How do you factor #18x^2-27x+4#?

2 Answers
Aug 9, 2015

Answer:

#(6x-1)(3x-4)#

Explanation:

One technique is to employ this handy formula:

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

You will get;

#x=1/6# and #x=4/3#

Rearrange them;

#6x-1=0# and #3x-4=0#

Multiply them;

#(6x-1)(3x-4)#

Voila!

I recommend this technique for large coefficients of #x^2#. In this case, it is 18. Because you have different possible ways to factor 18:

1 and 18, 2 and 9, 6 and 3.

This will save you the time from trial and error especially for even bigger coefficients.

Aug 9, 2015

Answer:

Solve #y = 18x^2 - 27x + 4#

Ans: (6x - 1)(3x - 4)

Explanation:

Use the new AC Method. y = 18(x - p)(x - q)
Converted trinomial: #y' = x^2 - 27x + 72 =# (x - p')(x - q').
p' and q' have same sign.
Factor pairs of (72) --> (2, 36)(3, 24) . This sum is 27 = -b.
Change the sum to b. Then, p' = -3 and q' = -24.
Therefor. #p = -3/18 = -1/6# and# q = -24/18 = -4/3#

#y = 18(x - 1/6)(x - 4/3) = (6x - 1)(3x - 4)#