How do you factor $200 x^{2} + 800 x + 800$ $200x^2 + 800x +800$ ?

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How do you factor $200 x^{2} + 800 x + 800$ $200x^2 + 800x +800$ ?

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Answer 1 · 2016-05-12T22:17:40

$200 x^{2} + 800 x + 800 = 200 {(x + 2)}^{2}$ $200x^2+800x+800 = 200(x+2)^2$

Explanation:

All of the terms are divisible by $200$ $200$ , so separate that out first:

$200 x^{2} + 800 x + 800 = 200 (x^{2} + 4 x + 4)$ $200x^2+800x+800 = 200(x^2+4x+4)$

The remaining quadratic factor is a perfect square trinomial:

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

Look at the pattern of the coefficients. It might remind you of this:

$144 = 12^{2}$ $144 = 12^2$

This is no coincidence. The coefficients are all small enough that no digits are carried during the multiplication of $12 \times 12$ $12xx12$ , so the digits $1, 4, 4$ $1,4,4$ of $144$ $144$ are the coefficients you get when you square $(x + 2)$ $(x+2)$ .

Putting it all together:

$200 x^{2} + 800 x + 800 = 200 {(x + 2)}^{2}$ $200x^2+800x+800 = 200(x+2)^2$

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How do you factor $200 x^{2} + 800 x + 800$ $200x^2 + 800x +800$ ?

1 Answer

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How do you factor $200 x^{2} + 800 x + 800$ $200x^2 + 800x +800$ ?