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

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

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...