Solve 3k^2+3k+1

$$3 k ^ { 2 } + 3 k + 1$$

Evaluate

$$3k^{2}+3k+1$$

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.This is achieved by dividing both sides of the equation by 3

$$x ^ 2 +1x +\frac{1}{3} = 0$$

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

$$r + s = -1 rs = \frac{1}{3}$$

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

$$r = -\frac{1}{2} - u s = -\frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{3}

$$(-\frac{1}{2} - u) (-\frac{1}{2} + u) = \frac{1}{3}$$

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

$$\frac{1}{4} - u^2 = \frac{1}{3}$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = \frac{1}{3}-\frac{1}{4} = \frac{1}{12}$$

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

$$u^2 = -\frac{1}{12} u = \pm\sqrt{-\frac{1}{12}} = \pm \frac{1}{\sqrt{12}}i$$

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

$$r =-\frac{1}{2} - \frac{1}{\sqrt{12}}i = -0.500 - 0.289i s = -\frac{1}{2} + \frac{1}{\sqrt{12}}i = -0.500 + 0.289i$$

Differentiate w.r.t. k

$$6k+3$$

Steps Using Derivative Rule for Sum

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$2\times 3k^{2-1}+3k^{1-1}$$

Multiply 2 times 3.

$$6k^{2-1}+3k^{1-1}$$

Subtract 1 from 2.

$$6k^{1}+3k^{1-1}$$

Subtract 1 from 1.

$$6k^{1}+3k^{0}$$

For any term t, t^{1}=t.

$$6k+3k^{0}$$

For any term t except 0, t^{0}=1.

$$6k+3\times 1$$

For any term t, t\times 1=t and 1t=t.

$$6k+3$$