Solve (n^2+n)

$$( n ^ { 2 } + n )$$

Factor

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

Solution Steps

Factor out n.

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

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$n^{2}+n=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$n=\frac{-1±\sqrt{1^{2}}}{2}$$

Take the square root of 1^{2}.

$$n=\frac{-1±1}{2}$$

Now solve the equation n=\frac{-1±1}{2} when ± is plus. Add -1 to 1.

$$n=\frac{0}{2}$$

Divide 0 by 2.

$$n=0$$

Now solve the equation n=\frac{-1±1}{2} when ± is minus. Subtract 1 from -1.

$$n=-\frac{2}{2}$$

Divide -2 by 2.

$$n=-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and -1 for x_{2}.

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

Simplify all the expressions of the form p-\left(-q\right) to p+q.

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

Evaluate

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