How do you factor #4x^2+16x+7#?

2 Answers
Aug 3, 2016

Answer:

#4x^2+16x+7=(2x+1)(2x+7)#

Explanation:

#4x^2+16x+7=(2x+1)(2x+7)#

Aug 3, 2016

Answer:

(2x + 1)(2x + 7)

Explanation:

Use the new AC Method (Socratic Search)
#y = 4x^2 + 16x + 7 =# (x + p)(x + q)
Converted trinomial: #y' = x^2 + 16x + 28 =# (x + p')(x + q')
p' and q' have same sign (ac > 0). Compose factor pairs of (ac = 28) -->(2, 14). This sum is 16 = b. Then, p' = 2 and q' = 14.
Back to y, #p = (p')/a = 2/4 = 1/2# and #q = (q')/a = 14/4 = 7/2#.
Factor form pf y:
#y = 4(x + 1/2)(x + 7/2) = (2x + 1)(2x + 7)#