Solve 2x^5-26x^3+72x

$$2 { x }^{ 5 } -26 { x }^{ 3 } +72x$$

Factor

$$2x\left(x-3\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)$$

Solution Steps

Factor out 2.

$$2\left(x^{5}-13x^{3}+36x\right)$$

Consider x^{5}-13x^{3}+36x. Factor out x.

$$x\left(x^{4}-13x^{2}+36\right)$$

Consider x^{4}-13x^{2}+36. Find one factor of the form x^{k}+m, where x^{k} divides the monomial with the highest power x^{4} and m divides the constant factor 36. One such factor is x^{2}-9. Factor the polynomial by dividing it by this factor.

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

Consider x^{2}-9. Rewrite x^{2}-9 as x^{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).

$$\left(x-3\right)\left(x+3\right)$$

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

$$\left(x-2\right)\left(x+2\right)$$

Rewrite the complete factored expression.

$$2x\left(x-3\right)\left(x+3\right)\left(x-2\right)\left(x+2\right)$$

Evaluate

$$2x\left(x^{4}-13x^{2}+36\right)$$