Solve 2x^8-x^5*2x^3*x^4

$$2 x ^ { 8 } - x ^ { 5 } \times 2 x ^ { 3 } \times x ^ { 4 }$$

Evaluate

$$2x^{8}-2x^{12}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 5 and 3 to get 8.

$$2x^{8}-x^{8}\times 2x^{4}$$

To multiply powers of the same base, add their exponents. Add 8 and 4 to get 12.

$$2x^{8}-x^{12}\times 2$$

Multiply -1 and 2 to get -2.

$$2x^{8}-2x^{12}$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(x^{8}-x^{5}x^{3}x^{4}\right)$$

Consider x^{8}-x^{12}. Factor out x^{8}.

$$x^{8}\left(1-x^{4}\right)$$

Consider 1-x^{4}. Rewrite 1-x^{4} as 1^{2}-\left(-x^{2}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Reorder the terms.

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

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

$$\left(1-x\right)\left(1+x\right)$$

Reorder the terms.

$$\left(-x+1\right)\left(x+1\right)$$

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

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