Solve 4x^2+4x

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

Factor

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

Solution Steps

Factor out 4.

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

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

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Take the square root of 4^{2}.

$$x=\frac{-4±4}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-4±4}{8}$$

Now solve the equation x=\frac{-4±4}{8} when ± is plus. Add -4 to 4.

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

Divide 0 by 8.

$$x=0$$

Now solve the equation x=\frac{-4±4}{8} when ± is minus. Subtract 4 from -4.

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

Divide -8 by 8.

$$x=-1$$

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 -1 for x_{2}.

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

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

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

Evaluate

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