Solve 125x^2+2x

$$125 x ^ { 2 } + 2 x$$

Factor

$$x\left(125x+2\right)$$

Solution Steps

Factor out x.

$$x\left(125x+2\right)$$

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.

$$125x^{2}+2x=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}}}{2\times 125}$$

Take the square root of 2^{2}.

$$x=\frac{-2±2}{2\times 125}$$

Multiply 2 times 125.

$$x=\frac{-2±2}{250}$$

Now solve the equation x=\frac{-2±2}{250} when ± is plus. Add -2 to 2.

$$x=\frac{0}{250}$$

Divide 0 by 250.

$$x=0$$

Now solve the equation x=\frac{-2±2}{250} when ± is minus. Subtract 2 from -2.

$$x=-\frac{4}{250}$$

Reduce the fraction \frac{-4}{250} to lowest terms by extracting and canceling out 2.

$$x=-\frac{2}{125}$$

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}{125} for x_{2}.

$$125x^{2}+2x=125x\left(x-\left(-\frac{2}{125}\right)\right)$$

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

$$125x^{2}+2x=125x\left(x+\frac{2}{125}\right)$$

Add \frac{2}{125} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$125x^{2}+2x=125x\times \frac{125x+2}{125}$$

Cancel out 125, the greatest common factor in 125 and 125.

$$125x^{2}+2x=x\left(125x+2\right)$$

Evaluate

$$x\left(125x+2\right)$$