# How do you factor 121-33n^2+9n^4?

##### 2 Answers
Oct 16, 2016

This trinomial is not factorable.

#### Explanation:

This trinomial is not factorable. The only factor pairs for $121$ are $1$ & $121$ and $11$ & $11$. The factor pairs for $9 {n}^{4}$ are $1 {n}^{2}$ & $9 {n}^{2}$ and $3 {n}^{2}$ & $3 {n}^{2}$. There is no way to place these factor pairs into two binomials so that the sum of the products of the inside terms and outside terms is $- 33 {n}^{2}$.

Oct 19, 2016

$9 \left({n}^{2} + \sqrt{\frac{11}{3}} n + \frac{11}{3}\right) \left({n}^{2} - \sqrt{\frac{11}{3}} n + \frac{11}{3}\right)$

#### Explanation:

Despite that the linear factors are complex, they are conjugate pairs.

So, the product of two conjugate linear factors is real.

And so, the biquadratic can be

factorized into two real quadratics..

Let $9 \left({n}^{4} - \frac{11}{3} {n}^{2} + \frac{121}{9}\right)$

=9(n^2 +an+c)((n^2+bn+d)

Comparing coefficients of ${n}^{3} \mathmr{and} n$ with 0,,

$b = - a \mathmr{and} d = c$

Comparing others,

$a = \pm \sqrt{\frac{11}{3}} , c = \frac{11}{3}$.

Now, the factors are $9 \left({n}^{2} + \sqrt{\frac{11}{3}} n + \frac{11}{3}\right) \left({n}^{2} - \sqrt{\frac{11}{3}} n + \frac{11}{3}\right)$

As a matter of fact, I succeeded only in this edition, to get a bug-free

answer. This is my experience. in this arithmetic game...