Solve m^3+m^2+5m^2+5m+6m+6

$$m ^ { 3 } + m ^ { 2 } + 5 m ^ { 2 } + 5 m + 6 m + 6$$

Evaluate

$$\left(m+1\right)\left(m+2\right)\left(m+3\right)$$

Solution Steps

Combine m^{2} and 5m^{2} to get 6m^{2}.

$$m^{3}+6m^{2}+5m+6m+6$$

Combine 5m and 6m to get 11m.

$$m^{3}+6m^{2}+11m+6$$

Factor

$$\left(m+1\right)\left(m+2\right)\left(m+3\right)$$

Solution Steps

Multiply and combine like terms.

$$m^{3}+6m^{2}+11m+6$$

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 1. One such root is -3. Factor the polynomial by dividing it by m+3.

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

Consider m^{2}+3m+2. Factor the expression by grouping. First, the expression needs to be rewritten as m^{2}+am+bm+2. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\times 2=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

Rewrite m^{2}+3m+2 as \left(m^{2}+m\right)+\left(2m+2\right).

$$\left(m^{2}+m\right)+\left(2m+2\right)$$

Factor out m in the first and 2 in the second group.

$$m\left(m+1\right)+2\left(m+1\right)$$

Factor out common term m+1 by using distributive property.

$$\left(m+1\right)\left(m+2\right)$$

Rewrite the complete factored expression.

$$\left(m+1\right)\left(m+2\right)\left(m+3\right)$$