How do you factor $x^{3} + 6 x^{2} - 5 x - 30$ $x^3+6x^2-5x-30$ ?

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How do you factor $x^{3} + 6 x^{2} - 5 x - 30$ $x^3+6x^2-5x-30$ ?

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Answer 1 · 2018-04-15T16:09:23

$x = - 6, x =$ $x= -6, x=$ $\pm$ $+-$ $\sqrt{}$ $sqrt$ 5

Explanation:

rearrange the equation
$x^{3} - 5 x + 6 x^{2} - 30$ $x^3 - 5x + 6x^2 - 30$

take out the common factors
$x (x^{2} - 5) + 6 (x^{2} - 5)$ $x(x^2 -5) +6(x^2 - 5)$

compress the terms
$(x + 6) (x^{2} - 5)$ $(x+6)(x^2 - 5)$

solve for $x$ $x$
$x = - 6, x =$ $x= -6, x=$ $\pm$ $+-$ $\sqrt{}$ $sqrt$ 5

Answer 2 · 2018-04-15T16:26:55

$(x + 6) (x - \sqrt{5}) (x + \sqrt{5})$ $(x+6)(x-sqrt5)(x+sqrt5)$

Explanation:

$factor by grouping the terms$ $color(blue)"factor by grouping the terms"$

$= x^{2} (x + 6) - 5 (x + 6)$ $=color(red)(x^2)(x+6)color(red)(-5)(x+6)$

$take out the common factor (x + 6)$ $"take out the "color(blue)"common factor "(x+6)$

$= (x + 6) (x^{2} - 5)$ $=(x+6)(color(red)(x^2-5))$

$x^{2} - 5 can be factored using difference of squares$ $x^2-5" can be factored using "color(blue)"difference of squares"$

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2=(a-b)(a+b)$

$x^{2} - 5 = x^{2} - {(\sqrt{5})}^{2}$ $x^2-5=x^2-(sqrt5)^2$

$with a = x and b = \sqrt{5}$ $"with "a=x" and "b=sqrt5$

$\Rightarrow x^{2} - 5 = (x - \sqrt{5}) (x + \sqrt{5})$ $rArrx^2-5=(x-sqrt5)(x+sqrt5)$

$\Rightarrow x^{3} + 6 x^{2} - 5 x - 30 = (x + 6) (x - \sqrt{5}) (x + \sqrt{5})$ $rArrx^3+6x^2-5x-30=(x+6)(x-sqrt5)(x+sqrt5)$

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How do you factor $x^{3} + 6 x^{2} - 5 x - 30$ $x^3+6x^2-5x-30$ ?

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How do you factor $x^{3} + 6 x^{2} - 5 x - 30$ $x^3+6x^2-5x-30$ ?