Solve x^2+5x-104

$$x ^ { 2 } + 5 x - 104$$

Factor

$$\left(x-8\right)\left(x+13\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-104. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-104\right)=-104$$

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 -104.

$$-1,104 -2,52 -4,26 -8,13$$

Calculate the sum for each pair.

$$-1+104=103 -2+52=50 -4+26=22 -8+13=5$$

The solution is the pair that gives sum 5.

$$a=-8 b=13$$

Rewrite x^{2}+5x-104 as \left(x^{2}-8x\right)+\left(13x-104\right).

$$\left(x^{2}-8x\right)+\left(13x-104\right)$$

Factor out x in the first and 13 in the second group.

$$x\left(x-8\right)+13\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x+13\right)$$

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.

$$x^{2}+5x-104=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-104\right)}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-104\right)}}{2}$$

Multiply -4 times -104.

$$x=\frac{-5±\sqrt{25+416}}{2}$$

Add 25 to 416.

$$x=\frac{-5±\sqrt{441}}{2}$$

Take the square root of 441.

$$x=\frac{-5±21}{2}$$

Now solve the equation x=\frac{-5±21}{2} when ± is plus. Add -5 to 21.

$$x=\frac{16}{2}$$

Divide 16 by 2.

$$x=8$$

Now solve the equation x=\frac{-5±21}{2} when ± is minus. Subtract 21 from -5.

$$x=-\frac{26}{2}$$

Divide -26 by 2.

$$x=-13$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 8 for x_{1} and -13 for x_{2}.

$$x^{2}+5x-104=\left(x-8\right)\left(x-\left(-13\right)\right)$$

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

$$x^{2}+5x-104=\left(x-8\right)\left(x+13\right)$$

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.

$$x ^ 2 +5x -104 = 0$$

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

$$r + s = -5 rs = -104$$

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

$$r = -\frac{5}{2} - u s = -\frac{5}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -104

$$(-\frac{5}{2} - u) (-\frac{5}{2} + u) = -104$$

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

$$\frac{25}{4} - u^2 = -104$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = -104-\frac{25}{4} = -\frac{441}{4}$$

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

$$u^2 = \frac{441}{4} u = \pm\sqrt{\frac{441}{4}} = \pm \frac{21}{2}$$

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

$$r =-\frac{5}{2} - \frac{21}{2} = -13 s = -\frac{5}{2} + \frac{21}{2} = 8$$

Evaluate

$$\left(x-8\right)\left(x+13\right)$$