Solve (4x^2+12x)=

$$( 4 x ^ { 2 } + 12 x ) =$$

Factor

$$4x\left(x+3\right)$$

Solution Steps

Factor out 4.

$$4\left(x^{2}+3x\right)$$

Consider x^{2}+3x. Factor out x.

$$x\left(x+3\right)$$

Rewrite the complete factored expression.

$$4x\left(x+3\right)$$

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.

$$4x^{2}+12x=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}}}{2\times 4}$$

Take the square root of 12^{2}.

$$x=\frac{-12±12}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-12±12}{8}$$

Now solve the equation x=\frac{-12±12}{8} when ± is plus. Add -12 to 12.

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

Divide 0 by 8.

$$x=0$$

Now solve the equation x=\frac{-12±12}{8} when ± is minus. Subtract 12 from -12.

$$x=-\frac{24}{8}$$

Divide -24 by 8.

$$x=-3$$

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 -3 for x_{2}.

$$4x^{2}+12x=4x\left(x-\left(-3\right)\right)$$

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

$$4x^{2}+12x=4x\left(x+3\right)$$

Evaluate

$$4x\left(x+3\right)$$