Is it possible to factor #y= 6x^2 - 27x - 15 #? If so, what are the factors?

Answer 1

Isaiah Anthony

#y = 6x^2-27x-15 = 3(2x+1)(x-5)#

Separate out the common factor #3#, then factor by grouping as follows:

#y = 6x^2-27x-15#

#=3(2x^2-9x-5)#

#=3(2x^2-10x+x-5)#

#=3(2x(x-5)+1(x-5))#

#=3(2x+1)(x-5)#

Note that I found the split of the middle term #-9x = -10x + x# using an AC method: Look for a pair of factors of #AC = 2*5 = 10# which differ by #B=9#. The pair #10#, #1# works since #10xx1 = 10# and #10-1 = 9#.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Eva Adamson

Yes, it is possible to factor the quadratic equation y = 6x^2 - 27x - 15. Factoring it, we get y = (2x - 5)(3x + 3).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.