Solve 3t^2+12t

$$3 t ^ { 2 } + 12 t$$

Factor

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

Solution Steps

Factor out 3.

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

Consider t^{2}+4t. Factor out t.

$$t\left(t+4\right)$$

Rewrite the complete factored expression.

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

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.

$$3t^{2}+12t=0$$

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.

$$t=\frac{-12±\sqrt{12^{2}}}{2\times 3}$$

Take the square root of 12^{2}.

$$t=\frac{-12±12}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{-12±12}{6}$$

Now solve the equation t=\frac{-12±12}{6} when ± is plus. Add -12 to 12.

$$t=\frac{0}{6}$$

Divide 0 by 6.

$$t=0$$

Now solve the equation t=\frac{-12±12}{6} when ± is minus. Subtract 12 from -12.

$$t=-\frac{24}{6}$$

Divide -24 by 6.

$$t=-4$$

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 -4 for x_{2}.

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

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

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

Evaluate

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