How do you factor 12c^2+23c-9?

1 Answer
George C.
Sep 23, 2016

12c^2+23c-9 = (3c-1)(4c+9)

Explanation:

12c^2+23c-9

Use an AC method:

Find a pair of factors of AC = 12*9 = 108 = 2^2*3^3 which differ by 23

Since 23 is not divisible by 3 or 2, one of the factors is divisible by 3^3 = 27 and one factor (possibly the same one) is divisible by 2^2 = 4. Note that 27-4 = 23, so we have found the required pair in 27, 4

Use this pair to split the middle term and factor by grouping:

12c^2+23c-9 = (12c^2+27c)-(4c+9)

color(white)(12c^2+23c-9) = 3c(4c+9)-1(4c+9)

color(white)(12c^2+23c-9) = (3c-1)(4c+9)

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