Each term has a common factor of t, so we can factor and use the zero product principle. Rewrite each term as the product of the GCF and the remaining terms.

[latex]\begin{array}{c}-t^2=t\left(-t\right)\\t=t\left(1\right)\end{array}[/latex]

Rewrite the polynomial equation using the factored terms in place of the original terms.

[latex]\begin{array}{c}-t^2+t=0\\t\left(-t\right)+t\left(1\right)\\t\left(-t+1\right)=0\end{array}[/latex]

Now we have a product on one side and zero on the other, so we can set each factor equal to zero using the zero product principle.

[latex]\begin{array}{c}t=0\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,\,\,-t+1=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-1}\,\,\,\underline{-1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-t=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{-t}{-1}=\frac{-1}{-1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=1\end{array}[/latex]

Answer

[latex]t=0\text{ OR }t=1[/latex]

First, move all the terms to one side.  The goal is to try and see if we can use the zero product principle, since that is the only tool we know for solving polynomial equations.

[latex]\begin{array}{c}\,\,\,\,\,\,\,s^2-4s=5\\\,\,\,\,\,\,\,s^2-4s-5=0\\\end{array}[/latex]

We now have all the terms on the left side, and zero on the other side. The polynomial[latex]s^2-4s-5[/latex] factors nicely, which makes this equation a good candidate for the zero product principle. (imagine that)

[latex]\begin{array}{c}s^2-4s-5=0\\\left(s+1\right)\left(s-5\right)=0\end{array}[/latex]

We separate our factors into two linear equations using the principle of zero products.

[latex]\begin{array}{c}\left(s-5\right)=0\\s-5=0\\\,\,\,\,\,\,\,\,\,s=5\end{array}[/latex]

OR

[latex]\begin{array}{c}\left(s+1\right)=0\\s+1=0\\s=-1\end{array}[/latex]

Answer

[latex]s=-1\text{ OR }s=5[/latex]

We can solve this in one of two ways.  One way is to eliminate the fractions like you may have done when solving linear equations, and the second is to find a common denominator and factor fractions. Eliminating fractions is easier, so we will show that way.

Start by multiplying the whole equation by 2 to eliminate fractions:

[latex]\begin{array}{ccc}2\left(y^2-5=-\frac{7}{2}y+\frac{5}{2}\right)\\\,\,\,\,\,\,2(y^2)+2(-5)=2\left(-\frac{7}{2}\right)+2\left(-\frac{5}{2}\right)\\2y^2-10=-7y+5\end{array}[/latex]

Now we can move all the terms to one side and see if this will factor so we can use the principle of zero products.

[latex]\begin{array}{c}2y^2-10=-7y-5\\2y^2-10+7y-5=0\\2y^2-15+7y=0\\2y^2+7y-15=0\end{array}[/latex]

We can now check whether this polynomial will factor, using a table we can list factors until we find two numbers with a product of [latex]2\cdot-15=-30[/latex] and a sum of 7.

Factors of [latex]2\cdot-15=-30[/latex]	Sum of Factors
[latex]1,-30[/latex]	[latex]-29[/latex]
[latex]-1,30[/latex]	29
[latex]2,-15[/latex]	[latex]-13[/latex]
[latex]-2,15[/latex]	13
[latex]3,-10[/latex]	[latex]-7[/latex]
[latex]-3,10[/latex]	7

We have found the factors that will produce the middle term we want,[latex]-3,10[/latex]. We need to place the factors in a way that will lead to a term of [latex]10y[/latex]:

[latex]\left(2y-3\right)\left(y+5\right)=0[/latex]

Now we can set each factor equal to zero and solve:

[latex]\begin{array}{ccc}\left(2y-3\right)=0\text{ OR }\left(y+5\right)=0\\2y=3\text{ OR }y=-5\\y=\frac{3}{2}\text{ OR }y=-5\end{array}[/latex]

You can always check your work to make sure your solutions are correct:

Check [latex]y=\frac{3}{2}[/latex]

[latex]\begin{array}{ccc}\left(\frac{3}{2}\right)^2-5=-\frac{7}{2}\left(\frac{3}{2}\right)+\frac{5}{2}\\\frac{9}{4}-5=-\frac{21}{4}+\frac{5}{2}\\\text{ common denominator = 4}\\\frac{9}{4}-\frac{20}{4}=-\frac{21}{4}+\frac{10}{4}\\-\frac{11}{4}=-\frac{11}{4}\end{array}[/latex]

[latex]y=\frac{3}{2}[/latex] is indeed a solution, now check [latex]y=-5[/latex]

[latex]\begin{array}{ccc}\left(-5\right)^2-5=-\frac{7}{2}\left(-5\right)+\frac{5}{2}\\25-5=\frac{35}{2}+\frac{5}{2}\\20=\frac{40}{2}\\20=20\end{array}[/latex]

[latex]y=-5[/latex] is also a solution, so we must have done something right!

 Answer

[latex]y=\frac{3}{2}\text{ OR }y=-5[/latex]

We are given that the height of the rocket is 4 feet from the ground on it’s way back down. We want to know how long it has taken the rocket to get to that point in it’s path, we are going to solve for t.

Substitute [latex]h(t) = 4[/latex] into the formula for height, and try to get zero on one side since we know we can use the zero product principle to solve polynomials.

