Solve n^{4}+6n^3+11n^2+6n+4n^3+24n^2+44n+24

$$n ^ { 4 } + 6 n ^ { 3 } + 11 n ^ { 2 } + 6 n + 4 n ^ { 3 } + 24 n ^ { 2 } + 44 n + 24$$

Evaluate

$$\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)$$

Solution Steps

Combine 6n^{3} and 4n^{3} to get 10n^{3}.

$$n^{4}+10n^{3}+11n^{2}+6n+24n^{2}+44n+24$$

Combine 11n^{2} and 24n^{2} to get 35n^{2}.

$$n^{4}+10n^{3}+35n^{2}+6n+44n+24$$

Combine 6n and 44n to get 50n.

$$n^{4}+10n^{3}+35n^{2}+50n+24$$

Factor

$$\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)$$

Solution Steps

Multiply and combine like terms.

$$n^{4}+10n^{3}+35n^{2}+50n+24$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 24 and q divides the leading coefficient 1. One such root is -4. Factor the polynomial by dividing it by n+4.

$$\left(n+4\right)\left(n^{3}+6n^{2}+11n+6\right)$$

Consider n^{3}+6n^{2}+11n+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 n+3.

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

Consider n^{2}+3n+2. Factor the expression by grouping. First, the expression needs to be rewritten as n^{2}+an+bn+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 n^{2}+3n+2 as \left(n^{2}+n\right)+\left(2n+2\right).

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

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

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

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

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

Rewrite the complete factored expression.

$$\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)$$