How do you factor #9q ^ { 2} - 7q - 18#?

3 Answers
Jul 16, 2017

#9q^2-7q-18#

Well, #7# is a prime number, so we can't factor any of the coefficients (#9, -7, -18#)

What we need to do is find two numbers that multiply to #-162# (#9 xx -18#) and adds to #-7#:

#color(white)(0)xx# #162# let's leave the sign out of it for now
#color(white)(0)+# #7#
............
#1 xx 162#
#2 xx 81#
#3 xx 54#
#6 xx 27#
#9 xx 18#

Well, that didn't work. We should check to make sure we even can factor this.

To do that, we need to graph the equation and see the roots (#x#-intercepts)

graph{y=9x^2-7x-18}

So this does have #x#-intercepts, but the roots are pretty weird numbers, with crazy decimals. That means we can't factor it.
However, we can still find solutions, using the quadratic formula:

#-b/(2a) +- (sqrt(b^2 - 4 a c))/(2a)#

#color(red)(a) = color(red)(9)#

#color(blue)(b) = color(blue)(-7)#

#color(green)(c) = color(green)(-18)#

#(-(-7))/(2(9)) +- (sqrt((-7)^2 - 4 xx (9) xx (-18)))/(2(9))#

#7/18 +- (sqrt(49 - 4 xx (9) xx (-18)))/18#

#7/18 +- sqrt(49- -648)/18#

#7/18 +- sqrt(697)/18#

#(7 +- sqrt697)/18#

We can't simplify #sqrt(697)#, because it simplifies to #17 xx 41#, both of which are prime.

So, we have two solutions:

number #1#

#7 + sqrt(697)/18 ~~ 18.47#

number #2#

#7 - sqrt(697)/18 ~~ 15.53#

Jul 16, 2017

can't be factored.

Explanation:

#y = 9q^2 - 7q - 18#
#D = b^2 - 4ac = 49 + 648 = 697#
Since D is not a perfect square, this trinomial can't be factored.

Jul 16, 2017

#9q^2-7q-18 = 1/36(18q-7-sqrt(697))(18q-7+sqrt(697))#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Use this with #a=(18q-7)# and #b=sqrt(697)# as follows:

#36(9q^2-7q-18) = 324q^2-252q-648#

#color(white)(36(9q^2-7q-18)) = (18q)^2-2(18q)7+49-697#

#color(white)(36(9q^2-7q-18)) = (18q-7)^2-(sqrt(697))^2#

#color(white)(36(9q^2-7q-18)) = ((18q-7)-sqrt(697))((18q-7)+sqrt(697))#

#color(white)(36(9q^2-7q-18)) = (18q-7-sqrt(697))(18q-7+sqrt(697))#

So:

#9q^2-7q-18 = 1/36(18q-7-sqrt(697))(18q-7+sqrt(697))#

Note that #sqrt(697)# does not simplify, since #697=17*41# has no square factors.