6.5 Polynomial Equations

$\begin{array}{cccccc}& & & & & 169x^2=49\hfill \\ \text{Write the quadratic equation in standard form.}\hfill & & & & & 169x^2-49=0\hfill \\ \text{Factor. It is a difference of squares.}\hfill & & & & & \left(13x-7\right)\left(13x+7\right)=0\hfill \\ \begin{array}{c}\text{Use the Zero Product Property to set each factor to}0.\hfill \\ \text{Solve each equation.}\hfill \\ \\ \\ \\ \\ \\ \end{array}\hfill & & & & & \begin{array}{ccccccccc}\hfill 13x-7& =\hfill & 0\hfill & & & & \hfill 13x+7& =\hfill & 0\hfill \\ \hfill 13x& =\hfill & 7\hfill & & & & \hfill 13x& =\hfill & \mathrm{-7}\hfill \\ \hfill x& =\hfill & \frac{7}{13}\hfill & & & & \hfill x& =\hfill & -\frac{7}{13}\hfill \end{array}\hfill \end{array}$

Check:

We leave the check up to you.

$\begin{array}{cccccc}& & & & & \left(3x-8\right)\left(x-1\right)=3x\hfill \\ \text{Multiply the binomials.}\hfill & & & & & 3x^2-11x+8=3x\hfill \\ \text{Write the quadratic equation in standard form.}\hfill & & & & & 3x^2-14x+8=0\hfill \\ \text{Factor the trinomial.}\hfill & & & & & \left(3x-2\right)\left(x-4\right)=0\hfill \\ \begin{array}{c}\text{Use the Zero Product Property to set each factor to 0.}\hfill \\ \text{Solve each equation.}\hfill \end{array}\hfill & & & & & \begin{array}{ccccccccc}\hfill 3x-2& =\hfill & 0\hfill & & & & \hfill x-4& =\hfill & 0\hfill \\ \hfill 3x& =\hfill & 2\hfill & & & & \hfill x& =\hfill & 4\hfill \end{array}\hfill \\ & & & & & x=\frac{2}{3}\hfill \\ \text{Check your answers.}\hfill & & & & & \text{The check is left to you.}\hfill \end{array}$

$\begin{array}{cccccc}& & & & & 3x^2=12x+63\hfill \\ \text{Write the quadratic equation in standard form.}\hfill & & & & & 3x^2-12x-63=0\hfill \\ \text{Factor the greatest common factor first.}\hfill & & & & & 3\left(x^2-4x-21\right)=0\hfill \\ \text{Factor the trinomial.}\hfill & & & & & 3\left(x-7\right)\left(x+3\right)=0\hfill \\ \begin{array}{c}\text{Use the Zero Product Property to set each factor to 0.}\hfill \\ \text{Solve each equation.}\hfill \end{array}\hfill & & & & & \begin{array}{ccccccccccccc}3\ne 0\hfill & & & & \hfill x-7& =\hfill & 0\hfill & & & & \hfill x+3& =\hfill & 0\hfill \\ 3\ne 0\hfill & & & & \hfill x& =\hfill & 7\hfill & & & & \hfill x& =\hfill & \mathrm{-3}\hfill \end{array}\hfill \\ \text{Check your answers.}\hfill & & & & & \text{The check is left to you.}\hfill \end{array}$

$\begin{array}{cccccc}& & & & & 9m^3+100m=60m^2\hfill \\ \text{Bring all the terms to one side so that the other side is zero.}\hfill & & & & & 9m^3-60m^2+100m=0\hfill \\ \text{Factor the greatest common factor first.}\hfill & & & & & m\left(9m^2-60m+100\right)=0\hfill \\ \text{Factor the trinomial.}\hfill & & & & & m(3m-10)(3m-10)=0\hfill \\ \begin{array}{c}\text{Use the Zero Product Property to set each factor to 0.}\hfill \\ \text{Solve each equation.}\hfill \end{array}\hfill & & & & & \begin{array}{ccccccccccccc}m=0\hfill & & & & \hfill 3m-10& =\hfill & 0\hfill & & & & \hfill 3m-10& =\hfill & 0\hfill \\ m=0\hfill & & & & \hfill m& =\hfill & \frac{10}{3}\hfill & & & & \hfill m& =\hfill & \frac{10}{3}\hfill \end{array}\hfill \\ \text{Check your answers.}\hfill & & & & & \text{The check is left to you.}\hfill \end{array}$

