Solve 6+8k+2k^2

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

Factor

$$2\left(k+1\right)\left(k+3\right)$$

Solution Steps

Factor out 2.

$$2\left(3+4k+k^{2}\right)$$

Consider 3+4k+k^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$k^{2}+4k+3$$

Factor the expression by grouping. First, the expression needs to be rewritten as k^{2}+ak+bk+3. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=3$$

Rewrite k^{2}+4k+3 as \left(k^{2}+k\right)+\left(3k+3\right).

$$\left(k^{2}+k\right)+\left(3k+3\right)$$

Factor out k in the first and 3 in the second group.

$$k\left(k+1\right)+3\left(k+1\right)$$

Factor out common term k+1 by using distributive property.

$$\left(k+1\right)\left(k+3\right)$$

Rewrite the complete factored expression.

$$2\left(k+1\right)\left(k+3\right)$$

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.

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

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.

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

Square 8.

$$k=\frac{-8±\sqrt{64-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$k=\frac{-8±\sqrt{64-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$k=\frac{-8±\sqrt{64-48}}{2\times 2}$$

Add 64 to -48.

$$k=\frac{-8±\sqrt{16}}{2\times 2}$$

Take the square root of 16.

$$k=\frac{-8±4}{2\times 2}$$

Multiply 2 times 2.

$$k=\frac{-8±4}{4}$$

Now solve the equation k=\frac{-8±4}{4} when ± is plus. Add -8 to 4.

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

Divide -4 by 4.

$$k=-1$$

Now solve the equation k=\frac{-8±4}{4} when ± is minus. Subtract 4 from -8.

$$k=-\frac{12}{4}$$

Divide -12 by 4.

$$k=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1 for x_{1} and -3 for x_{2}.

$$2k^{2}+8k+6=2\left(k-\left(-1\right)\right)\left(k-\left(-3\right)\right)$$

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

$$2k^{2}+8k+6=2\left(k+1\right)\left(k+3\right)$$

Evaluate

$$2\left(k+1\right)\left(k+3\right)$$