How do you factor #y=2x^4+21x^3+49x^2 #?

2 Answers
Apr 12, 2016

Answer:

#y = x^2(2x+7)(x+7)#

Explanation:

First identify the factor common to all the terms which is #x^2#.
Thus
#y = x^2(2x^2+21x+49)#

The bracketed term can then be factored as
#y = x^2(2x+7)(x+7)#

Apr 12, 2016

Answer:

#y = x^2(2x + 7)(x + 7)#

Explanation:

#y = x^2f(x) = x^2(2x^2 + 21x + 49)#
Factor f(x) by the systematic new AC Method (Socratic Search)
#f(x) = 2x^2 + 21x + 49 =# 2(x + p)(x + q)
Converted trinomial: #f'(x) = x^2 + 21x + 98 =# (x + p')(x + q')
p' and q' have same sign because ac > 0.
Factor pairs of (ac = 98) --> (2, 49)(7, 14). This sum is 21 = b.
Then, p' = 7 and q' = 14.
Back to f(x) --> 3p = (p')/a = 7/2# and #q = (q')/a = 14/2 = 7.# Factored form: #f(x) = 2(x + 7/2)(x + 7) = (2x + 7)(x + 7).# Therefor: #y = x^2f(x) = x^2(2x + 7)(x + 7)#