Solve 460x^2-5x+40

$$460 x ^ { 2 } - 5 x + 40$$

Factor

$$5\left(92x^{2}-x+8\right)$$

Solution Steps

Factor out 5. Polynomial 92x^{2}-x+8 is not factored since it does not have any rational roots.

$$5\left(92x^{2}-x+8\right)$$

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.

$$460x^{2}-5x+40=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 460\times 40}}{2\times 460}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 460\times 40}}{2\times 460}$$

Multiply -4 times 460.

$$x=\frac{-\left(-5\right)±\sqrt{25-1840\times 40}}{2\times 460}$$

Multiply -1840 times 40.

$$x=\frac{-\left(-5\right)±\sqrt{25-73600}}{2\times 460}$$

Add 25 to -73600.

$$x=\frac{-\left(-5\right)±\sqrt{-73575}}{2\times 460}$$

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

$$460x^{2}-5x+40$$

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 460

$$x ^ 2 -\frac{1}{92}x +\frac{2}{23} = 0$$

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

$$r + s = \frac{1}{92} rs = \frac{2}{23}$$

Two numbers r and s sum up to \frac{1}{92} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{92} = \frac{1}{184}. 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>

$$r = \frac{1}{184} - u s = \frac{1}{184} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{23}

$$(\frac{1}{184} - u) (\frac{1}{184} + u) = \frac{2}{23}$$

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

$$\frac{1}{33856} - u^2 = \frac{2}{23}$$

Simplify the expression by subtracting \frac{1}{33856} on both sides

$$-u^2 = \frac{2}{23}-\frac{1}{33856} = \frac{2943}{33856}$$

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

$$u^2 = -\frac{2943}{33856} u = \pm\sqrt{-\frac{2943}{33856}} = \pm \frac{\sqrt{2943}}{184}i$$

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

$$r =\frac{1}{184} - \frac{\sqrt{2943}}{184}i = 0.005 - 0.295i s = \frac{1}{184} + \frac{\sqrt{2943}}{184}i = 0.005 + 0.295i$$

Evaluate

$$460x^{2}-5x+40$$