Solve k^2+2k-8

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

Factor

$$\left(k-2\right)\left(k+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as k^{2}+ak+bk-8. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-8\right)=-8$$

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 -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=-2 b=4$$

Rewrite k^{2}+2k-8 as \left(k^{2}-2k\right)+\left(4k-8\right).

$$\left(k^{2}-2k\right)+\left(4k-8\right)$$

Factor out k in the first and 4 in the second group.

$$k\left(k-2\right)+4\left(k-2\right)$$

Factor out common term k-2 by using distributive property.

$$\left(k-2\right)\left(k+4\right)$$

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.

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

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.

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

Square 2.

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

Multiply -4 times -8.

$$k=\frac{-2±\sqrt{4+32}}{2}$$

Add 4 to 32.

$$k=\frac{-2±\sqrt{36}}{2}$$

Take the square root of 36.

$$k=\frac{-2±6}{2}$$

Now solve the equation k=\frac{-2±6}{2} when ± is plus. Add -2 to 6.

$$k=\frac{4}{2}$$

Divide 4 by 2.

$$k=2$$

Now solve the equation k=\frac{-2±6}{2} when ± is minus. Subtract 6 from -2.

$$k=-\frac{8}{2}$$

Divide -8 by 2.

$$k=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and -4 for x_{2}.

$$k^{2}+2k-8=\left(k-2\right)\left(k-\left(-4\right)\right)$$

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

$$k^{2}+2k-8=\left(k-2\right)\left(k+4\right)$$

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.

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

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

$$r + s = -2 rs = -8$$

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

$$r = -1 - u s = -1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -8

$$(-1 - u) (-1 + u) = -8$$

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

$$1 - u^2 = -8$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -8-1 = -9$$

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

$$u^2 = 9 u = \pm\sqrt{9} = \pm 3$$

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

$$r =-1 - 3 = -4 s = -1 + 3 = 2$$

Evaluate

$$\left(k-2\right)\left(k+4\right)$$