Solve -16t^2+96t-108

$$- 16 t ^ { 2 } + 96 t - 108$$

Factor

$$4\left(3-2t\right)\left(2t-9\right)$$

Solution Steps

Factor out 4.

$$4\left(-4t^{2}+24t-27\right)$$

Consider -4t^{2}+24t-27. Factor the expression by grouping. First, the expression needs to be rewritten as -4t^{2}+at+bt-27. To find a and b, set up a system to be solved.

$$a+b=24 ab=-4\left(-27\right)=108$$

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 108.

$$1,108 2,54 3,36 4,27 6,18 9,12$$

Calculate the sum for each pair.

$$1+108=109 2+54=56 3+36=39 4+27=31 6+18=24 9+12=21$$

The solution is the pair that gives sum 24.

$$a=18 b=6$$

Rewrite -4t^{2}+24t-27 as \left(-4t^{2}+18t\right)+\left(6t-27\right).

$$\left(-4t^{2}+18t\right)+\left(6t-27\right)$$

Factor out -2t in the first and 3 in the second group.

$$-2t\left(2t-9\right)+3\left(2t-9\right)$$

Factor out common term 2t-9 by using distributive property.

$$\left(2t-9\right)\left(-2t+3\right)$$

Rewrite the complete factored expression.

$$4\left(2t-9\right)\left(-2t+3\right)$$

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.

$$-16t^{2}+96t-108=0$$

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.

$$t=\frac{-96±\sqrt{96^{2}-4\left(-16\right)\left(-108\right)}}{2\left(-16\right)}$$

Square 96.

$$t=\frac{-96±\sqrt{9216-4\left(-16\right)\left(-108\right)}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-96±\sqrt{9216+64\left(-108\right)}}{2\left(-16\right)}$$

Multiply 64 times -108.

$$t=\frac{-96±\sqrt{9216-6912}}{2\left(-16\right)}$$

Add 9216 to -6912.

$$t=\frac{-96±\sqrt{2304}}{2\left(-16\right)}$$

Take the square root of 2304.

$$t=\frac{-96±48}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-96±48}{-32}$$

Now solve the equation t=\frac{-96±48}{-32} when ± is plus. Add -96 to 48.

$$t=-\frac{48}{-32}$$

Reduce the fraction \frac{-48}{-32} to lowest terms by extracting and canceling out 16.

$$t=\frac{3}{2}$$

Now solve the equation t=\frac{-96±48}{-32} when ± is minus. Subtract 48 from -96.

$$t=-\frac{144}{-32}$$

Reduce the fraction \frac{-144}{-32} to lowest terms by extracting and canceling out 16.

$$t=\frac{9}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{2} for x_{1} and \frac{9}{2} for x_{2}.

$$-16t^{2}+96t-108=-16\left(t-\frac{3}{2}\right)\left(t-\frac{9}{2}\right)$$

Subtract \frac{3}{2} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-16t^{2}+96t-108=-16\times \frac{-2t+3}{-2}\left(t-\frac{9}{2}\right)$$

Subtract \frac{9}{2} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-16t^{2}+96t-108=-16\times \frac{-2t+3}{-2}\times \frac{-2t+9}{-2}$$

Multiply \frac{-2t+3}{-2} times \frac{-2t+9}{-2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-16t^{2}+96t-108=-16\times \frac{\left(-2t+3\right)\left(-2t+9\right)}{-2\left(-2\right)}$$

Multiply -2 times -2.

$$-16t^{2}+96t-108=-16\times \frac{\left(-2t+3\right)\left(-2t+9\right)}{4}$$

Cancel out 4, the greatest common factor in -16 and 4.

$$-16t^{2}+96t-108=-4\left(-2t+3\right)\left(-2t+9\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -6x +\frac{27}{4} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 6 rs = \frac{27}{4}$$

Two numbers r and s sum up to 6 exactly when the average of the two numbers is \frac{1}{2}*6 = 3. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 3 - u s = 3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{27}{4}

$$(3 - u) (3 + u) = \frac{27}{4}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$9 - u^2 = \frac{27}{4}$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = \frac{27}{4}-9 = -\frac{9}{4}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =3 - \frac{3}{2} = 1.500 s = 3 + \frac{3}{2} = 4.500$$

Evaluate

$$-16t^{2}+96t-108$$