Is it possible to factor $y = 3 x^{2} + 9 x - 6$ $y=3x^2 + 9x-6$ ? If so, what are the factors?

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Is it possible to factor $y = 3 x^{2} + 9 x - 6$ $y=3x^2 + 9x-6$ ? If so, what are the factors?

1 Answer

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Answer 1 · 2017-02-22T00:08:26

$3 x^{2} + 9 x - 6 = 3 (x + \frac{3}{2} - \frac{\sqrt{17}}{2}) (x + \frac{3}{2} + \frac{\sqrt{17}}{2})$ $3x^2+9x-6 = 3(x+3/2-sqrt(17)/2)(x+3/2+sqrt(17)/2)$

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 $a = (2 x + 3)$ $a=(2x+3)$ and $b = \sqrt{17}$ $b=sqrt(17)$ as follows:

$3 x^{2} + 9 x - 6 = \frac{3}{4} (4 x^{2} + 12 x - 8)$ $3x^2+9x-6 = 3/4(4x^2+12x-8)$

$3 x^{2} + 9 x - 6 = \frac{3}{4} ({(2 x)}^{2} + 2 (2 x) (3) + 3^{2} - 17)$ $color(white)(3x^2+9x-6) = 3/4((2x)^2+2(2x)(3)+3^2-17)$

$3 x^{2} + 9 x - 6 = \frac{3}{4} ({(2 x + 3)}^{2} - {(\sqrt{17})}^{2})$ $color(white)(3x^2+9x-6) = 3/4((2x+3)^2-(sqrt(17))^2)$

$3 x^{2} + 9 x - 6 = \frac{3}{4} (2 x + 3 - \sqrt{17}) (2 x + 3 + \sqrt{17})$ $color(white)(3x^2+9x-6) = 3/4(2x+3-sqrt(17))(2x+3+sqrt(17))$

$3 x^{2} + 9 x - 6 = 3 (x + \frac{3}{2} - \frac{\sqrt{17}}{2}) (x + \frac{3}{2} + \frac{\sqrt{17}}{2})$ $color(white)(3x^2+9x-6) = 3(x+3/2-sqrt(17)/2)(x+3/2+sqrt(17)/2)$

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Is it possible to factor $y = 3 x^{2} + 9 x - 6$ $y=3x^2 + 9x-6$ ? If so, what are the factors?

1 Answer

Explanation:

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Is it possible to factor $y = 3 x^{2} + 9 x - 6$ $y=3x^2 + 9x-6$ ? If so, what are the factors?