and worked from there.

When looking for solutions of polynomials in one variable, the highest order term becomes dominant quite quickly. If all the coefficients are integers, and the highest order coefficient is 1 then any rational solution is an integer, and any integer solution must be a factor of the constant term. In the case of $x^{3} + 6 x^{2} - x - 30$ $x^3+6x^2-x-30$ the constant term $30$ $30$ has quite a few factors you might choose, but only -6, -5, -3, -2, -1, 0, 1, 2, 3, 5 and 6 would probably make sense to try.

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How do you factor by grouping $x^{3} + 6 x^{2} - x - 30$ $x^3+6x^2-x-30$ ?

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How do you factor by grouping $x^{3} + 6 x^{2} - x - 30$ $x^3+6x^2-x-30$ ?