How do you factor $8 x^{3} + 4 x^{2} - 18 x - 9$ $8x^3+4x^2-18x-9$ ?

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How do you factor $8 x^{3} + 4 x^{2} - 18 x - 9$ $8x^3+4x^2-18x-9$ ?

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Answer 1 · 2016-06-22T21:56:37

$8 x^{3} + 4 x^{2} - 18 x - 9 = (2 x - 3) (2 x + 3) (2 x + 1)$ $8x^3+4x^2-18x-9=(2x-3)(2x+3)(2x+1)$

Explanation:

Notice that the ratio between the first and second terms is the same as that between the third and fourth terms. So we can factor this cubic by grouping:

$8 x^{3} + 4 x^{2} - 18 x - 9$ $8x^3+4x^2-18x-9$

$= (8 x^{3} + 4 x^{2}) - (18 x + 9)$ $=(8x^3+4x^2)-(18x+9)$

$= 4 x^{2} (2 x + 1) - 9 (2 x + 1)$ $=4x^2(2x+1)-9(2x+1)$

$= (4 x^{2} - 9) (2 x + 1)$ $=(4x^2-9)(2x+1)$

$= ({(2 x)}^{2} - 3^{2}) (2 x + 1)$ $=((2x)^2-3^2)(2x+1)$

$= (2 x - 3) (2 x + 3) (2 x + 1)$ $=(2x-3)(2x+3)(2x+1)$

Note that we also used the difference of squares identity:

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

with $a = 2 x$ $a=2x$ and $b = 3$ $b=3$

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How do you factor $8 x^{3} + 4 x^{2} - 18 x - 9$ $8x^3+4x^2-18x-9$ ?

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How do you factor $8 x^{3} + 4 x^{2} - 18 x - 9$ $8x^3+4x^2-18x-9$ ?