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Mathematics 15 Online
OpenStudy (anonymous):

HELP PLEASEE! Simplify (x^2-5x+16)/(15xy^2)/(2x^2-7x+3)/(5x^2y)

OpenStudy (anonymous):

This is what it looks like.

OpenStudy (anonymous):

I know you flip the denomintor and mulitply, but im struggling?

OpenStudy (anonymous):

you need to solve equations (x^2-5x+16) and (2x^2-7x+3) and write them as (x-x1)(x-x2) where x1 and x2 are solutions of first equation (do the same for another equation)

OpenStudy (anonymous):

notice that in second equation you have 2x^2 so (x-x1)(x-x2) in this case will be 2*(x-x1)(x-x2)

OpenStudy (anonymous):

OpenStudy (anonymous):

Those are the answers i can choose from, i think i'm doing something wrong though,

OpenStudy (anonymous):

i keep getting the last one, and i don't think that's it

OpenStudy (anonymous):

First, I flipped it then multiply so; ((x^2 - 5x + 16) (5x^2 y))/ ((15x y^2) (2x^2 -7x +3)) (x(x^2 - 5x + 16))/ (3y (2x^2 - 7x + 3)) ((x (x - 8) (x - 2)) +5x) /(3y (2x - 1)(x-3)) from the numerator factor out the x so; (x ((x-8)(x-2) +5)) / (3y (2x - 1)(x-3)) (x (x^2 -10x +16 + 5))/ (3y (2x - 1)(x-3)) (x(x^2 - 10x + 21)) / (3y (2x - 1)(x-3)) Factor (x^2 - 10x + 2) so = (x (x -7) (x - 3)) / (3y (2x - 1)(x-3)) you cancel out (x-3) so the answer is = (x(x-7)) / (3y(2x-1))

OpenStudy (anonymous):

Is your problem this Simplify (x^2-5x+16)/(15xy^2)/(2x^2-7x+3)/(5x^2y) or this (x^2-5x+6)/(15xy^2)/(2x^2-7x+3)/(5x^2y)

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