Solve 4x+20x^2

$$4 x + 20 x ^ { 2 }$$

Factor

$$4x\left(5x+1\right)$$

Solution Steps

Factor out 4.

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

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

$$x\left(1+5x\right)$$

Rewrite the complete factored expression.

$$4x\left(5x+1\right)$$

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.

$$20x^{2}+4x=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}}}{2\times 20}$$

Take the square root of 4^{2}.

$$x=\frac{-4±4}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-4±4}{40}$$

Now solve the equation x=\frac{-4±4}{40} when ± is plus. Add -4 to 4.

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

Divide 0 by 40.

$$x=0$$

Now solve the equation x=\frac{-4±4}{40} when ± is minus. Subtract 4 from -4.

$$x=-\frac{8}{40}$$

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

$$x=-\frac{1}{5}$$

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{1}{5} for x_{2}.

$$20x^{2}+4x=20x\left(x-\left(-\frac{1}{5}\right)\right)$$

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

$$20x^{2}+4x=20x\left(x+\frac{1}{5}\right)$$

Add \frac{1}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}+4x=20x\times \frac{5x+1}{5}$$

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

$$20x^{2}+4x=4x\left(5x+1\right)$$

Evaluate

$$4x\left(5x+1\right)$$