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}\,\,\,\,\,\,\,6t=3t^2-12t\\\underline{-6t}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-6t}\\\,\,\,\,\,\,\,0=3t^2-18t\\\end{array}[/latex]

We now have all the terms on the right side, and zero on the other side. Each term has a common factor of [latex]3t[/latex], so we can factor and use the zero product principle. If you are uncertain how we found the common factor [latex]3t[/latex], review the section on greatest common factor.

Rewrite each term as the product of the GCF and the remaining terms. Note how we leave the negative sign on the [latex]6[/latex] when we factor [latex]3t[/latex] out of [latex]-18t[/latex].

[latex]\begin{array}{c}3t^2=3t\left(t\right)\\-18t=3t\left(-6\right)\end{array}[/latex]

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

[latex]\begin{array}{c}0=3t^2-18t\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=3t\left(t\right)+3t\left(-6\right)\\0=3t\left(t-6\right)\end{array}[/latex]

Solve the two equations.

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=t-6\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,0=3t\\\,\,\,\,\,\,\,\,\,\underline{+6}\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{+6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{0}{3}=\frac{3t}{3}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,6=t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,0=t\end{array}[/latex]

Answer

[latex]t=6\text{ OR }t=0[/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. We will work through the second way.

First, we will find a common denominator, then factor fractions.

[latex]\begin{array}{l}\frac{1}{2}y=-4y-\frac{1}{2}y^2\\0=-\frac{1}{2}y-4y-\frac{1}{2}y^2\end{array}[/latex]

[latex]2[/latex] is a common denominator for  [latex]-\frac{1}{2}y\text{ and }-4y[/latex]

[latex]\frac{2}{2}\cdot{-4y}=-\frac{8y}{2}[/latex]

Rewrite the equation with the common denominator, then combine like terms.

[latex]\begin{array}{l}0=-\frac{1}{2}y-4y-\frac{1}{2}y^2\\\text{ }\\0=-\frac{1}{2}y-\frac{8y}{2}-\frac{1}{2}y^2\\\text{}\\0=-\frac{9}{2}y-\frac{1}{2}y^2\end{array}[/latex]

Find the greatest common factor of the terms of the polynomial:

Factors of [latex]-\frac{9}{2}y[/latex]

[latex]-\frac{1}{2}\cdot{3}\cdot{3}\cdot{y}[/latex]

Factors of [latex]-\frac{1}{2}y^2[/latex]

[latex]-\frac{1}{2}\cdot{y}\cdot{y}[/latex]

Both terms have [latex]-\frac{1}{2}\text{ and }y[/latex] in common.

Rewrite each term as the product of the GCF and the remaining terms.

[latex]-\frac{9}{2}y=-\frac{1}{2}y\left(3\cdot{3}\right)=-\frac{1}{2}y\left(9\right)[/latex]

[latex]-\frac{1}{2}y^2=-\frac{1}{2}y\left(y\right)[/latex]

 Rewrite the polynomial equation using the factored terms in place of the original terms. Pay attention to signs when we factor. Notice that we end up with a sum as a factor because the common factor is a negative number. [latex]\left(9+y\right)[/latex]

[latex]\begin{array}{l}0=-\frac{9}{2}y-\frac{1}{2}y^2\\\text{}\\-\frac{1}{2}y\left(9\right)-\frac{1}{2}y\left(y\right)\\\text{}\\-\frac{1}{2}y\left(9+y\right)\end{array}[/latex]

Solve the two equations.

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=9+y\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,0=-\frac{1}{2}y\\\,\,\,\,\,\,\,\,\,\underline{-9}\,\,\,\,\,\,\underline{-9}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\,\,\,\,\,\,\,\,\,-9=y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,0=y\end{array}[/latex]

Answer

[latex]y=-9\text{ OR }y=0[/latex]