Solve (3x^5-x^2+4x-7)+(x^3-3x^2+2x+1)

$$(3 { x }^{ 5 } - { x }^{ 2 } +4x-7)+( { x }^{ 3 } -3 { x }^{ 2 } +2x+1)$$

Evaluate

$$3x^{5}+x^{3}-4x^{2}+6x-6$$

Solution Steps

Combine -x^{2} and -3x^{2} to get -4x^{2}.

$$3x^{5}-4x^{2}+4x-7+x^{3}+2x+1$$

Combine 4x and 2x to get 6x.

$$3x^{5}-4x^{2}+6x-7+x^{3}+1$$

Add -7 and 1 to get -6.

$$3x^{5}-4x^{2}+6x-6+x^{3}$$

Factor

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

Solution Steps

Multiply and combine like terms.

$$3x^{5}+x^{3}-4x^{2}+6x-6$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 3. One such root is 1. Factor the polynomial by dividing it by x-1. Polynomial 3x^{4}+3x^{3}+4x^{2}+6 is not factored since it does not have any rational roots.

$$\left(x-1\right)\left(3x^{4}+3x^{3}+4x^{2}+6\right)$$