Solve f(x)=16x^2+192x+568

$$f ( x ) = 16 x ^ { 2 } + 192 x + 568$$

Factor

$$16\left(x-\left(-\frac{\sqrt{2}}{2}-6\right)\right)\left(x-\left(\frac{\sqrt{2}}{2}-6\right)\right)$$

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.

$$16x^{2}+192x+568=0$$

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.

$$x=\frac{-192±\sqrt{192^{2}-4\times 16\times 568}}{2\times 16}$$

Square 192.

$$x=\frac{-192±\sqrt{36864-4\times 16\times 568}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-192±\sqrt{36864-64\times 568}}{2\times 16}$$

Multiply -64 times 568.

$$x=\frac{-192±\sqrt{36864-36352}}{2\times 16}$$

Add 36864 to -36352.

$$x=\frac{-192±\sqrt{512}}{2\times 16}$$

Take the square root of 512.

$$x=\frac{-192±16\sqrt{2}}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-192±16\sqrt{2}}{32}$$

Now solve the equation x=\frac{-192±16\sqrt{2}}{32} when ± is plus. Add -192 to 16\sqrt{2}.

$$x=\frac{16\sqrt{2}-192}{32}$$

Divide -192+16\sqrt{2} by 32.

$$x=\frac{\sqrt{2}}{2}-6$$

Now solve the equation x=\frac{-192±16\sqrt{2}}{32} when ± is minus. Subtract 16\sqrt{2} from -192.

$$x=\frac{-16\sqrt{2}-192}{32}$$

Divide -192-16\sqrt{2} by 32.

$$x=-\frac{\sqrt{2}}{2}-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -6+\frac{\sqrt{2}}{2} for x_{1} and -6-\frac{\sqrt{2}}{2} for x_{2}.

$$16x^{2}+192x+568=16\left(x-\left(\frac{\sqrt{2}}{2}-6\right)\right)\left(x-\left(-\frac{\sqrt{2}}{2}-6\right)\right)$$

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 16

$$x ^ 2 +12x +\frac{71}{2} = 0$$

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

$$r + s = -12 rs = \frac{71}{2}$$

Two numbers r and s sum up to -12 exactly when the average of the two numbers is \frac{1}{2}*-12 = -6. 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>

$$r = -6 - u s = -6 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{71}{2}

$$(-6 - u) (-6 + u) = \frac{71}{2}$$

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

$$36 - u^2 = \frac{71}{2}$$

Simplify the expression by subtracting 36 on both sides

$$-u^2 = \frac{71}{2}-36 = -\frac{1}{2}$$

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

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

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

$$r =-6 - \frac{1}{\sqrt{2}} = -6.707 s = -6 + \frac{1}{\sqrt{2}} = -5.293$$

Evaluate

$$16x^{2}+192x+568$$