Example

Solve: [latex-display]-t^2+t=0[/latex-display]

Answer: Each term has a common factor of [latex]t[/latex], so we can factor and use the zero product principle. Rewrite each term as the product of the GCF and the remaining terms. [latex-display]\begin{array}{c}-t^2=t\left(-t\right)\\t=t\left(1\right)\end{array}[/latex-display] 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-display]t=0\text{ OR }t=1[/latex-display]

Example

Solve: [latex]s^2-4s=5[/latex]

Answer: 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-display]\begin{array}{c}\left(s-5\right)=0\\s-5=0\\\,\,\,\,\,\,\,\,\,s=5\end{array}[/latex-display] OR [latex-display]\begin{array}{c}\left(s+1\right)=0\\s+1=0\\s=-1\end{array}[/latex-display]

Answer

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

Example

Solve [latex]y^2-5=-\frac{7}{2}y+\frac{5}{2}[/latex]

Answer: 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 [latex]2[/latex] to eliminate fractions: [latex-display]\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-display] 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-display]\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-display] 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]	[latex]29[/latex]
[latex]2,-15[/latex]	[latex]-13[/latex]
[latex]-2,15[/latex]	[latex]13[/latex]
[latex]3,-10[/latex]	[latex]-7[/latex]
[latex]-3,10[/latex]	[latex]7[/latex]

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-display]\left(2y-3\right)\left(y+5\right)=0[/latex-display] Now we can set each factor equal to zero and solve: [latex-display]\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-display] You can always check your work to make sure your solutions are correct: Check [latex]y=\frac{3}{2}[/latex] [latex-display]\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-display] [latex-display]y=\frac{3}{2}[/latex] is indeed a solution, now check [latex]y=-5[/latex-display] [latex-display]\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-display] [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]