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

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

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Answer 1 · 2018-05-31T13:17:06

$y = \frac{1}{2} {(x + 4)}^{2}$ $y=1/2(x+4)^2$

Explanation:

$take out a common factor \frac{1}{2}$ $"take out a "color(blue)"common factor "1/2$

$y = \frac{1}{2} (x^{2} + 8 x + 16)$ $y=1/2(x^2+8x+16)$

$x^{2} + 8 x + 16 is a perfect square$ $x^2+8x+16" is a "color(blue)"perfect square"$

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

$y = \frac{1}{2} x^{2} + 4 x + 8 = \frac{1}{2} {(x + 4)}^{2}$ $y=1/2x^2+4x+8=1/2(x+4)^2$

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

1 Answer

Explanation:

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