Home › STAAR EOC Algebra I › Mathematics › Factoring › Using the ac method, factor: 18n^2 - 37n + 15.
Using the ac method, factor: 18n^2 - 37n + 15.
A(2n-5)(9n-3)
B(n-3)(18n-5)
C(2n-3)(9n-5)
D(9n-1)(2n-15)
Answer & Solution
Correct answer: C. (2n-3)(9n-5)
1. Multiply a and c: 18 times 15 equals 270.
2. Find two numbers that multiply to 270 and add to -37: -10 and -27 work, since -10 times -27 is 270 and -10 plus -27 is -37.
3. Split the middle term using -10n and -27n, then factor by grouping: 18n squared - 10n - 27n + 15 groups into 2n(9n - 5) - 3(9n - 5).
4. Factor out the shared binomial to get (2n - 3)(9n - 5).
5. Check by multiplying: 18n squared - 10n - 27n + 15 gives back 18n squared - 37n + 15.
_Source: OpenStax Elementary Algebra (CC BY 4.0), Ch 7 "Factoring", section 7.3 Factor Quadratic Trinomials with Leading Coefficient Other than 1_
Related questions
Factor: x^3 + 64.According to the Sum and Difference of Cubes Pattern, how does a^3+b^3 factor?Factor completely: 36x^2y - 48xy + 16y.Factor completely: 8x^2y - 98y.Factor completely: 6x^2 + 96.What happens when you try to factor a sum of squares like a^2 + b^2?What is the completely factored form of x^4 - y^4?Factor: 100 - h^2.