(-62y^(3)+10y^(2)+32y)/(2y)

Mathematics

1 answer:

sattari [20]2 years ago

6 0

Divide each term in the numerator by 2y:-

= -31y^2 + 5y + 16 is the answer.

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$\displaystyle\ \large \boldsymbol{} 2t^2+5t+2=0 \\\\2t^2+\underbrace{4t+t}_{5t}+2=0 \\\\ (2t^2+4t)+(t+2)=0 \\\\ 2t(\underline{t+2})+(\underline{t+2})=0 \\\\ \boxed{(2t+1) (t+2)=0 }$

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