1. rational expressions #2

i cannot finish 1 c) and d)... that is how far i got.

i am having some problems with turning trinomials with no common factors into binomials

thank you

2. Re: rational expressions #2

In (c) just find the roots of the quadratic equation you're trying to factor, which are 4 and $\frac{5}{2}$. So one factor would have to be $(x-4)$ and the other would have to be $2z-5$. Try something similar to the other problem you're working on.

3. Re: rational expressions #2

Hello, Mathnood768!

You need more Factoring practice.

$(c)\;\frac{4z^2-25}{2z^2 - 13z + 20} \times \frac{z-4}{4z+10}$

Factor and reduce: . $\frac{({\color{blue}\rlap{//////}}2z+5)({\color{red}\rlap{//////}}2z-5)}{({\color{red}\rlap{//////}}2z-5)({\color{green}\rlap{/////}}z-4)} \times \frac{{\color{green}\rlap{/////}}z-4}{2({\color{blue}\rlap{//////}}2z+5)} \;\;=\;\;\frac{1}{2}$

$(d)\;\frac{8y^2-2y-3}{y^2-1} \div \frac{2y^2-3y-2}{2y-2} \div \frac{3-4y}{y+1}$

$\frac{8y^2-2y-3}{y^2-1} \times \frac{2y-2}{2y^2-3y-2} \times \frac{y+1}{-(4y-3)}$

. . $=\;\frac{({\color{blue}\rlap{//////}}2y+1)({\color{red}\rlap{//////}}4y-3)}{(\rlap{/////}y+1)({\color{green}\rlap{/////}}y-1)} \times \frac{2({\color{green}\rlap{/////}}y-1)}{(y-2)({\color{blue}\rlap{//////}}2y+1)} \times -\frac{\rlap{/////}y+1}{{\color{red}\rlap{//////}}4y-3}$

. . $=\;-\frac{2}{y-2} \;=\;\frac{2}{2-y}$

thank you!

5. Re: rational expressions #2

why is it not 2/y - 2 though? why do you say its negative?

6. Re: rational expressions #2

The minus comes from the 3- 4y in the original problem. In order to be able to cancel with the 4y- 3 which is a factor $8y^2- 2y- 3= (2y+ 1)(4y- 3)$, Soroban factored out a "-1" writing it as -(4y- 3).