Solve 4x^2+x+3

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

Evaluate

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

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 4

$$x ^ 2 +\frac{1}{4}x +\frac{3}{4} = 0$$

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

$$r + s = -\frac{1}{4} rs = \frac{3}{4}$$

Two numbers r and s sum up to -\frac{1}{4} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{4} = -\frac{1}{8}. 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>

$$r = -\frac{1}{8} - u s = -\frac{1}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{4}

$$(-\frac{1}{8} - u) (-\frac{1}{8} + u) = \frac{3}{4}$$

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

$$\frac{1}{64} - u^2 = \frac{3}{4}$$

Simplify the expression by subtracting \frac{1}{64} on both sides

$$-u^2 = \frac{3}{4}-\frac{1}{64} = \frac{47}{64}$$

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

$$u^2 = -\frac{47}{64} u = \pm\sqrt{-\frac{47}{64}} = \pm \frac{\sqrt{47}}{8}i$$

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

$$r =-\frac{1}{8} - \frac{\sqrt{47}}{8}i = -0.125 - 0.857i s = -\frac{1}{8} + \frac{\sqrt{47}}{8}i = -0.125 + 0.857i$$

Differentiate w.r.t. x

$$8x+1$$

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}.

$$2\times 4x^{2-1}+x^{1-1}$$

Multiply 2 times 4.

$$8x^{2-1}+x^{1-1}$$

Subtract 1 from 2.

$$8x^{1}+x^{1-1}$$

Subtract 1 from 1.

$$8x^{1}+x^{0}$$

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

$$8x+x^{0}$$

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

$$8x+1$$