Solve h(t)=-16t^2+160t+1536

$$h ( t ) = - 16 t ^ { 2 } + 160 t + 1536$$

Factor

$$-16\left(t-16\right)\left(t+6\right)$$

Solution Steps

Factor out 16.

$$16\left(-t^{2}+10t+96\right)$$

Consider -t^{2}+10t+96. Factor the expression by grouping. First, the expression needs to be rewritten as -t^{2}+at+bt+96. To find a and b, set up a system to be solved.

$$a+b=10 ab=-96=-96$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -96.

$$-1,96 -2,48 -3,32 -4,24 -6,16 -8,12$$

Calculate the sum for each pair.

$$-1+96=95 -2+48=46 -3+32=29 -4+24=20 -6+16=10 -8+12=4$$

The solution is the pair that gives sum 10.

$$a=16 b=-6$$

Rewrite -t^{2}+10t+96 as \left(-t^{2}+16t\right)+\left(-6t+96\right).

$$\left(-t^{2}+16t\right)+\left(-6t+96\right)$$

Factor out -t in the first and -6 in the second group.

$$-t\left(t-16\right)-6\left(t-16\right)$$

Factor out common term t-16 by using distributive property.

$$\left(t-16\right)\left(-t-6\right)$$

Rewrite the complete factored expression.

$$16\left(t-16\right)\left(-t-6\right)$$

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.

$$-16t^{2}+160t+1536=0$$

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.

$$t=\frac{-160±\sqrt{160^{2}-4\left(-16\right)\times 1536}}{2\left(-16\right)}$$

Square 160.

$$t=\frac{-160±\sqrt{25600-4\left(-16\right)\times 1536}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-160±\sqrt{25600+64\times 1536}}{2\left(-16\right)}$$

Multiply 64 times 1536.

$$t=\frac{-160±\sqrt{25600+98304}}{2\left(-16\right)}$$

Add 25600 to 98304.

$$t=\frac{-160±\sqrt{123904}}{2\left(-16\right)}$$

Take the square root of 123904.

$$t=\frac{-160±352}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-160±352}{-32}$$

Now solve the equation t=\frac{-160±352}{-32} when ± is plus. Add -160 to 352.

$$t=\frac{192}{-32}$$

Divide 192 by -32.

$$t=-6$$

Now solve the equation t=\frac{-160±352}{-32} when ± is minus. Subtract 352 from -160.

$$t=-\frac{512}{-32}$$

Divide -512 by -32.

$$t=16$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -6 for x_{1} and 16 for x_{2}.

$$-16t^{2}+160t+1536=-16\left(t-\left(-6\right)\right)\left(t-16\right)$$

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

$$-16t^{2}+160t+1536=-16\left(t+6\right)\left(t-16\right)$$

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.

$$x ^ 2 -10x -96 = 0$$

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

$$r + s = 10 rs = -96$$

Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. 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>

$$r = 5 - u s = 5 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -96

$$(5 - u) (5 + u) = -96$$

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

$$25 - u^2 = -96$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = -96-25 = -121$$

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

$$u^2 = 121 u = \pm\sqrt{121} = \pm 11$$

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

$$r =5 - 11 = -6 s = 5 + 11 = 16$$

Evaluate

$$-16\left(t-16\right)\left(t+6\right)$$