factor. 13+14m+m^2
\[(m+1) (m+13) \]
oh this is easy dude what adds to make 14 and multiplies to make 13?
robot you aren't supposed to just give them the answer lol
How do you determine which numbers you add and which one you multiple?
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 \[m ^{2} +14m + 13 then look for \]
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... :-)
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"
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
make sense? Have any others you'd like to try the rules with?
yes, make plenty sense. I have another one for you.
2x^3-18x^2+40x
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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