Question

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

Answer 1

`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.