How do you factor $72 x^{3} - 228 x^{2} + 140 x$ $72x^3-228x^2+140x$ ?

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How do you factor $72 x^{3} - 228 x^{2} + 140 x$ $72x^3-228x^2+140x$ ?

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Answer 1 · 2017-02-18T08:15:02

$72 x^{3} - 228 x^{2} + 140 x = 4 x (3 x - 7) (6 x - 5)$ $72x^3-228x^2+140x = 4x(3x-7)(6x-5)$

Explanation:

Complete the square then use the difference of squares identity:

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

with $a = (12 x - 19)$ $a=(12x-19)$ and $b = 9$ $b=9$ as follows:

$72 x^{3} - 228 x^{2} + 140 x = \frac{1}{2} x (144 x^{2} - 456 x + 280)$ $72x^3-228x^2+140x = 1/2x(144x^2-456x+280)$

$72 x^{3} - 228 x^{2} + 140 x = \frac{1}{2} x (144 x^{2} - 456 x + 361 - 81)$ $color(white)(72x^3-228x^2+140x) = 1/2x(144x^2-456x+361-81)$

$72 x^{3} - 228 x^{2} + 140 x = \frac{1}{2} x ({(12 x)}^{2} - 2 (12 x) (19) + 19^{2} - 81)$ $color(white)(72x^3-228x^2+140x) = 1/2x((12x)^2-2(12x)(19)+19^2-81)$

$72 x^{3} - 228 x^{2} + 140 x = \frac{1}{2} x ({(12 x - 19)}^{2} - 9^{2})$ $color(white)(72x^3-228x^2+140x) = 1/2x((12x-19)^2-9^2)$

$72 x^{3} - 228 x^{2} + 140 x = \frac{1}{2} x ((12 x - 19) - 9) ((12 x - 19) + 9)$ $color(white)(72x^3-228x^2+140x) = 1/2x((12x-19)-9)((12x-19)+9)$

$72 x^{3} - 228 x^{2} + 140 x = \frac{1}{2} x (12 x - 28) (12 x - 10)$ $color(white)(72x^3-228x^2+140x) = 1/2x(12x-28)(12x-10)$

$72 x^{3} - 228 x^{2} + 140 x = \frac{1}{2} x (4 (3 x - 7)) (2 (6 x - 5))$ $color(white)(72x^3-228x^2+140x) = 1/2x(4(3x-7))(2(6x-5))$

$72 x^{3} - 228 x^{2} + 140 x = 4 x (3 x - 7) (6 x - 5)$ $color(white)(72x^3-228x^2+140x) = 4x(3x-7)(6x-5)$

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How do you factor $72 x^{3} - 228 x^{2} + 140 x$ $72x^3-228x^2+140x$ ?

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How do you factor $72 x^{3} - 228 x^{2} + 140 x$ $72x^3-228x^2+140x$ ?