# How do you find all the zeros of F(x) = 9x^4 - 37x^2 + 4?

The zeroes are $\frac{1}{3} , - \frac{1}{3} , 2 , - 2$

#### Explanation:

We notice that

F(x)=9x^4-37x^2+4=9x^4-36x^2+4-x^2= 9x^2(x^2-4)-(x^2-4)=(9x^2-1)(x^2-4)= (3x-1)*(3x+1)(x+2)(x-2)

Hence the zeroes are the values for which

$F \left(x\right) = 0 \implies \left(3 x - 1\right) \left(3 x + 1\right) \left(x + 2\right) \left(x - 2\right) = 0$

which are ${x}_{1} = \frac{1}{3} , {x}_{2} = - \frac{1}{3} , {x}_{3} = 2 , {x}_{4} = - 2$