Solve f(x)=5x^6-4x^4-45x^2+36

$$f ( x ) = 5 x ^ { 6 } - 4 x ^ { 4 } - 45 x ^ { 2 } + 36$$

Factor

$$\left(5x^{2}-4\right)\left(x^{2}-3\right)\left(x^{2}+3\right)$$

Solution Steps

Do the grouping 5x^{6}-4x^{4}-45x^{2}+36=\left(5x^{6}-4x^{4}\right)+\left(-45x^{2}+36\right), and factor out x^{4} in the first and -9 in the second group.

$$x^{4}\left(5x^{2}-4\right)-9\left(5x^{2}-4\right)$$

Factor out common term 5x^{2}-4 by using distributive property.

$$\left(5x^{2}-4\right)\left(x^{4}-9\right)$$

Consider x^{4}-9. Rewrite x^{4}-9 as \left(x^{2}\right)^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x^{2}-3\right)\left(x^{2}+3\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: 5x^{2}-4,x^{2}-3,x^{2}+3.

$$\left(5x^{2}-4\right)\left(x^{2}-3\right)\left(x^{2}+3\right)$$

Evaluate

$$\left(5x^{2}-4\right)\left(x^{4}-9\right)$$