How do you factor #5n ^ { 2} - 50n + 125#?

1 Answer
Jackson H.
May 19, 2017

#5(n-5)^2#

Explanation:

First, let's get rid of the 5.

#5n^2 - 50n + 125 = 5(n^2 - 10n +25)#

Now, when we factor the function in the parenthesis, we can ignore the 5 on the outside.

#(n^2 - 10n +25)#

Don't forget to multiply it back in at the end!

First, I look at the constant. There are a few ways to multiply to get to the number 25.

#(5*5) = 25#
#(25*1) = 25#
#(-5*-5) = 25#

Since we have a negative number (-10n) let's just go with the negative factors for now.

We split the equation into it's factors now.

#(n^2 - 10n +25) = (n-5)(n-5)#

We know that -5 and -5 add up to -10, which is the second factor in our equation, and -5 times -5 multiplies to 25, so this equation looks good!

Don't forget to multiply in that 5 we factored out earlier!

#5n^2 - 50n + 125 = 5(n^2 - 10n +25)#

#5(n^2 - 10n +25) = 5(n-5)(n-5)#

#5(n-5)(n-5) = 5(n-5)^2#

Bonus: A really good way to check is to plug a random number into the original equation and in to your answer. Since the equations are equal, any number plugged into both equations should produce the same result.

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