Is it possible to factor $y = 2 x^{2} + 8 x + 4$ $y=2x^2 + 8x+4$ ? If so, what are the factors?

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Is it possible to factor $y = 2 x^{2} + 8 x + 4$ $y=2x^2 + 8x+4$ ? If so, what are the factors?

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Answer 1 · 2015-12-08T19:25:21

First you can take out the factor $2$ $2$

Explanation:

$\to y = 2 (x^{2} + 4 x + 2)$ $->y=2(x^2+4x+2)$

This cannot be factored any further without using radical expressions.

Answer 2 · 2015-12-08T19:33:01

$2 x^{2} + 8 x + 4 = 2 (x^{2} + 4 x + 2)$ $2x^2+8x+4 = 2(x^2+4x+2)$

$= 2 (x + 2 + \sqrt{2}) (x + 2 - \sqrt{2})$ $= 2(x+2+sqrt(2))(x+2-sqrt(2))$

Explanation:

Given $y = 2 x^{2} + 8 x + 4$ $y = 2x^2+8x+4$ first separate out the common scalar factor $2$ $2$ to get:

$y = 2 (x^{2} + 4 x + 2)$ $y = 2(x^2+4x+2)$

The quadratic factor is in the form $a x^{2} + b x + c$ $ax^2+bx+c$ with $a = 1$ $a=1$ , $b = 4$ $b=4$ , $c = 2$ $c=2$ . This has zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- 4 \pm \sqrt{{(- 4)}^{2} - (4 \times 1 \times 2)}}{2 \times 1}$ $x = (-b+-sqrt(b^2-4ac))/(2a) = (-4+-sqrt((-4)^2-(4xx1xx2)))/(2xx1)$

$= \frac{- 4 \pm \sqrt{8}}{2} = \frac{- 4 \pm \sqrt{2^{2} \cdot 2}}{2} = \frac{- 4 \pm 2 \sqrt{2}}{2} = - 2 \pm \sqrt{2}$ $=(-4+-sqrt(8))/2 = (-4+-sqrt(2^2*2))/2 = (-4+-2sqrt(2))/2 = -2+-sqrt(2)$

These zeros correspond to factors $(x + 2 + \sqrt{2})$ $(x+2+sqrt(2))$ and $(x + 2 - \sqrt{2})$ $(x+2-sqrt(2))$

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Another way of finding this is by completing the square, then using the difference of squares identity:

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

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

$y = 2 x^{2} + 8 x + 4$ $y = 2x^2+8x+4$

$= 2 (x^{2} + 4 x + 2)$ $=2(x^2+4x+2)$

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

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

$= 2 ({(x + 2)}^{2} - {(\sqrt{2})}^{2})$ $=2((x+2)^2-(sqrt(2))^2)$

$= 2 ((x + 2) - \sqrt{2}) ((x + 2) + \sqrt{2})$ $=2((x+2)-sqrt(2))((x+2)+sqrt(2))$

$= 2 (x + 2 - \sqrt{2}) (x + 2 + \sqrt{2})$ $=2(x+2-sqrt(2))(x+2+sqrt(2))$

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Is it possible to factor $y = 2 x^{2} + 8 x + 4$ $y=2x^2 + 8x+4$ ? If so, what are the factors?

2 Answers

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Is it possible to factor $y = 2 x^{2} + 8 x + 4$ $y=2x^2 + 8x+4$ ? If so, what are the factors?