Solve n^4-2n^3-3n^2

$$n ^ { 4 } - 2 n ^ { 3 } - 3 n ^ { 2 }$$

Factor

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

Solution Steps

Factor out n^{2}.

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

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

$$a+b=-2 ab=1\left(-3\right)=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-3 b=1$$

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

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

Factor out n in n^{2}-3n.

$$n\left(n-3\right)+n-3$$

Factor out common term n-3 by using distributive property.

$$\left(n-3\right)\left(n+1\right)$$

Rewrite the complete factored expression.

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

Evaluate

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