How do you factor $20 x^{4} + 16 x^{3} - 5 x - 4$ $20x^4+16x^3-5x-4$ ?

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How do you factor $20 x^{4} + 16 x^{3} - 5 x - 4$ $20x^4+16x^3-5x-4$ ?

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Answer 1 · 2016-10-02T12:45:43

$20 x^{4} + 16 x^{3} - 5 x - 4 = (4 x^{3} - 1) (5 x + 4)$ $20x^4+16x^3-5x-4 = (4x^3-1)(5x+4)$

$20 x^{4} + 16 x^{3} - 5 x - 4 = (\sqrt[3]{4} x - 1) (2 \sqrt[3]{2} x^{2} + \sqrt[3]{4} x + 1) (5 x + 4)$ $color(white)(20x^4+16x^3-5x-4) = (root(3)(4)x-1)(2root(3)(2)x^2+root(3)(4)x+1)(5x+4)$

Explanation:

Notice that the ratio of the first and second terms is the same as that of the third and fourth terms. So this quadrinomial will factor by grouping:

$20 x^{4} + 16 x^{3} - 5 x - 4 = (20 x^{4} + 16 x^{3}) - (5 x + 4)$ $20x^4+16x^3-5x-4 = (20x^4+16x^3)-(5x+4)$

$20 x^{4} + 16 x^{3} - 5 x - 4 = 4 x^{3} (5 x + 4) - 1 (5 x + 4)$ $color(white)(20x^4+16x^3-5x-4) = 4x^3(5x+4)-1(5x+4)$

$20 x^{4} + 16 x^{3} - 5 x - 4 = (4 x^{3} - 1) (5 x + 4)$ $color(white)(20x^4+16x^3-5x-4) = (4x^3-1)(5x+4)$

We can factor $4 x^{3} - 1$ $4x^3-1$ as a difference of cubes using irrational coefficients:

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3 = (a-b)(a^2+ab+b^2)$

So:

$4 x^{3} - 1 = {(\sqrt[3]{4} x)}^{3} - 1^{3}$ $4x^3-1 = (root(3)(4)x)^3-1^3$

$4 x^{3} - 1 = (\sqrt[3]{4} x - 1) ({(\sqrt[3]{4} x)}^{2} + \sqrt[3]{4} x + 1)$ $color(white)(4x^3-1) = (root(3)(4)x-1)((root(3)(4)x)^2+root(3)(4)x+1)$

$4 x^{3} - 1 = (\sqrt[3]{4} x - 1) (\sqrt[3]{16} x^{2} + \sqrt[3]{4} x + 1)$ $color(white)(4x^3-1) = (root(3)(4)x-1)(root(3)(16)x^2+root(3)(4)x+1)$

$4 x^{3} - 1 = (\sqrt[3]{4} x - 1) (2 \sqrt[3]{2} x^{2} + \sqrt[3]{4} x + 1)$ $color(white)(4x^3-1) = (root(3)(4)x-1)(2root(3)(2)x^2+root(3)(4)x+1)$

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How do you factor $20 x^{4} + 16 x^{3} - 5 x - 4$ $20x^4+16x^3-5x-4$ ?

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