ⓐ
$\begin{array}{cccccc}& & & & & f\left(x\right)=x^2+2x-2\hfill \\ \text{Substitute 6 for}f\left(x\right).\hfill & & & & & 6=x^2+2x-2\hfill \\ \text{Put the quadratic in standard form.}\hfill & & & & & x^2+2x-8=0\hfill \\ \text{Factor the trinomial.}\hfill & & & & & \left(x+4\right)\left(x-2\right)=0\hfill \\ \begin{array}{c}\text{Use the zero product property.}\hfill \\ \text{Solve.}\hfill \end{array}\hfill & & & & & \begin{array}{ccccccccccc}\hfill x+4& =\hfill & 0\hfill & & & \text{or}\hfill & & & \hfill x-2& =\hfill & 0\hfill \\ \hfill x& =\hfill & \mathrm{-4}\hfill & & & \text{or}\hfill & & & \hfill x& =\hfill & 2\hfill \end{array}\hfill \\ \text{Check:}\hfill & & & & & \\ \\ \\ \\ \\ \\ \\ \\ \\ \begin{array}{ccccc}f\left(x\right)=x^2+2x-2\hfill & & & & f\left(x\right)=x^2+2x-2\hfill \\ f\left(\text{−}4\right)={\left(\text{−}4\right)}^2+2\left(\text{−}4\right)-2\hfill & & & & f\left(2\right)=2^2+2·2-2\hfill \\ f\left(\text{−}4\right)=16-8-2\hfill & & & & f\left(2\right)=4+4-2\hfill \\ f\left(\text{−}4\right)=6✓\hfill & & & & f\left(2\right)=6✓\hfill \end{array}\hfill & & & & & \end{array}$

ⓑ Since $f\left(\mathrm{-4}\right)=6$ and $f\left(2\right)=6,$ the points $\left(\mathrm{-4},6\right)$ and $\left(2,6\right)$ lie on the graph of the function.

ⓐ To find the zeros of the function, we need to find when the function value is 0.
$\begin{array}{cccccc}& & & & & f\left(x\right)=3x^2+10x-8\hfill \\ \text{Substitute 0 for}f\left(x\right).\hfill & & & & & 0=3x^2+10x-8\hfill \\ \text{Factor the trinomial.}\hfill & & & & & \left(x+4\right)\left(3x-2\right)=0\hfill \\ \begin{array}{c}\text{Use the zero product property.}\hfill \\ \text{Solve.}\hfill \end{array}\hfill & & & & & \begin{array}{ccccccccccc}\hfill x+4& =\hfill & 0\hfill & & & \text{or}\hfill & & & \hfill 3x-2& =\hfill & 0\hfill \\ \hfill x& =\hfill & \mathrm{-4}\hfill & & & \text{or}\hfill & & & \hfill x& =\hfill & \frac{2}{3}\hfill \end{array}\hfill \end{array}$

ⓑ An x-intercept occurs when $y=0.$ Since $f\left(\mathrm{-4}\right)=0$ and $f\left(\frac{2}{3}\right)=0,$ the points $\left(\mathrm{-4},0\right)$ and $\left(\frac{2}{3},0\right)$ lie on the graph. These points are x-intercepts of the function.


ⓒ A y-intercept occurs when $x=0.$ To find the y-intercepts we need to find $f\left(0\right).$
$\begin{array}{cccccc}& & & & & f\left(x\right)=3x^2+10x-8\hfill \\ \text{Find}f\left(0\right)\text{by substituting 0 for}x.\hfill & & & & & f\left(0\right)=3·0^2+10·0-8\hfill \\ \text{Simplify.}\hfill & & & & & f\left(0\right)=\mathrm{-8}\hfill \end{array}$
Since $f\left(0\right)=\mathrm{-8},$ the point $\left(0,\mathrm{-8}\right)$ lies on the graph. This point is the y-intercept of the function.

