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Mathematics 23 Online
OpenStudy (anonymous):

Need help with dividing polynomials 2x^2+16x+30/5x^2+13x-6

OpenStudy (anonymous):

(2x+6)(x+5) the dashed line is the fraction, this may make it easier ------------ (5x-2)(x+3)

OpenStudy (anonymous):

my problem is more of a "its hard to factor with so many numbers...

OpenStudy (anonymous):

"

OpenStudy (anonymous):

why is it harder to factor to factor with so many numbers? both the numerator and denominator are quadratics. do you knpw how to factor a quadratic?

OpenStudy (niel):

its really simple actually one you get past factoring but that got me too

OpenStudy (niel):

sorry its actually my question but i didnt have an account until a minute ago

OpenStudy (niel):

its just confusing thats all

OpenStudy (anonymous):

so did you get the answer or not?

OpenStudy (niel):

still looking at it trying to piece together what i know with what i learned in class

OpenStudy (anonymous):

no:(

OpenStudy (anonymous):

okay, do you know how to factor quadratic polynomials?

OpenStudy (niel):

yes ... ok so is it 2 (x+5)/5x-2?

OpenStudy (niel):

i just hav trouble when the leading coefficient is >1

OpenStudy (anonymous):

okay, notice that 2x^2+16x+30 has a common factor of 2 so take the 2 out. your expression become 2(x^2+8x+15) leading coefficients also work the same way as normal quadratics. http://www.youtube.com/watch?v=eF6zYNzlZKQ

OpenStudy (niel):

but what about the denominator

OpenStudy (niel):

sorry i meant when there is no gcf other than 1

OpenStudy (anonymous):

5x^2+13x-6 = 5x^2 +15x-2x -6 = 5x(x+3)-2(x+3)=(5x-2)(x+3) multiply the end terms. in this case, they are 5 and -6, giving -30. split the middle term such that the product of the split is the same as the product of the end term so we split 13x as 15x-2x. because 15 *(-2) = -30

OpenStudy (niel):

so 5*-6=-30 and just guess and check for the factors of -30?

OpenStudy (anonymous):

(2x+6)(x+5) ------------ (5x-2)(x+3)

OpenStudy (anonymous):

@niel, that is right

OpenStudy (niel):

ok well thanks... ill come back if i need help again. this is an amazing site, glad hayley showed me

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