How do you factor $8 x^{3} + 28 x^{2} + 24 x$ $8x^3+28x^2+24x$ ?

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How do you factor $8 x^{3} + 28 x^{2} + 24 x$ $8x^3+28x^2+24x$ ?

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Answer 1 · 2015-05-10T23:22:29

$8 x^{3} + 28 x^{2} + 24 x = 4 x (2 x^{2} + 7 x + 6) = 4 x (2 x + 3) (x + 2)$ $8x^3+28x^2+24x = 4x(2x^2+7x+6) = 4x(2x+3)(x+2)$ .

To find this, first note that all the terms are divisible by $4$ $4$ and by $x$ $x$ . Dividing through by $4 x$ $4x$ yields the quadratic $2 x^{2} + 7 x + 6$ $2x^2+7x+6$ .

If this has linear factors with integer coefficients they must be of the form $(2 x + a)$ $(2x+a)$ and $(x + b)$ $(x+b)$ in order that the product starts with $2 x^{2}$ $2x^2$ , where $a b = 6$ $ab=6$ and $2 b + a = 7$ $2b+a=7$ . From this, it's easy to see that $a = 3$ $a=3$ and $b = 2$ $b=2$ , so $2 x^{2} + 7 x + 6 = (2 x + 3) (x + 2)$ $2x^2+7x+6 = (2x+3)(x+2)$ .

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How do you factor $8 x^{3} + 28 x^{2} + 24 x$ $8x^3+28x^2+24x$ ?

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