Solve f(x)=2x^2+16x+29

$$f ( x ) = 2 x ^ { 2 } + 16 x + 29$$

Factor

$$2\left(x-\left(-\frac{\sqrt{6}}{2}-4\right)\right)\left(x-\left(\frac{\sqrt{6}}{2}-4\right)\right)$$

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.

$$2x^{2}+16x+29=0$$

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.

$$x=\frac{-16±\sqrt{16^{2}-4\times 2\times 29}}{2\times 2}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 2\times 29}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-16±\sqrt{256-8\times 29}}{2\times 2}$$

Multiply -8 times 29.

$$x=\frac{-16±\sqrt{256-232}}{2\times 2}$$

Add 256 to -232.

$$x=\frac{-16±\sqrt{24}}{2\times 2}$$

Take the square root of 24.

$$x=\frac{-16±2\sqrt{6}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-16±2\sqrt{6}}{4}$$

Now solve the equation x=\frac{-16±2\sqrt{6}}{4} when ± is plus. Add -16 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}-16}{4}$$

Divide -16+2\sqrt{6} by 4.

$$x=\frac{\sqrt{6}}{2}-4$$

Now solve the equation x=\frac{-16±2\sqrt{6}}{4} when ± is minus. Subtract 2\sqrt{6} from -16.

$$x=\frac{-2\sqrt{6}-16}{4}$$

Divide -16-2\sqrt{6} by 4.

$$x=-\frac{\sqrt{6}}{2}-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4+\frac{\sqrt{6}}{2} for x_{1} and -4-\frac{\sqrt{6}}{2} for x_{2}.

$$2x^{2}+16x+29=2\left(x-\left(\frac{\sqrt{6}}{2}-4\right)\right)\left(x-\left(-\frac{\sqrt{6}}{2}-4\right)\right)$$

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 2

$$x ^ 2 +8x +\frac{29}{2} = 0$$

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

$$r + s = -8 rs = \frac{29}{2}$$

Two numbers r and s sum up to -8 exactly when the average of the two numbers is \frac{1}{2}*-8 = -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>

$$r = -4 - u s = -4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{29}{2}

$$(-4 - u) (-4 + u) = \frac{29}{2}$$

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

$$16 - u^2 = \frac{29}{2}$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = \frac{29}{2}-16 = -\frac{3}{2}$$

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

$$u^2 = \frac{3}{2} u = \pm\sqrt{\frac{3}{2}} = \pm \frac{\sqrt{3}}{\sqrt{2}}$$

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

$$r =-4 - \frac{\sqrt{3}}{\sqrt{2}} = -5.225 s = -4 + \frac{\sqrt{3}}{\sqrt{2}} = -2.775$$

Evaluate

$$2x^{2}+16x+29$$