How do you factor #200x^2 + 800x +800#?

Question

Eli Assouline · Accepted Answer

#200x^2+800x+800 = 200(x+2)^2#

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

#200x^2+800x+800 = 200(x^2+4x+4)#

The remaining quadratic factor is a perfect square trinomial:

#x^2+4x+4 = (x+2)^2#

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

#144 = 12^2#

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

Putting it all together:

#200x^2+800x+800 = 200(x+2)^2#

Genesis Allen · Answer

undefined To factor the expression $200x^2 + 800x + 800$, you first need to find the greatest common factor (GCF) of all terms, which is 200. Then, you can factor out 200 from each term to simplify the expression. After factoring out the GCF, you'll have $200(x^2 + 4x + 4)$. Now, you can factor the quadratic expression inside the parentheses using the appropriate method, such as factoring by grouping or using the quadratic formula. In this case, the quadratic expression $x^2 + 4x + 4$ factors to $(x + 2)^2$. Therefore, the fully factored expression is $200(x + 2)^2$.

Julia Adams · Answer

undefined To factor the quadratic expression $200x^2 + 800x + 800$, you can start by finding the greatest common factor (GCF) of all the terms, if any. In this case, the GCF is 200.

After factoring out the GCF, you'll have $200(x^2 + 4x + 4)$.

Then, you need to factor the quadratic expression $x^2 + 4x + 4$ further. This can be done by finding two numbers that multiply to 4 (the constant term) and add to 4 (the coefficient of the linear term). These numbers are 2 and 2.

Therefore, $x^2 + 4x + 4$ can be factored as $(x + 2)(x + 2)$, or $(x + 2)^2$.

Putting it all together, the factored form of $200x^2 + 800x + 800$ is $200(x + 2)^2$.