2.5 Quadratic equations  (Page 2/14)

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

The last pair, $3\cdot \left(\mathrm{-2}\right)$ 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 $x=2$ and $x=\mathrm{-3.}$ We can see how the solutions relate to the graph in [link] . The solutions are the x- intercepts of $x^2+x-6=0.$

Find two numbers whose product equals $15$ and whose sum equals $8.$ List the factors of $15.$

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 $x=\mathrm{-3}$ and $x=\mathrm{-5.}$

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 $x=3$ and $x=\mathrm{-3.}$

First, multiply $ac:4\left(9\right)=36.$ Then list the factors of $36.$

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

Solve using the zero-product property.

The solutions are $x=-\frac{3}{4},$ $x=\mathrm{-3.}$ See [link] .