Solve f(x)=x^2-10x-96?

$$f ( x ) = x ^ { 2 } - 10 x - 96 ?$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-96. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\left(-96\right)=-96$$

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

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

Factor out x in the first and 6 in the second group.

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

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

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

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.

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

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-96\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-96\right)}}{2}$$

Multiply -4 times -96.

$$x=\frac{-\left(-10\right)±\sqrt{100+384}}{2}$$

Add 100 to 384.

$$x=\frac{-\left(-10\right)±\sqrt{484}}{2}$$

Take the square root of 484.

$$x=\frac{-\left(-10\right)±22}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±22}{2}$$

Now solve the equation x=\frac{10±22}{2} when ± is plus. Add 10 to 22.

$$x=\frac{32}{2}$$

Divide 32 by 2.

$$x=16$$

Now solve the equation x=\frac{10±22}{2} when ± is minus. Subtract 22 from 10.

$$x=-\frac{12}{2}$$

Divide -12 by 2.

$$x=-6$$

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

$$x^{2}-10x-96=\left(x-16\right)\left(x-\left(-6\right)\right)$$

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

$$x^{2}-10x-96=\left(x-16\right)\left(x+6\right)$$

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

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