First, note that it is critical that the equation is equal to zero.

Since each term has a common factor of [latex]t[/latex], we can factor and use the zero product principle. Recall that we typically choose to also factor out the negative in the case of a negative leading coefficient, so we will factor out [latex]-t[/latex].

[latex]-t^2+t=-t(t-1)[/latex]

Rewrite the polynomial equation using the factored form of the polynomial.

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

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

[latex]\begin{array}{c}{\displaystyle\frac{-t}{-1}=\frac{0}{-1}}\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,\,\,t-1=0\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{+1}\,\,\,\,\,\underline{+1}\\ \,\,\, \,\,\,\,\,\,t=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,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 to use the zero product principle since that is the only tool we currently know for solving polynomial equations of degree larger than 1.

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

We now have all the terms on the same side and zero on the other side. The polynomial [latex]s^2-4s-5[/latex] factors nicely using the product and sum technique.

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

We now apply the zero product principle, separating our factors into two linear equations.

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

OR

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

Answer

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

First notice that there is a GCF of 2 that we can factor out before applying the difference of squares.

[latex]2(4x^2-25)=0[/latex]

[latex]2(2x+5)(2x-5)[/latex]

Often students get confused by a constant like 2 in front. We could try setting this factor to zero, but 2 can never equal zero so this makes no sense. Another way of looking at this is that we are solving for the variable (in this problem, x) and there is no x in this factor, so we cannot obtain a solution from it.

However, we still have two other factors to apply the zero product principle to.

[latex]2x+5=0[/latex]

[latex]x=-\frac{5}{2}[/latex]

OR

[latex]2x-5=0[/latex]

[latex]x=\frac{5}{2}[/latex]

Answer

[latex]x=-\frac{5}{2} \mbox{ OR } x=\frac{5}{2}[/latex]

Like the previous example, there is a GCF. However, this time the GCF includes a variable. After factoring out the common factor of [latex]3z[/latex], we can then apply the product and sum method to factor the resulting trinomial.

[latex]3z(z^2-3z-10)=0[/latex]

[latex]3z(z-5)(z+2)=0[/latex]

Setting the first factor of [latex]3z[/latex] equal to zero will result in a solution!

[latex]3z=0[/latex]

[latex]z=0[/latex]

OR

[latex]z-5=0[/latex]

[latex]z=5[/latex]

OR

[latex]z+2=0[/latex]

[latex]z=-2[/latex]

Answer

[latex]z=0 \mbox{ OR } z=5 \mbox{ OR } z=-2[/latex]

An all-too common error is to set each factor equal to 8, but it is absolutely not true that if the product of two numbers is 8, one of them must equal 8. This is (one reason) 0 is such a special number.

So first, we must “unfactor” (multiply out the left side of) this problem so that we can move the 8 over and start fresh.

[latex](x-3)(x+4)=8[/latex]

[latex]x^2+4x-3x-12=8[/latex]

[latex]x^2+x-12=8[/latex]

[latex]x^2+x-20=0[/latex]

Now that we have a polynomial equal to zero, we can factor and apply the zero product principle.

[latex](x+5)(x-4)=0[/latex]

Setting each to zero yields

[latex]x+5=0[/latex]

[latex]x=-5[/latex]

OR

[latex]x-4=0[/latex]

[latex]x=4[/latex]

Answer

[latex]x=-5 \mbox{ OR } x=4[/latex]