Solve 4x^2+2x+32

$$4 x ^ { 2 } + 2 x + 32$$

Factor

$$2\left(2x^{2}+x+16\right)$$

Solution Steps

Factor out 2. Polynomial 2x^{2}+x+16 is not factored since it does not have any rational roots.

$$2\left(2x^{2}+x+16\right)$$

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.

$$4x^{2}+2x+32=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}-4\times 4\times 32}}{2\times 4}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 4\times 32}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-2±\sqrt{4-16\times 32}}{2\times 4}$$

Multiply -16 times 32.

$$x=\frac{-2±\sqrt{4-512}}{2\times 4}$$

Add 4 to -512.

$$x=\frac{-2±\sqrt{-508}}{2\times 4}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$4x^{2}+2x+32$$

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

$$x ^ 2 +\frac{1}{2}x +8 = 0$$

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

$$r + s = -\frac{1}{2} rs = 8$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = 8

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

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

$$\frac{1}{16} - u^2 = 8$$

Simplify the expression by subtracting \frac{1}{16} on both sides

$$-u^2 = 8-\frac{1}{16} = \frac{127}{16}$$

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

$$u^2 = -\frac{127}{16} u = \pm\sqrt{-\frac{127}{16}} = \pm \frac{\sqrt{127}}{4}i$$

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

$$r =-\frac{1}{4} - \frac{\sqrt{127}}{4}i = -0.250 - 2.817i s = -\frac{1}{4} + \frac{\sqrt{127}}{4}i = -0.250 + 2.817i$$

Evaluate

$$2\left(2x^{2}+x+16\right)$$