How do you factor the expression #42x^2+77x+21#?

1 Answer
Mar 10, 2016

Answer:

y = 7(3x + 1)(2x + 3)

Explanation:

Use the systematic new AC Method to factor trinomials (Google, Yahoo)
#y = 42x^2 + 77x + 21 =# 42(x + p)(x + q)
Converted trinomial: #y' = x^2 + 77x + 882 = #(x + p')(x + q').
p' and q' have same sign, since ac > 0.
Compose factor pairs of (ac = 882) --> ...(9, 98)(14, 63). This sum is
77 = b. Then p' = 14 and q' = 63.
Back to original trinomial: #p = (p')/a = 14/42 = 1/3# and #q = (q')/a = = 63/42 = 9/6 = 3/2.#
Factored form #y = 42(x + 1/3)(x + 3/2) = 7(3x + 1)(2x + 3)#