Solve 3t^2-15t

$$3 t ^ { 2 } - 15 t$$

Factor

$$3t\left(t-5\right)$$

Solution Steps

Factor out 3.

$$3\left(t^{2}-5t\right)$$

Consider t^{2}-5t. Factor out t.

$$t\left(t-5\right)$$

Rewrite the complete factored expression.

$$3t\left(t-5\right)$$

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.

$$3t^{2}-15t=0$$

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.

$$t=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}}}{2\times 3}$$

Take the square root of \left(-15\right)^{2}.

$$t=\frac{-\left(-15\right)±15}{2\times 3}$$

The opposite of -15 is 15.

$$t=\frac{15±15}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{15±15}{6}$$

Now solve the equation t=\frac{15±15}{6} when ± is plus. Add 15 to 15.

$$t=\frac{30}{6}$$

Divide 30 by 6.

$$t=5$$

Now solve the equation t=\frac{15±15}{6} when ± is minus. Subtract 15 from 15.

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

Divide 0 by 6.

$$t=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and 0 for x_{2}.

$$3t^{2}-15t=3\left(t-5\right)t$$

Evaluate

$$3t\left(t-5\right)$$