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

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

Factor

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

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 positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=3$$

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

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

Factor out n in the first and 3 in the second group.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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