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

Apr 12, 2016

$y = {x}^{2} \left(2 x + 7\right) \left(x + 7\right)$

#### Explanation:

First identify the factor common to all the terms which is ${x}^{2}$.
Thus
$y = {x}^{2} \left(2 {x}^{2} + 21 x + 49\right)$

The bracketed term can then be factored as
$y = {x}^{2} \left(2 x + 7\right) \left(x + 7\right)$

Apr 12, 2016

$y = {x}^{2} \left(2 x + 7\right) \left(x + 7\right)$

#### Explanation:

$y = {x}^{2} f \left(x\right) = {x}^{2} \left(2 {x}^{2} + 21 x + 49\right)$
Factor f(x) by the systematic new AC Method (Socratic Search)
$f \left(x\right) = 2 {x}^{2} + 21 x + 49 =$ 2(x + p)(x + q)
Converted trinomial: $f ' \left(x\right) = {x}^{2} + 21 x + 98 =$ (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$\mathmr{and}$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)#