Solve k^2-8k-65

$$k ^ { 2 } - 8 k - 65$$

Factor

$$\left(k-13\right)\left(k+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as k^{2}+ak+bk-65. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\left(-65\right)=-65$$

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 -65.

$$1,-65 5,-13$$

Calculate the sum for each pair.

$$1-65=-64 5-13=-8$$

The solution is the pair that gives sum -8.

$$a=-13 b=5$$

Rewrite k^{2}-8k-65 as \left(k^{2}-13k\right)+\left(5k-65\right).

$$\left(k^{2}-13k\right)+\left(5k-65\right)$$

Factor out k in the first and 5 in the second group.

$$k\left(k-13\right)+5\left(k-13\right)$$

Factor out common term k-13 by using distributive property.

$$\left(k-13\right)\left(k+5\right)$$

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.

$$k^{2}-8k-65=0$$

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.

$$k=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-65\right)}}{2}$$

Square -8.

$$k=\frac{-\left(-8\right)±\sqrt{64-4\left(-65\right)}}{2}$$

Multiply -4 times -65.

$$k=\frac{-\left(-8\right)±\sqrt{64+260}}{2}$$

Add 64 to 260.

$$k=\frac{-\left(-8\right)±\sqrt{324}}{2}$$

Take the square root of 324.

$$k=\frac{-\left(-8\right)±18}{2}$$

The opposite of -8 is 8.

$$k=\frac{8±18}{2}$$

Now solve the equation k=\frac{8±18}{2} when ± is plus. Add 8 to 18.

$$k=\frac{26}{2}$$

Divide 26 by 2.

$$k=13$$

Now solve the equation k=\frac{8±18}{2} when ± is minus. Subtract 18 from 8.

$$k=-\frac{10}{2}$$

Divide -10 by 2.

$$k=-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 13 for x_{1} and -5 for x_{2}.

$$k^{2}-8k-65=\left(k-13\right)\left(k-\left(-5\right)\right)$$

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

$$k^{2}-8k-65=\left(k-13\right)\left(k+5\right)$$

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.

$$x ^ 2 -8x -65 = 0$$

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

$$r + s = 8 rs = -65$$

Two numbers r and s sum up to 8 exactly when the average of the two numbers is \frac{1}{2}*8 = 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>

$$r = 4 - u s = 4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -65

$$(4 - u) (4 + u) = -65$$

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

$$16 - u^2 = -65$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = -65-16 = -81$$

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

$$u^2 = 81 u = \pm\sqrt{81} = \pm 9$$

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

$$r =4 - 9 = -5 s = 4 + 9 = 13$$

Evaluate

$$\left(k-13\right)\left(k+5\right)$$