Solve (13n^2+11n-2n^4)+(-13n^2-3n-6n^4

$$( 13 n ^ { 2 } + 11 n - 2 n ^ { 4 } ) + ( - 13 n ^ { 2 } - 3 n - 6 n ^ { 4 }$$

Evaluate

$$8n-8n^{4}$$

Solution Steps

Combine 13n^{2} and -13n^{2} to get 0.

$$11n-2n^{4}-3n-6n^{4}$$

Combine 11n and -3n to get 8n.

$$8n-2n^{4}-6n^{4}$$

Combine -2n^{4} and -6n^{4} to get -8n^{4}.

$$8n-8n^{4}$$

Factor

$$8n\left(n-1\right)\left(-n^{2}-n-1\right)$$

Solution Steps

Factor out n.

$$n\left(8-8n^{3}\right)$$

Consider 13n+11-2n^{3}-13n-3-6n^{3}. Multiply and combine like terms.

$$-8n^{3}+8$$

Consider -8n^{3}+8. Factor out 8.

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

Consider -n^{3}+1. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient -1. One such root is 1. Factor the polynomial by dividing it by n-1.

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

Rewrite the complete factored expression. Polynomial -n^{2}-n-1 is not factored since it does not have any rational roots.

$$8n\left(n-1\right)\left(-n^{2}-n-1\right)$$