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]\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]

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

This equation is already in standard form, so we will proceed with factoring the left side of the equation.

To factor [latex]{x}^{2}+x - 6=0[/latex], we look for two numbers whose product equals [latex]-6[/latex] and whose sum equals [latex]1[/latex]. Begin by looking at the possible factors of [latex]-6[/latex].

The last pair, [latex]3\cdot \left(-2\right)[/latex] sums to [latex]1[/latex], so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.

To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.

The two solutions are [latex]x=2[/latex] and [latex]x=-3[/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.

[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]

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

We can see that this equation is already in standard for, so we will proceed with factoring.

Because this equation has a leading coefficient not equal to 1, we will factor by grouping:

multiply [latex]ac:4\left(9\right)=36[/latex]. Then list the factors of [latex]36[/latex].

The only pair of factors that sums to [latex]15[/latex] is [latex]3+12[/latex]. Rewrite the equation replacing the b term, [latex]15x[/latex], with two terms using [latex]3[/latex] and [latex]12[/latex] as coefficients of x. Factor the first two terms, and then factor the last two terms.

Solve using the zero-product property.

[latex]\begin{array}{lll}&\left(4x+3\right)\left(x+3\right)=0 & \hfill \\ \left(4x+3\right)\hfill=0 & & \left(x+3\right)=0 \hfill \\ x=-\frac{3}{4} & & x=-3 \hfill \end{array}[/latex]

The solutions are [latex]x=-\frac{3}{4}[/latex], [latex]x=-3[/latex].

Notice that when we graph the equation  [latex]y=4{x}^{2}+15x+9[/latex] we can see that the x-intercepts are [latex](-3,0)[/latex] and [latex](-\frac{3}{4},0)[/latex]. This agrees with our solutions!

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]\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}y\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.

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 middle term of [latex]7y[/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!

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

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

[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 [latex]45[/latex] 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 [latex]-45[/latex] and a sum of [latex]-4[/latex]:

There are more factors that will give [latex]-45[/latex], but we have found the ones that sum to [latex]-4[/latex], 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]k=9\text{ and }k=-5[/latex]

Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property.

[latex]\begin{array}{lll}& {x}^{2}-9=0 & \hfill \\ \left(x - 3\right)\hfill=0 & & \left(x+3\right)=0 \hfill \\ x=3 & & x=-3 \hfill \end{array}[/latex]

The solutions are [latex]x=3[/latex] and [latex]x=-3[/latex].

This equation does not look like a quadratic, as the highest power is [latex]3[/latex], not [latex]2[/latex]. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out [latex]-x[/latex] from all of the terms and then proceed with grouping.

Use grouping on the expression in parentheses.

Now, we use the zero-product property. Notice that we have three factors.

[latex]\begin{array}{ccc}-x=0&\left(3x+2\right)=0&\left(x+1\right)=0 \hfill \\ x=0 & x=-\frac{2}{3} & x=-1 \hfill \end{array}[/latex]

The solutions are [latex]x=0[/latex], [latex]x=-\frac{2}{3}[/latex], and [latex]x=-1[/latex].