How do you factor the trinomial #x^2-3x-18#?

2 Answers
Jun 28, 2018

Use sum/product

Explanation:

We have a trinomial with a one in front of #x^2#. This tells us that we can use the #color(blue)("sum/product")# method here. To use this method, let's review the steps:

The general formula for a quadratic equation is #color(red)("ax^2+bx+c)#

#-># Find two numbers that both #"add"# to the b term (-3) and #"multiply"# to the c term (-18). So, what two numbers can I add to get -3 and multiply together to get -18?

#color(red)(-6 and 3)#

#color(blue)("-6+3= -3")#
#color(blue)("-6x3= -18")#

#-># We can now rewrite the original equation using our two new factors:
#color(red)("(x-6)(x+3)")#

You can check that this is correct by the distributive property, and you should find that it matches the original equation.

#-># Now, we set both binomials equal to 0.

#color(blue)("x-6=0")#
#color(blue)("x+3=0")#

#-># Solve for x in both new equations.

You get: #color(red)(x=6 and x=-3)#

Sometimes, this type of factoring will be given in terms of real life situations. In that case, you will most likely reject the negative factor, and only use the positive factor. But for this type of question, both factors can work.

Be sure to plug in your factors into the original equation to check your work.

Jun 28, 2018

#(x+3)(x-6)#

Explanation:

#x^2-3x-18#

#"Look at 18". "you got 2 choices", 2xx 9 or 3xx6#
By just looking at the options you can see that 3 and 6 will be the
right option.

#(x+3)(x-6)#