Solve -t^2-3t+88

$$- t ^ { 2 } - 3 t + 88$$

Factor

$$\left(8-t\right)\left(t+11\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -t^{2}+at+bt+88. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-88=-88$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -88.

$$1,-88 2,-44 4,-22 8,-11$$

Calculate the sum for each pair.

$$1-88=-87 2-44=-42 4-22=-18 8-11=-3$$

The solution is the pair that gives sum -3.

$$a=8 b=-11$$

Rewrite -t^{2}-3t+88 as \left(-t^{2}+8t\right)+\left(-11t+88\right).

$$\left(-t^{2}+8t\right)+\left(-11t+88\right)$$

Factor out t in the first and 11 in the second group.

$$t\left(-t+8\right)+11\left(-t+8\right)$$

Factor out common term -t+8 by using distributive property.

$$\left(-t+8\right)\left(t+11\right)$$

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.

$$-t^{2}-3t+88=0$$

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.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\times 88}}{2\left(-1\right)}$$

Square -3.

$$t=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 88}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{-\left(-3\right)±\sqrt{9+4\times 88}}{2\left(-1\right)}$$

Multiply 4 times 88.

$$t=\frac{-\left(-3\right)±\sqrt{9+352}}{2\left(-1\right)}$$

Add 9 to 352.

$$t=\frac{-\left(-3\right)±\sqrt{361}}{2\left(-1\right)}$$

Take the square root of 361.

$$t=\frac{-\left(-3\right)±19}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$t=\frac{3±19}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{3±19}{-2}$$

Now solve the equation t=\frac{3±19}{-2} when ± is plus. Add 3 to 19.

$$t=\frac{22}{-2}$$

Divide 22 by -2.

$$t=-11$$

Now solve the equation t=\frac{3±19}{-2} when ± is minus. Subtract 19 from 3.

$$t=-\frac{16}{-2}$$

Divide -16 by -2.

$$t=8$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -11 for x_{1} and 8 for x_{2}.

$$-t^{2}-3t+88=-\left(t-\left(-11\right)\right)\left(t-8\right)$$

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

$$-t^{2}-3t+88=-\left(t+11\right)\left(t-8\right)$$

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.

$$x ^ 2 +3x -88 = 0$$

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

$$r + s = -3 rs = -88$$

Two numbers r and s sum up to -3 exactly when the average of the two numbers is \frac{1}{2}*-3 = -\frac{3}{2}. 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>

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

To solve for unknown quantity u, substitute these in the product equation rs = -88

$$(-\frac{3}{2} - u) (-\frac{3}{2} + u) = -88$$

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

$$\frac{9}{4} - u^2 = -88$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = -88-\frac{9}{4} = -\frac{361}{4}$$

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

$$u^2 = \frac{361}{4} u = \pm\sqrt{\frac{361}{4}} = \pm \frac{19}{2}$$

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

$$r =-\frac{3}{2} - \frac{19}{2} = -11 s = -\frac{3}{2} + \frac{19}{2} = 8$$

Evaluate

$$\left(8-t\right)\left(t+11\right)$$