Solve 8n+20n^2

$$8 n + 20 n ^ { 2 }$$

Factor

$$4n\left(5n+2\right)$$

Solution Steps

Factor out 4.

$$4\left(2n+5n^{2}\right)$$

Consider 2n+5n^{2}. Factor out n.

$$n\left(2+5n\right)$$

Rewrite the complete factored expression.

$$4n\left(5n+2\right)$$

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.

$$20n^{2}+8n=0$$

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.

$$n=\frac{-8±\sqrt{8^{2}}}{2\times 20}$$

Take the square root of 8^{2}.

$$n=\frac{-8±8}{2\times 20}$$

Multiply 2 times 20.

$$n=\frac{-8±8}{40}$$

Now solve the equation n=\frac{-8±8}{40} when ± is plus. Add -8 to 8.

$$n=\frac{0}{40}$$

Divide 0 by 40.

$$n=0$$

Now solve the equation n=\frac{-8±8}{40} when ± is minus. Subtract 8 from -8.

$$n=-\frac{16}{40}$$

Reduce the fraction \frac{-16}{40} to lowest terms by extracting and canceling out 8.

$$n=-\frac{2}{5}$$

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{2}{5} for x_{2}.

$$20n^{2}+8n=20n\left(n-\left(-\frac{2}{5}\right)\right)$$

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

$$20n^{2}+8n=20n\left(n+\frac{2}{5}\right)$$

Add \frac{2}{5} to n by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20n^{2}+8n=20n\times \frac{5n+2}{5}$$

Cancel out 5, the greatest common factor in 20 and 5.

$$20n^{2}+8n=4n\left(5n+2\right)$$

Evaluate

$$4n\left(5n+2\right)$$