Solve Q(x)=8-2x^2+6x^4-9x^5-8+x^2

$$Q ( x ) = 8 - 2 x ^ { 2 } + 6 x ^ { 4 } - 9 x ^ { 5 } - 8 + x ^ { 2 }$$

Evaluate

$$x^{2}\left(-9x^{3}+6x^{2}-1\right)$$

Solution Steps

Subtract 8 from 8 to get 0.

$$-2x^{2}+6x^{4}-9x^{5}+x^{2}$$

Combine -2x^{2} and x^{2} to get -x^{2}.

$$-x^{2}+6x^{4}-9x^{5}$$

Factor

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

Solution Steps

Multiply and combine like terms.

$$-9x^{5}+6x^{4}-x^{2}$$

Factor out x^{2}.

$$x^{2}\left(-9x^{3}+6x^{2}-1\right)$$

Consider -9x^{3}+6x^{2}-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 -9. One such root is -\frac{1}{3}. Factor the polynomial by dividing it by 3x+1.

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

Rewrite the complete factored expression. Polynomial -3x^{2}+3x-1 is not factored since it does not have any rational roots.

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