## Intermediate Algebra: Connecting Concepts through Application

$\color{blue}{\left\{4, 7\right\}}$
Add $7$ to both sides to obtain: $t^2-11t+28=0$ RECALL: A trinomial of the form $x^2+bx+c$ can be factored if there are integers $d$ and $e$ such that $c=de$ and $b=d+e$. The trinomial's factored form will be: $x^2+bx+c=(x+d)(x+e)$ The trinomial in the equation has $b=-11$ and $c=28$. Note that $28=-7(-4)$ and $-11= -7+(-4)$. This means that $d=-7$ and $e=-4$. Thus, the factored form of the trinomial is: $[t+(-t)][t+(-4)]=(t-7)(t-4)$ The equation above can be written as: $(t-7)(t-4)=0$ Use the Zero-Product Property by equating each factor to zero. Then, solve each equation to obtain: \begin{array}{ccc} &t-7 = 0 &\text{ or } &t-4=0 \\&t=7&\text{ or } &t=4 \end{array} Checking: If t=7: $7^2-11(7)+21=-7 \\49-77+21=-7 \\-28+27=-7 \\-7=-7$ If $t=4$: $4^2-11(4)+21=-7 \\16-44+21=-7 \\-28+21=-7 \\-7=-7$ Thus, the solution set is $\color{blue}{\left\{4, 7\right\}}$.