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Mathematics 24 Online
OpenStudy (bbb911):

x^2 -15x =0

OpenStudy (anonymous):

\(x^2-15x=0\) Looks like a factoring question to me. What can you divide both terms by?

OpenStudy (bbb911):

the 15 by 3 and 5 right?

OpenStudy (anonymous):

No, what is common to both terms? One term is \(x^2\) and the other is \(-15x\). A term is the coefficient, variable and exponent.

OpenStudy (bbb911):

they both have x's right

OpenStudy (bbb911):

i wasn't so sure if this would be it. (x ) (x ) to start with

OpenStudy (anonymous):

Right, so divide both terms by x and you get \(\large{x(x-15)}\). If there was a constant term (other than 0), then you would have used the reverse FOIL form. (x )(x )

OpenStudy (bbb911):

so the x would be on the outside right? then the answer would be x=15 right

OpenStudy (anonymous):

No. You would still set each = to 0 to solve.

OpenStudy (anonymous):

\(x = 0\) and \( x-15 = 0\)

OpenStudy (bbb911):

so would it be15x=0 but because the the negative now turns into a postive

OpenStudy (anonymous):

I shouldn't have said no... \(x=15\) is one answer.

OpenStudy (anonymous):

This is a second degree polynomial and therefore has two answers.

OpenStudy (bbb911):

oh so there's more than one answer?

OpenStudy (anonymous):

Just the same as if it had factored to this format: \((x-2)(x+5)\) \(x-2=0\) and \(x+5=0\)

OpenStudy (anonymous):

Only in this case it factors to: \(x\) and \(x-15\)

OpenStudy (bbb911):

oh ok thank ou so uch :)

OpenStudy (anonymous):

Your welcome.

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