Solve 4x^2+8x+5

$$4 x ^ { 2 } + 8 x + 5$$

Evaluate

$$4x^{2}+8x+5$$

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

$$x ^ 2 +2x +\frac{5}{4} = 0$$

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

$$r + s = -2 rs = \frac{5}{4}$$

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 = \frac{5}{4}

$$(-1 - u) (-1 + u) = \frac{5}{4}$$

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

$$1 - u^2 = \frac{5}{4}$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = \frac{5}{4}-1 = \frac{1}{4}$$

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

$$u^2 = -\frac{1}{4} u = \pm\sqrt{-\frac{1}{4}} = \pm \frac{1}{2}i$$

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

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

Differentiate w.r.t. x

$$8\left(x+1\right)$$

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}.

$$2\times 4x^{2-1}+8x^{1-1}$$

Multiply 2 times 4.

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

Subtract 1 from 2.

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

Subtract 1 from 1.

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

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

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

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

$$8x+8\times 1$$

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

$$8x+8$$