Question

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

Answer 1

`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)`

Answer 2

`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)#