Solve k^2+6k+8

$$k ^ { 2 } + 6 k + 8$$

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.

$$a+b=6 ab=1\times 8=8$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 6.

$$a=2 b=4$$

Rewrite k^{2}+6k+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.

$$k^{2}+6k+8=0$$

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.

$$k=\frac{-6±\sqrt{6^{2}-4\times 8}}{2}$$

Square 6.

$$k=\frac{-6±\sqrt{36-4\times 8}}{2}$$

Multiply -4 times 8.

$$k=\frac{-6±\sqrt{36-32}}{2}$$

Add 36 to -32.

$$k=\frac{-6±\sqrt{4}}{2}$$

Take the square root of 4.

$$k=\frac{-6±2}{2}$$

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

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

Divide -4 by 2.

$$k=-2$$

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

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

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}.

$$k^{2}+6k+8=\left(k-\left(-2\right)\right)\left(k-\left(-4\right)\right)$$

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

$$k^{2}+6k+8=\left(k+2\right)\left(k+4\right)$$

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.

$$x ^ 2 +6x +8 = 0$$

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

$$r + s = -6 rs = 8$$

Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -3. 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>

$$r = -3 - u s = -3 + u$$

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

$$(-3 - u) (-3 + u) = 8$$

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

$$9 - u^2 = 8$$

Simplify the expression by subtracting 9 on both sides

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

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

$$u^2 = 1 u = \pm\sqrt{1} = \pm 1$$

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

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

Evaluate

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