Solve 4n^2+2n

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

Factor

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

Solution Steps

Factor out 2.

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

Consider 2n^{2}+n. Factor out n.

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

Rewrite the complete factored expression.

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

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.

$$4n^{2}+2n=0$$

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.

$$n=\frac{-2±\sqrt{2^{2}}}{2\times 4}$$

Take the square root of 2^{2}.

$$n=\frac{-2±2}{2\times 4}$$

Multiply 2 times 4.

$$n=\frac{-2±2}{8}$$

Now solve the equation n=\frac{-2±2}{8} when ± is plus. Add -2 to 2.

$$n=\frac{0}{8}$$

Divide 0 by 8.

$$n=0$$

Now solve the equation n=\frac{-2±2}{8} when ± is minus. Subtract 2 from -2.

$$n=-\frac{4}{8}$$

Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

$$n=-\frac{1}{2}$$

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 -\frac{1}{2} for x_{2}.

$$4n^{2}+2n=4n\left(n-\left(-\frac{1}{2}\right)\right)$$

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

$$4n^{2}+2n=4n\left(n+\frac{1}{2}\right)$$

Add \frac{1}{2} to n by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4n^{2}+2n=4n\times \frac{2n+1}{2}$$

Cancel out 2, the greatest common factor in 4 and 2.

$$4n^{2}+2n=2n\left(2n+1\right)$$

Evaluate

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