Write and Solve:

[latex]\begin{array}{l}h(t)=4=−2t^{2}+7t+4\\4=-2t^2+7t+4\\\underline{-4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-4}\\0=-2t^2+7t\end{array}[/latex]

Now we can factor out a t from each term:

[latex]0=t\left(-2t+7\right)[/latex]

Solve each equation for t using the zero product principle:

[latex]\begin{array}{l}t=0\text{ OR }-2t+7=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-7}\,\,\,\,\,\,\,\underline{-7}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{-2t}{-2}=\frac{-7}{-2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=\frac{7}{2}=3.5\end{array}[/latex]

It doesn’t make sense for us to choose t=0 because we are interested in the amount of time that has passed when the projectile is 4 feet from hitting the ground on it’s way back down. We will choose t=3.5

Check the answer on your own for practice.

Answer

[latex]t=3.5\text{ seconds }[/latex]

Read and understand: We know the lengths of all the sides of a triangle in terms of one side. We also know that the Pythagorean theorem will give us a relationship between the side lengths of a right triangle.

Translate: 

[latex]\begin{array}{l}a^2+b^2=c^2\\x^2+\left(x+2\right)^2=\left(x+4\right)^2\end{array}[/latex]

Solve:  If we can move all the terms to one side and factor, we can use the zero product principle to solve.  Since this is the only method we know – let’s hope it works!

First, multiply the binomials and simplify so we can see what we are working with.

[latex]\begin{array}{l}x^2+\left(x+2\right)^2=\left(x+4\right)^2\\x^2+x^2+4x+4=x^2+8x+16\\2x^2+4x+4=x^2+8x+16\end{array}[/latex]

Now move all the terms to one side and see if we can factor.

[latex]\begin{array}{l}2x^2+4x+4=x^2+8x+16\\\underline{-x^2}\,\,\,\underline{-8x}\,\,\,\underline{-16}\,\,\,\,\,\underline{-x^2}\,\,\,\underline{-8x}\,\,\,\underline{-16}\\x^2-4x-12=0\end{array}[/latex]

This went from a messy looking problem to something promising. We can factor using the shortcut:

[latex]-6\cdot{2}=-12,\text{ and }-6+2=-4[/latex]

So we can build our binomial factors with -6 and 2:

[latex]\left(x-6\right)\left(x+2\right)=0[/latex]

Set each factor equal to zero:

[latex]x-6=0, x=6[/latex]

[latex]x+2=0, x=-2[/latex]

Interpret: Ok, it doesn’t make sense to have a length equal to -2, so we can safely throw that solution out.  The lengths of the sides are as follows:

[latex]x=6[/latex]

[latex]x+2=6+2=8[/latex]

[latex]x+4=6+4=10[/latex]

Check: Since we know the relationship between the sides of a right triangle we can check that we are correct. Sometimes it helps to draw a picture

.

We know that [latex]a^2+b^2=c^2[/latex], so we can substitute the values we found:

[latex]\begin{array}{l}6^2+8^2=10^2\\36+64=100\\100=100\end{array}[/latex]

Our solution checks out.

Answer

The lengths of the sides of the right triangle are 6, 8, and 10

Read and Understand: Profit is the difference between revenue and cost, so we will need to define P = R – C for the company.

Define and Translate: [latex]\begin{array}{c}R=-0.09x^2+7000x\\C=2000x+750,000\end{array}[/latex]

Write and Solve: Substitute the expressions for R and C into the Profit equation.

[latex]\begin{array}{c}P=R-C\\=-0.09x^2+7000x-\left(2000x+750,000\right)\\=-0.09x^2+7000x-2000x-750,000\\=-0.09x^2+5000x-750,000\end{array}[/latex]

Remember that when you subtract a polynomial, you have to subtract every term of the polynomial.

Answer

[latex]P=-0.09x^2+5000x-750,000[/latex]

Read and Understand: The profit polynomial defined in the previous example,[latex]P=-0.09x^2+5000x-750,000[/latex], gives profit based on x number of phones manufactured.  We need to substitute the given numbers of phones manufactured into the equation, then try to understand what our answer means in terms of profit and number of phones manufactured.

We will move straight into write and solve since we already have our polynomial. It is probably easiest to use a calculator since the numbers in this problem are so large.

Write and Solve: 

Substitute x = 100

[latex]\begin{array}{c}P=-0.09x^2+5000x-750,000\\=-0.09\left(100\right)^2+5000\left(100\right)-750,000\\=-900+500,000-750,000\\=-250,900\end{array}[/latex]

Interpret: When the number of phones manufactured is 100, the profit for the business is $-250,000.  This is not what we want!  The company must produce more than 100 phones to make a profit.

Write and Solve: 

Substitute x = 25,000

[latex]\begin{array}{c}P=-0.09x^2+5000x-750,000\\=-0.09\left(25000\right)^2+5000\left(25000\right)-750,000\\=-6120000+125,000,000-750,000\\=118,130,000\end{array}[/latex]

Interpret: When the number of phones manufactured is 25,000, the profit for the business is $118,130,000.  This is more like it!  If the company makes 25,000 phones it will make a profit after it pays all it’s bills.

 If this is true, then the company should make even more phones so it can make even more money, right?  Actually, something different happens as the number of items manufactured increases without bound.

Write and Solve: 

Substitute x = 60,000

[latex]\begin{array}{c}P=-0.09x^2+5000x-750,000\\=-0.09\left(60000\right)^2+5000\left(60000\right)-750,000\\=-324,000,000+300,000,000-750,000\\=-24,750,000\end{array}[/latex]

Interpret: When the number of phones manufactured is 60,000, the profit for the business is $-24,750,000.  Wait a minute! If the company makes 60,000 phones it will lose money, what happened? At some point, the cost to manufacture the phones will overcome the amount of profit that the business can make.  If this is interesting to you, you may enjoy reading about Economics and Business models.