Solve -16t^2+104t+560

$$- 16 t ^ { 2 } + 104 t + 560$$

Factor

$$8\left(10-t\right)\left(2t+7\right)$$

Solution Steps

Factor out 8.

$$8\left(-2t^{2}+13t+70\right)$$

Consider -2t^{2}+13t+70. Factor the expression by grouping. First, the expression needs to be rewritten as -2t^{2}+at+bt+70. To find a and b, set up a system to be solved.

$$a+b=13 ab=-2\times 70=-140$$

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 -140.

$$-1,140 -2,70 -4,35 -5,28 -7,20 -10,14$$

Calculate the sum for each pair.

$$-1+140=139 -2+70=68 -4+35=31 -5+28=23 -7+20=13 -10+14=4$$

The solution is the pair that gives sum 13.

$$a=20 b=-7$$

Rewrite -2t^{2}+13t+70 as \left(-2t^{2}+20t\right)+\left(-7t+70\right).

$$\left(-2t^{2}+20t\right)+\left(-7t+70\right)$$

Factor out 2t in the first and 7 in the second group.

$$2t\left(-t+10\right)+7\left(-t+10\right)$$

Factor out common term -t+10 by using distributive property.

$$\left(-t+10\right)\left(2t+7\right)$$

Rewrite the complete factored expression.

$$8\left(-t+10\right)\left(2t+7\right)$$

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.

$$-16t^{2}+104t+560=0$$

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.

$$t=\frac{-104±\sqrt{104^{2}-4\left(-16\right)\times 560}}{2\left(-16\right)}$$

Square 104.

$$t=\frac{-104±\sqrt{10816-4\left(-16\right)\times 560}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$t=\frac{-104±\sqrt{10816+64\times 560}}{2\left(-16\right)}$$

Multiply 64 times 560.

$$t=\frac{-104±\sqrt{10816+35840}}{2\left(-16\right)}$$

Add 10816 to 35840.

$$t=\frac{-104±\sqrt{46656}}{2\left(-16\right)}$$

Take the square root of 46656.

$$t=\frac{-104±216}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-104±216}{-32}$$

Now solve the equation t=\frac{-104±216}{-32} when ± is plus. Add -104 to 216.

$$t=\frac{112}{-32}$$

Reduce the fraction \frac{112}{-32} to lowest terms by extracting and canceling out 16.

$$t=-\frac{7}{2}$$

Now solve the equation t=\frac{-104±216}{-32} when ± is minus. Subtract 216 from -104.

$$t=-\frac{320}{-32}$$

Divide -320 by -32.

$$t=10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{7}{2} for x_{1} and 10 for x_{2}.

$$-16t^{2}+104t+560=-16\left(t-\left(-\frac{7}{2}\right)\right)\left(t-10\right)$$

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

$$-16t^{2}+104t+560=-16\left(t+\frac{7}{2}\right)\left(t-10\right)$$

Add \frac{7}{2} to t by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-16t^{2}+104t+560=-16\times \frac{-2t-7}{-2}\left(t-10\right)$$

Cancel out 2, the greatest common factor in -16 and 2.

$$-16t^{2}+104t+560=8\left(-2t-7\right)\left(t-10\right)$$

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.

$$x ^ 2 -\frac{13}{2}x -35 = 0$$

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

$$r + s = \frac{13}{2} rs = -35$$

Two numbers r and s sum up to \frac{13}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{2} = \frac{13}{4}. 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>

$$r = \frac{13}{4} - u s = \frac{13}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -35

$$(\frac{13}{4} - u) (\frac{13}{4} + u) = -35$$

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

$$\frac{169}{16} - u^2 = -35$$

Simplify the expression by subtracting \frac{169}{16} on both sides

$$-u^2 = -35-\frac{169}{16} = -\frac{729}{16}$$

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

$$u^2 = \frac{729}{16} u = \pm\sqrt{\frac{729}{16}} = \pm \frac{27}{4}$$

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

$$r =\frac{13}{4} - \frac{27}{4} = -3.500 s = \frac{13}{4} + \frac{27}{4} = 10$$

Evaluate

$$560+104t-16t^{2}$$