Solve F(x)=x^6+9x^3+0+8

$$F ( x ) = x ^ { 6 } + 9 x ^ { 3 } + 0 + 8$$

Factor

$$\left(x+1\right)\left(x+2\right)\left(x^{2}-x+1\right)\left(x^{2}-2x+4\right)$$

Solution Steps

Find one factor of the form x^{k}+m, where x^{k} divides the monomial with the highest power x^{6} and m divides the constant factor 8. One such factor is x^{3}+8. Factor the polynomial by dividing it by this factor.

$$\left(x^{3}+8\right)\left(x^{3}+1\right)$$

Consider x^{3}+8. Rewrite x^{3}+8 as x^{3}+2^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

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

Consider x^{3}+1. Rewrite x^{3}+1 as x^{3}+1^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

$$\left(x+1\right)\left(x^{2}-x+1\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: x^{2}-x+1,x^{2}-2x+4.

$$\left(x^{2}-x+1\right)\left(x+1\right)\left(x+2\right)\left(x^{2}-2x+4\right)$$

Evaluate

$$x^{6}+9x^{3}+8$$

Solution Steps

Add 0 and 8 to get 8.

$$x^{6}+9x^{3}+8$$