Solve f(x)=-32x^2+x+5

$$f ( x ) = - 32 x ^ { 2 } + x + 5$$

Factor

$$-32\left(x-\frac{1-\sqrt{641}}{64}\right)\left(x-\frac{\sqrt{641}+1}{64}\right)$$

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.

$$-32x^{2}+x+5=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-32\right)\times 5}}{2\left(-32\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-32\right)\times 5}}{2\left(-32\right)}$$

Multiply -4 times -32.

$$x=\frac{-1±\sqrt{1+128\times 5}}{2\left(-32\right)}$$

Multiply 128 times 5.

$$x=\frac{-1±\sqrt{1+640}}{2\left(-32\right)}$$

Add 1 to 640.

$$x=\frac{-1±\sqrt{641}}{2\left(-32\right)}$$

Multiply 2 times -32.

$$x=\frac{-1±\sqrt{641}}{-64}$$

Now solve the equation x=\frac{-1±\sqrt{641}}{-64} when ± is plus. Add -1 to \sqrt{641}.

$$x=\frac{\sqrt{641}-1}{-64}$$

Divide -1+\sqrt{641} by -64.

$$x=\frac{1-\sqrt{641}}{64}$$

Now solve the equation x=\frac{-1±\sqrt{641}}{-64} when ± is minus. Subtract \sqrt{641} from -1.

$$x=\frac{-\sqrt{641}-1}{-64}$$

Divide -1-\sqrt{641} by -64.

$$x=\frac{\sqrt{641}+1}{64}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1-\sqrt{641}}{64} for x_{1} and \frac{1+\sqrt{641}}{64} for x_{2}.

$$-32x^{2}+x+5=-32\left(x-\frac{1-\sqrt{641}}{64}\right)\left(x-\frac{\sqrt{641}+1}{64}\right)$$

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.

$$x ^ 2 -\frac{1}{32}x -\frac{5}{32} = 0$$

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

$$r + s = \frac{1}{32} rs = -\frac{5}{32}$$

Two numbers r and s sum up to \frac{1}{32} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{32} = \frac{1}{64}. 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>

$$r = \frac{1}{64} - u s = \frac{1}{64} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{32}

$$(\frac{1}{64} - u) (\frac{1}{64} + u) = -\frac{5}{32}$$

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

$$\frac{1}{4096} - u^2 = -\frac{5}{32}$$

Simplify the expression by subtracting \frac{1}{4096} on both sides

$$-u^2 = -\frac{5}{32}-\frac{1}{4096} = -\frac{641}{4096}$$

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

$$u^2 = \frac{641}{4096} u = \pm\sqrt{\frac{641}{4096}} = \pm \frac{\sqrt{641}}{64}$$

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

$$r =\frac{1}{64} - \frac{\sqrt{641}}{64} = -0.380 s = \frac{1}{64} + \frac{\sqrt{641}}{64} = 0.411$$

Evaluate

$$5+x-32x^{2}$$