Solve (2z^3+9z^2+6z+9)+(z+3)

$$( 2 z ^ { 3 } + 9 z ^ { 2 } + 6 z + 9 ) + ( z + 3 )$$

Evaluate

$$\left(z+4\right)\left(2z^{2}+z+3\right)$$

Solution Steps

Combine 6z and z to get 7z.

$$2z^{3}+9z^{2}+7z+9+3$$

Add 9 and 3 to get 12.

$$2z^{3}+9z^{2}+7z+12$$

Factor

$$\left(z+4\right)\left(2z^{2}+z+3\right)$$

Solution Steps

Multiply and combine like terms.

$$2z^{3}+9z^{2}+7z+12$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 12 and q divides the leading coefficient 2. One such root is -4. Factor the polynomial by dividing it by z+4. Polynomial 2z^{2}+z+3 is not factored since it does not have any rational roots.

$$\left(z+4\right)\left(2z^{2}+z+3\right)$$