Example

Solve for k: [latex]-2k^2+90=-8k[/latex]

Answer: We need to move all the terms to one side so we can use the zero product principle.

[latex]\begin{array}{l}-2k^2+90=-8k\\\underline{+8k}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{+8k}\\-2k^2+8k+90=0\end{array}[/latex]

You will either need to try to factor out a -2, or use the method where we multiply [latex]-2\cdot{90}[/latex] and find factors that sum to 8. Each term is divisible by 2, so we can factor out -2.

[latex]-2\left(k^2-4k-45\right)=0[/latex]

Note how we changed the signs when we factored out a negative number. If we can factor the polynomial, we will be able to solve.   Using the shortcut for factoring we will start with the variable and place a plus and a minus sign in the binomials. We do this because 45 is negative and the only way to get a product that is negative is if one of the factors is negative.

[latex]-2\left(k-\,\,\,\right)\left(k+\,\,\,\right)=0[/latex]

We want our factors to have a product of -45 and a sum of -4:

Factors whose product is -45	Sum of the factors
[latex]1\cdot-45=-45[/latex]	[latex]1-45=-44[/latex]
[latex]3\cdot-15=-45[/latex]	[latex]3-15=-12[/latex]
[latex]5\cdot-9=-45[/latex]	[latex]5-9=-4[/latex]

There are more factors that will give -45, but we have found the ones that sum to -4, so we will stop. Fill in the rest of the binomials with the factors we found.

[latex]-2\left(k-9\right)\left(k+5\right)=0[/latex]

Now we can set each factor equal to zero using the zero product rule.

[latex]-2=0[/latex] This solution is nonsense so we discard it.

[latex]k-9=0, k=9[/latex]

[latex]k+5=0, k=-5[/latex]

Answers

[latex-display]k=9\text{ and }k=-5[/latex-display]

Example

A right triangle has one leg with length x, another whose length is greater by two,  and the length of the hypotenuse is greater by four.  Find the lengths of the sides of the triangle. Use the image below.

Answer: 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

Example

Use the formula for the height of the rocket in the previous example to find the time when the rocket is 4 feet from hitting the ground on it's way back down.  Refer to the image. [latex-display]h=−2t^{2}+7t+4[/latex-display]

Answer: Read and understand: 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. Translate: Substitute h = 4 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=−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]

Interpret: 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

Answer

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