Solve w+w^2

$$w + w ^ { 2 }$$

Factor

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

Solution Steps

Factor out w.

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

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.

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

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.

$$w=\frac{-1±\sqrt{1^{2}}}{2}$$

Take the square root of 1^{2}.

$$w=\frac{-1±1}{2}$$

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

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

Divide 0 by 2.

$$w=0$$

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

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

Divide -2 by 2.

$$w=-1$$

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}.

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

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

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

Evaluate

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