Solve {l}{6x^2+8x}{+64}

$$\left. \begin{array} { l } { 6 x ^ { 2 } + 8 x } \\ { + 64 } \end{array} \right.$$

Factor

$$2\left(3x^{2}+4x+32\right)$$

Solution Steps

Factor out 2. Polynomial 3x^{2}+4x+32 is not factored since it does not have any rational roots.

$$2\left(3x^{2}+4x+32\right)$$

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.

$$6x^{2}+8x+64=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}-4\times 6\times 64}}{2\times 6}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 6\times 64}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-8±\sqrt{64-24\times 64}}{2\times 6}$$

Multiply -24 times 64.

$$x=\frac{-8±\sqrt{64-1536}}{2\times 6}$$

Add 64 to -1536.

$$x=\frac{-8±\sqrt{-1472}}{2\times 6}$$

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

$$6x^{2}+8x+64$$

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 6

$$x ^ 2 +\frac{4}{3}x +\frac{32}{3} = 0$$

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

$$r + s = -\frac{4}{3} rs = \frac{32}{3}$$

Two numbers r and s sum up to -\frac{4}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{4}{3} = -\frac{2}{3}. 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>

$$r = -\frac{2}{3} - u s = -\frac{2}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{32}{3}

$$(-\frac{2}{3} - u) (-\frac{2}{3} + u) = \frac{32}{3}$$

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

$$\frac{4}{9} - u^2 = \frac{32}{3}$$

Simplify the expression by subtracting \frac{4}{9} on both sides

$$-u^2 = \frac{32}{3}-\frac{4}{9} = \frac{92}{9}$$

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

$$u^2 = -\frac{92}{9} u = \pm\sqrt{-\frac{92}{9}} = \pm \frac{\sqrt{92}}{3}i$$

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

$$r =-\frac{2}{3} - \frac{\sqrt{92}}{3}i = -0.667 - 3.197i s = -\frac{2}{3} + \frac{\sqrt{92}}{3}i = -0.667 + 3.197i$$

Evaluate

$$6x^{2}+8x+64$$