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
In (c) just find the roots of the quadratic equation you're trying to factor, which are 4 and $\displaystyle \frac{5}{2}$. So one factor would have to be $\displaystyle (x-4)$ and the other would have to be $\displaystyle 2z-5$. Try something similar to the other problem you're working on.
Hello, Mathnood768!
You need more Factoring practice.
$\displaystyle (c)\;\frac{4z^2-25}{2z^2 - 13z + 20} \times \frac{z-4}{4z+10}$
Factor and reduce: .$\displaystyle \frac{({\color{blue}\rlap{//////}}2z+5)({\color{red}\rlap{//////}}2z-5)}{({\color{red}\rlap{//////}}2z-5)({\color{green}\rlap{/////}}z-4)} \times \frac{{\color{green}\rlap{/////}}z-4}{2({\color{blue}\rlap{//////}}2z+5)} \;\;=\;\;\frac{1}{2}$
$\displaystyle (d)\;\frac{8y^2-2y-3}{y^2-1} \div \frac{2y^2-3y-2}{2y-2} \div \frac{3-4y}{y+1}$
$\displaystyle \frac{8y^2-2y-3}{y^2-1} \times \frac{2y-2}{2y^2-3y-2} \times \frac{y+1}{-(4y-3)}$
. . $\displaystyle =\;\frac{({\color{blue}\rlap{//////}}2y+1)({\color{red}\rlap{//////}}4y-3)}{(\rlap{/////}y+1)({\color{green}\rlap{/////}}y-1)} \times \frac{2({\color{green}\rlap{/////}}y-1)}{(y-2)({\color{blue}\rlap{//////}}2y+1)} \times -\frac{\rlap{/////}y+1}{{\color{red}\rlap{//////}}4y-3}$
. . $\displaystyle =\;-\frac{2}{y-2} \;=\;\frac{2}{2-y}$