How do you factor the trinomial 16a^2-22ab-3b^216a2−22ab−3b2?

1 Answer
George C.
Dec 13, 2015

Use an AC Method: Look for a pair of factors of 16xx3 = 4816×3=48 with difference 2222, hence find:

16a^2-22ab-3b^2 = (8a+b)(2a-3b)16a2−22ab−3b2=(8a+b)(2a−3b)

Explanation:

Multiply the first coefficient A=16A=16 by the third coefficient (ignoring sign) C=3C=3 to get 4848.

Then since the sign of the third coefficient is negative, look for a pair of factors of AC=48AC=48 whose difference is B=22B=22.

The pair (B1, B2) = (24, 2)(B1,B2)=(24,2) works, so use that to split the middle term:

16a^2-24ab+2ab-3b^216a2−24ab+2ab−3b2

We can then factor by grouping to find:

16a^2-24ab+2ab-3b^216a2−24ab+2ab−3b2

=(16a^2-24ab)+(2ab-3b^2)=(16a2−24ab)+(2ab−3b2)

=8a(2a-3b)+b(2a-3b)=8a(2a−3b)+b(2a−3b)

=(8a+b)(2a-3b)=(8a+b)(2a−3b)

Alternatively, just write down the pairs (A, B1)(A,B1) and (A, B2)(A,B2) with appropriate signs and divide by highest common factors to find pairs of coefficients as follows:

(A, B1) -> (16, -24) -> (2, -3) -> (2a-3b)(A,B1)→(16,−24)→(2,−3)→(2a−3b) (dividing through by 88)

(A, B2) -> (16, 2) -> (8, 1) -> (8a+b)(A,B2)→(16,2)→(8,1)→(8a+b) (dividing through by 22)

The slight complication is the choosing of signs for the second term so that the resulting +-B1 +- B2 = -22±B1±B2=−22

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