How do you factor #20x ^ { 2} - 43x - 33#?

2 Answers
Jul 23, 2017

#(5x+3)(4x-11)#

Explanation:

#20x^2-43x-33#

#:.=(5x+3)(4x-11)#

Jul 23, 2017

(5x + 3)(4x - 11)

Explanation:

Methodically factor the trinomial by the new AC method (Socratic, Google Search).
#f(x) = 20x^2 - 43x - 33 =# 20(x + p)(x + q)
Converted trinomial
#f'(x) = x^2 - 43x - 660 = #(x + p')(x + q').
Proceeding: find p' and q', then, divide them a = 20, to get p and q.
Find p' and q', that have opposite signs (ac < 0), knowing the sum (b = -43) and the product (ac = - 660). Compose factor pairs of (-660) --> .... (6, - 110)(12, -55). This last sum is (- 43 = b).
Then, p' = 12 and q' = - 55. Back to original f(x):
#p = (p')/a = 12/20 = 3/5#, and #q = (q')/a = - 55/20 = - 11/4#.
Factored form:
#f(x) = 20(x + p)(x + q) = (x + 3/5)(x - 11/4) = (5x + 3)(4x - 11)#