Solve 64x+128x^2

$$64 x + 128 x ^ { 2 }$$

Factor

$$64x\left(2x+1\right)$$

Solution Steps

Factor out 64.

$$64\left(x+2x^{2}\right)$$

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

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

Rewrite the complete factored expression.

$$64x\left(2x+1\right)$$

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.

$$128x^{2}+64x=0$$

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.

$$x=\frac{-64±\sqrt{64^{2}}}{2\times 128}$$

Take the square root of 64^{2}.

$$x=\frac{-64±64}{2\times 128}$$

Multiply 2 times 128.

$$x=\frac{-64±64}{256}$$

Now solve the equation x=\frac{-64±64}{256} when ± is plus. Add -64 to 64.

$$x=\frac{0}{256}$$

Divide 0 by 256.

$$x=0$$

Now solve the equation x=\frac{-64±64}{256} when ± is minus. Subtract 64 from -64.

$$x=-\frac{128}{256}$$

Reduce the fraction \frac{-128}{256} to lowest terms by extracting and canceling out 128.

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

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}.

$$128x^{2}+64x=128x\left(x-\left(-\frac{1}{2}\right)\right)$$

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

$$128x^{2}+64x=128x\left(x+\frac{1}{2}\right)$$

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

$$128x^{2}+64x=128x\times \frac{2x+1}{2}$$

Cancel out 2, the greatest common factor in 128 and 2.

$$128x^{2}+64x=64x\left(2x+1\right)$$

Evaluate

$$64x\left(2x+1\right)$$