Calculus with Applications (10th Edition)

$\displaystyle \frac{2y+1}{y+1}$
Factoring trinomials $ax^{2}+bx+c:$ search for integer factors of $a\times c$, whose sum is $b.$ If we find these factors, say m and n, rewrite the trinomial as $ax^{2}+mx+nx+c$ and factor the expression "in pairs". Factor the numerator: we search for integer factors of 6$\times$4=24, whose sum is 11 We find +8 and +3, $6y^{2}+11y+4$=$6y^{2}+3y+8y+4$ $=3y(2y+1)+4(2y+1)=(3y+4)(2y+1)$ Factor the denominator: we search for integer factors of 3$\times$4=12, whose sum is 7 We find +3 and +4, $3y^{2}+7y+4=3y^{2}+3y+4y+4$ $=3y(y+1)+4(y+1)=(3y+4)(y+1)$ $\displaystyle \frac{6y^{2}+11y+4}{3y^{2}+7y+4}=\frac{(3y+4)(2y+1)}{(3y+4)(y+1)}$ ... reduce $=\displaystyle \frac{2y+1}{y+1}$