Solve x^6+11x^3+24

$$x ^ { 6 } + 11 x ^ { 3 } + 24$$

Factor

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

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 24. One such factor is x^{3}+8. Factor the polynomial by dividing it by this factor.

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

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).

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

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

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

Evaluate

$$x^{6}+11x^{3}+24$$