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Given a quadratic equation with the leading coefficient of 1, factor it.
Factor and solve the equation: $\text{\hspace{0.17em}}{x}^{2}+x-6=0.$
To factor $\text{\hspace{0.17em}}{x}^{2}+x-6=0,$ we look for two numbers whose product equals $\text{\hspace{0.17em}}\mathrm{-6}\text{\hspace{0.17em}}$ and whose sum equals 1. Begin by looking at the possible factors of $\text{\hspace{0.17em}}\mathrm{-6.}$
The last pair, $\text{\hspace{0.17em}}3\cdot \left(\mathrm{-2}\right)\text{\hspace{0.17em}}$ sums to 1, 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 $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{-3.}\text{\hspace{0.17em}}$ We can see how the solutions relate to the graph in [link] . The solutions are the x- intercepts of $\text{\hspace{0.17em}}y={x}^{2}+x-6=0.$
Factor and solve the quadratic equation: $\text{\hspace{0.17em}}{x}^{2}-5x-6=0.$
$\left(x-6\right)\left(x+1\right)=0;x=6,x=-1$
Solve the quadratic equation by factoring: $\text{\hspace{0.17em}}{x}^{2}+8x+15=0.$
Find two numbers whose product equals $\text{\hspace{0.17em}}15\text{\hspace{0.17em}}$ and whose sum equals $\text{\hspace{0.17em}}8.\text{\hspace{0.17em}}$ List the factors of $\text{\hspace{0.17em}}15.$
The numbers that add to 8 are 3 and 5. Then, write the factors, set each factor equal to zero, and solve.
The solutions are $\text{\hspace{0.17em}}\mathrm{-3}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{-5.}$
Solve the quadratic equation by factoring: $\text{\hspace{0.17em}}{x}^{2}-4x-21=0.$
$\left(x\mathrm{-7}\right)\left(x+3\right)=0,$ $x=7,$ $x=\mathrm{-3.}$
Solve the difference of squares equation using the zero-product property: $\text{\hspace{0.17em}}{x}^{2}-9=0.$
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.
The solutions are $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{-3.}$
Solve by factoring: $\text{\hspace{0.17em}}{x}^{2}-25=0.$
$\left(x+5\right)\left(x\mathrm{-5}\right)=0,$ $x=\mathrm{-5},$ $x=5.$
When the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures:
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