Solve 2x^{6}*x^3-27x^4+25x^2*x^7

$$2 x ^ { 6 } \cdot x ^ { 3 } - 27 x ^ { 4 } + 25 x ^ { 2 } \cdot x ^ { 7 }$$

Evaluate

$$27x^{4}\left(x^{5}-1\right)$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 6 and 3 to get 9.

$$2x^{9}-27x^{4}+25x^{2}x^{7}$$

To multiply powers of the same base, add their exponents. Add 2 and 7 to get 9.

$$2x^{9}-27x^{4}+25x^{9}$$

Combine 2x^{9} and 25x^{9} to get 27x^{9}.

$$27x^{9}-27x^{4}$$

Factor

$$27\left(x-1\right)x^{4}\left(x^{4}+x^{3}+x^{2}+x+1\right)$$

Solution Steps

Factor out x^{2}.

$$x^{2}\left(2x^{4}x^{3}-27x^{2}+25x^{7}\right)$$

Consider 2x^{7}-27x^{2}+25x^{7}. Factor out x^{2}.

$$x^{2}\left(2x^{5}-27+25x^{5}\right)$$

Consider 2x^{5}-27+25x^{5}. Multiply and combine like terms.

$$27x^{5}-27$$

Consider 27x^{5}-27. Factor out 27.

$$27\left(x^{5}-1\right)$$

Consider x^{5}-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 x-1.

$$\left(x-1\right)\left(x^{4}+x^{3}+x^{2}+x+1\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: x^{4}+x^{3}+x^{2}+x+1,x^{2}.

$$27x^{2}x^{2}\left(x-1\right)\left(x^{4}+x^{3}+x^{2}+x+1\right)$$