$\begin{array}{cccccc}\mathbf{\text{Step 1. Read}}\text{the problem.}\hfill & & & & & \\ \mathbf{\text{Step 2. Identify}}\text{what we are looking for.}\hfill & & & & & \text{We are looking for two consecutive integers.}\hfill \\ \mathbf{\text{Step 3. Name}}\text{what we are looking for.}\hfill & & & & & \text{Let}n=\text{the first integer.}\hfill \\ & & & & & n+2=\text{next consecutive odd integer}\hfill \\ \begin{array}{c}\mathbf{\text{Step 4. Translate}}\text{into an equation. Restate the}\hfill \\ \text{problem in a sentence.}\hfill \end{array}\hfill & & & & & \begin{array}{c}\text{The product of the two consecutive odd}\hfill \\ \text{integers is 323.}\hfill \end{array}\hfill \\ & & & & & n\left(n+2\right)=323\hfill \\ \mathbf{\text{Step 5. Solve}}\text{the equation.}\hfill & & & & & n^2+2n=323\hfill \\ \text{Bring all the terms to one side.}\hfill & & & & & n^2+2n-323=0\hfill \\ \text{Factor the trinomial.}\hfill & & & & & \left(n-17\right)\left(n+19\right)=0\hfill \\ \begin{array}{c}\text{Use the Zero Product Property.}\hfill \\ \text{Solve the equations.}\hfill \end{array}\hfill & & & & & \begin{array}{ccccccccc}\hfill n-17& =\hfill & 0\hfill & & & & \hfill n+19& =\hfill & 0\hfill \\ \hfill n& =\hfill & 17\hfill & & & & \hfill n& =\hfill & \mathrm{-19}\hfill \end{array}\hfill \end{array}$
There are two values for n that are solutions to this problem. So there are two sets of consecutive odd integers that will work.
$\begin{array}{cccccc}\text{If the first integer is}n=17\hfill & & & & & \text{If the first integer is}n=\mathrm{-19}\hfill \\ \text{then the next odd integer is}\hfill & & & & & \text{then the next odd integer is}\hfill \\ n+2\hfill & & & & & n+2\hfill \\ 17+2\hfill & & & & & \text{−}19+2\hfill \\ 19\hfill & & & & & \text{−}17\hfill \\ 17,19\hfill & & & & & \text{−}17,\mathrm{-19}\hfill \\ \mathbf{\text{Step 6. Check}}\text{the answer.}\hfill & & & & & \\ \text{The results are consecutive odd integers}\hfill & & & & & \\ 17,19\text{and}\text{−}19,\mathrm{-17}.\hfill & & & & & \\ 17·19=323✓\text{−}19\left(\text{−}17\right)=323✓\hfill & & & & & \\ \text{Both pairs of consecutive integers are solutions.}\hfill & & & & & \\ \mathbf{\text{Step 7. Answer}}\text{the question}\hfill & & & & & \text{The consecutive integers are 17, 19 and}\text{−}19,\mathrm{-17}.\hfill \end{array}$

ⓐ The zeros of this function are found by solving $h\left(t\right)=0.$ This will tell us when the ball will hit the ground.
$\begin{array}{cccccc}& & & & & h\left(t\right)=0\hfill \\ \text{Substitute in the polynomial for}h\left(t\right).\hfill & & & & & \mathrm{-16}{t}^2+64t+80=0\hfill \\ \text{Factor the GCF,}\mathrm{-16}.\hfill & & & & & \mathrm{-16}(t^2-4t-5)=0\hfill \\ \text{Factor the trinomial.}\hfill & & & & & \mathrm{-16}\left(t-5\right)\left(t+1\right)=0\hfill \\ \begin{array}{c}\text{Use the Zero Product Property.}\hfill \\ \text{Solve.}\hfill \end{array}\hfill & & & & & \begin{array}{ccccccccccc}\hfill t-5& =\hfill & 0\hfill & & & & & & \hfill t+1& =\hfill & 0\hfill \\ \hfill t& =\hfill & 5\hfill & & & & & & \hfill t& =\hfill & \mathrm{-1}\hfill \end{array}\hfill \end{array}$

The result $t=5$ tells us the ball will hit the ground 5 seconds after it is thrown. Since time cannot be negative, the result $t=\mathrm{-1}$ is discarded.

ⓑ The ball will be 80 feet above the ground when $h\left(t\right)=80.$
$\begin{array}{cccccc}& & & & & h\left(t\right)=80\hfill \\ \text{Substitute in the polynomial for}h\left(t\right).\hfill & & & & & \mathrm{-16}{t}^2+64t+80=80\hfill \\ \text{Subtract 80 from both sides.}\hfill & & & & & \mathrm{-16}{t}^2+64t=0\hfill \\ \text{Factor the GCF,}\mathrm{-16}t.\hfill & & & & & \mathrm{-16}t(t-4)=0\hfill \\ \begin{array}{c}\text{Use the Zero Product Property.}\hfill \\ \text{Solve.}\hfill \end{array}\hfill & & & & & \begin{array}{ccccccccc}\hfill \mathrm{-16}t& =\hfill & 0\hfill & & & & \hfill t-4& =\hfill & 0\hfill \\ \hfill t& =\hfill & 0\hfill & & & & \hfill t& =\hfill & 4\hfill \end{array}\hfill \\ & & & & & \begin{array}{c}\text{The ball will be at 80 feet the moment Dennis}\hfill \\ \text{tosses the ball and then 4 seconds later, when}\hfill \\ \text{the ball is falling.}\hfill \end{array}\hfill \end{array}$

ⓒ To find the height ball at $t=2$ seconds we find $h\left(2\right).$
$\begin{array}{cccccc}& & & & & h\left(t\right)=\mathrm{-16}{t}^2+64t+80\hfill \\ \text{To find}h\left(2\right)\text{substitute 2 for}t.\hfill & & & & & h\left(2\right)=\mathrm{-16}{\left(2\right)}^2+64·2+80\hfill \\ \text{Simplify.}\hfill & & & & & h\left(2\right)=144\hfill \\ & & & & & \text{After 2 seconds, the ball will be at 144 feet.}\hfill \end{array}$