## Intermediate Algebra: Connecting Concepts through Application

$\left\{-\dfrac{7}{2}, -\dfrac{4}{3}\right\}$
Solve the equation using the $\text{ac-method}$. With $a=6$ and $c=28$, $ac= 6(28)=168$ Recall: $ax^2+bx+c$ can be factored if there exists $d$ and $e$ such that $ac=de$ and $d+e=b$. The trinomial can be factored by grouping by rewriting the trinomial as $ax^2+dx+ex+c$. Look for factors of $168$ whose sum is $29$ Note that $168=21(8)$ and $21+8=29$. This means that the trinomial is factorable. Solve the given equation by grouping: \begin{align*} 6w^2+29w+28&=0\\ 6w^2+21w+8w+28&=0\\ (6w^2+21w)+(8w+28)&=0\\ 3w(2w+7)+4(2w+7)&=0\\ (2w+7)(3w+4)&=0\end{align*} Use the Zero-Product Property by equating each factor to zero, then solving each equation to obtain: \begin{align*} 2w+7&=0 &\text{or}& &3w+4=0\\ 2w&=-7 &\text{or}& &3w=-4\\ w&=-\frac{7}{2} &\text{or}& &3w=-\frac{4}{3}\\ \end{align*}