How do you solve $x^{2} + 3 x + 2$ $x^2+3x+2$ by completing the square?

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Answer 1 · 2015-05-17T19:57:19

The answer is $x = - 1$ $x=-1$ and $x = - 2$ $x=-2$

$x^{2} + 4 x + 4 - x - 2 = 0 \Rightarrow {(x + 2)}^{2} - (x + 2) = 0$ $x^2+4x+4-x-2=0 => (x+2)^2-(x+2)=0$

So you know one answer is $x = - 2$ $x=-2$

Now you can divide the polynomial by its factor $(x + 2)$ $(x+2)$ and you have

$x + 2 - 1 = 0 \Rightarrow x = - 1$ $x+2 - 1 = 0 => x=-1$

How do you solve x2+3x+2x^2+3x+2 by completing the square?