Solve f(x)=4x^2-5x+20

$$f ( x ) = 4 x ^ { 2 } - 5 x + 20$$

Evaluate

$$4x^{2}-5x+20$$

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

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

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

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

Two numbers r and s sum up to \frac{5}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{4} = \frac{5}{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>

$$r = \frac{5}{8} - u s = \frac{5}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 5

$$(\frac{5}{8} - u) (\frac{5}{8} + u) = 5$$

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

$$\frac{25}{64} - u^2 = 5$$

Simplify the expression by subtracting \frac{25}{64} on both sides

$$-u^2 = 5-\frac{25}{64} = \frac{295}{64}$$

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

$$u^2 = -\frac{295}{64} u = \pm\sqrt{-\frac{295}{64}} = \pm \frac{\sqrt{295}}{8}i$$

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

$$r =\frac{5}{8} - \frac{\sqrt{295}}{8}i = 0.625 - 2.147i s = \frac{5}{8} + \frac{\sqrt{295}}{8}i = 0.625 + 2.147i$$

Differentiate w.r.t. x

$$8x-5$$

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}.

$$2\times 4x^{2-1}-5x^{1-1}$$

Multiply 2 times 4.

$$8x^{2-1}-5x^{1-1}$$

Subtract 1 from 2.

$$8x^{1}-5x^{1-1}$$

Subtract 1 from 1.

$$8x^{1}-5x^{0}$$

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

$$8x-5x^{0}$$

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

$$8x-5$$