How do you factor #15x^2-9x-14#?

1 Answer
George C.
Apr 16, 2017

#15x^2-9x-14 = 1/60(30x-9-sqrt(921))(30x-9+sqrt(921))#

Explanation:

Here's how you can do it by completing the square. I will use #60 = 15*2^2# as a multiplier to allow me to work mostly in integers. Then use the difference of squares identity:

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

with #a=(30x-9)# and #b=sqrt(921)# as follows:

#15x^2-9x-14 = 1/60(900x^2-540x-840)#

#color(white)(15x^2-9x-14) = 1/60((30x)^2-2(30x)(9)+9^2-921)#

#color(white)(15x^2-9x-14) = 1/60((30x-9)^2-(sqrt(921))^2)#

#color(white)(15x^2-9x-14) = 1/60((30x-9)-sqrt(921))((30x-9)+sqrt(921))#

#color(white)(15x^2-9x-14) = 1/60(30x-9-sqrt(921))(30x-9+sqrt(921))#

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