How do you factor #9q ^ { 2} - 7q - 18#?
3 Answers
Well,
What we need to do is find two numbers that multiply to
............
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 (
graph{y=9x^2-7x-18}
So this does have
However, we can still find solutions, using the quadratic formula:
We can't simplify
So, we have two solutions:
number
number
can't be factored.
Explanation:
Since D is not a perfect square, this trinomial can't be factored.
Explanation:
The difference of squares identity can be written:
#a^2-b^2 = (a-b)(a+b)#
Use this with
#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