Solve 36t-4t^2-72=

$$36 t - 4 t ^ { 2 } - 72 =$$

Factor

$$4\left(3-t\right)\left(t-6\right)$$

Solution Steps

Factor out 4.

$$4\left(9t-t^{2}-18\right)$$

Consider 9t-t^{2}-18. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-t^{2}+9t-18$$

Factor the expression by grouping. First, the expression needs to be rewritten as -t^{2}+at+bt-18. To find a and b, set up a system to be solved.

$$a+b=9 ab=-\left(-18\right)=18$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 18.

$$1,18 2,9 3,6$$

Calculate the sum for each pair.

$$1+18=19 2+9=11 3+6=9$$

The solution is the pair that gives sum 9.

$$a=6 b=3$$

Rewrite -t^{2}+9t-18 as \left(-t^{2}+6t\right)+\left(3t-18\right).

$$\left(-t^{2}+6t\right)+\left(3t-18\right)$$

Factor out -t in the first and 3 in the second group.

$$-t\left(t-6\right)+3\left(t-6\right)$$

Factor out common term t-6 by using distributive property.

$$\left(t-6\right)\left(-t+3\right)$$

Rewrite the complete factored expression.

$$4\left(t-6\right)\left(-t+3\right)$$

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.

$$-4t^{2}+36t-72=0$$

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.

$$t=\frac{-36±\sqrt{36^{2}-4\left(-4\right)\left(-72\right)}}{2\left(-4\right)}$$

Square 36.

$$t=\frac{-36±\sqrt{1296-4\left(-4\right)\left(-72\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$t=\frac{-36±\sqrt{1296+16\left(-72\right)}}{2\left(-4\right)}$$

Multiply 16 times -72.

$$t=\frac{-36±\sqrt{1296-1152}}{2\left(-4\right)}$$

Add 1296 to -1152.

$$t=\frac{-36±\sqrt{144}}{2\left(-4\right)}$$

Take the square root of 144.

$$t=\frac{-36±12}{2\left(-4\right)}$$

Multiply 2 times -4.

$$t=\frac{-36±12}{-8}$$

Now solve the equation t=\frac{-36±12}{-8} when ± is plus. Add -36 to 12.

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

Divide -24 by -8.

$$t=3$$

Now solve the equation t=\frac{-36±12}{-8} when ± is minus. Subtract 12 from -36.

$$t=-\frac{48}{-8}$$

Divide -48 by -8.

$$t=6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and 6 for x_{2}.

$$-4t^{2}+36t-72=-4\left(t-3\right)\left(t-6\right)$$

Evaluate

$$4\left(3-t\right)\left(t-6\right)$$