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Mathematics 26 Online
OpenStudy (anonymous):

factor. 13+14m+m^2

OpenStudy (anonymous):

\[(m+1) (m+13) \]

OpenStudy (anonymous):

oh this is easy dude what adds to make 14 and multiplies to make 13?

OpenStudy (anonymous):

robot you aren't supposed to just give them the answer lol

OpenStudy (anonymous):

How do you determine which numbers you add and which one you multiple?

OpenStudy (anonymous):

Kathy, when you factor (and first turn it around so the x^2 term is in the front, easy to do here with all + signs.... so \[m ^{2} +14m + 13 then look for \]

OpenStudy (anonymous):

two numbers that MULTIPLY to give teh back # (only options are 1 and 13) and ADD to give the middle term (or linear term)...1 +13 = 14, so no prob... :-)

OpenStudy (anonymous):

as for the signs, you look at the back sign to see if you will have like or unlike signs in your ( ) a + in back means "two of the same sign" and a - in the back means "one of each"

OpenStudy (anonymous):

and in this problem since both signs are the same and are + (look at the middle term to determine that) it doesn't matter which number goes in which () AS LONG AS THE LEADING COEFFICIENT IS 1... in this case 1m^2

OpenStudy (anonymous):

make sense? Have any others you'd like to try the rules with?

OpenStudy (anonymous):

yes, make plenty sense. I have another one for you.

OpenStudy (anonymous):

2x^3-18x^2+40x

OpenStudy (anonymous):

ok, back if you still need the help....now this one has an extra step....you need to first check the coefficients to see if there is a gcf.....in this case, there's a 2x for the gcf...factor it out and you get 2x(x^2 -9x + 20)

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