Solve P(x)=3x^2+240x-1800

$$P ( x ) = 3 x ^ { 2 } + 240 x - 1800$$

Factor

$$3\left(x-\left(-10\sqrt{22}-40\right)\right)\left(x-\left(10\sqrt{22}-40\right)\right)$$

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.

$$3x^{2}+240x-1800=0$$

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.

$$x=\frac{-240±\sqrt{240^{2}-4\times 3\left(-1800\right)}}{2\times 3}$$

Square 240.

$$x=\frac{-240±\sqrt{57600-4\times 3\left(-1800\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-240±\sqrt{57600-12\left(-1800\right)}}{2\times 3}$$

Multiply -12 times -1800.

$$x=\frac{-240±\sqrt{57600+21600}}{2\times 3}$$

Add 57600 to 21600.

$$x=\frac{-240±\sqrt{79200}}{2\times 3}$$

Take the square root of 79200.

$$x=\frac{-240±60\sqrt{22}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-240±60\sqrt{22}}{6}$$

Now solve the equation x=\frac{-240±60\sqrt{22}}{6} when ± is plus. Add -240 to 60\sqrt{22}.

$$x=\frac{60\sqrt{22}-240}{6}$$

Divide -240+60\sqrt{22} by 6.

$$x=10\sqrt{22}-40$$

Now solve the equation x=\frac{-240±60\sqrt{22}}{6} when ± is minus. Subtract 60\sqrt{22} from -240.

$$x=\frac{-60\sqrt{22}-240}{6}$$

Divide -240-60\sqrt{22} by 6.

$$x=-10\sqrt{22}-40$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -40+10\sqrt{22} for x_{1} and -40-10\sqrt{22} for x_{2}.

$$3x^{2}+240x-1800=3\left(x-\left(10\sqrt{22}-40\right)\right)\left(x-\left(-10\sqrt{22}-40\right)\right)$$

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 3

$$x ^ 2 +80x -600 = 0$$

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

$$r + s = -80 rs = -600$$

Two numbers r and s sum up to -80 exactly when the average of the two numbers is \frac{1}{2}*-80 = -40. 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>

$$r = -40 - u s = -40 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -600

$$(-40 - u) (-40 + u) = -600$$

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

$$1600 - u^2 = -600$$

Simplify the expression by subtracting 1600 on both sides

$$-u^2 = -600-1600 = -2200$$

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

$$u^2 = 2200 u = \pm\sqrt{2200} = \pm \sqrt{2200}$$

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

$$r =-40 - \sqrt{2200} = -86.904 s = -40 + \sqrt{2200} = 6.904$$

Evaluate

$$3\left(x^{2}+80x-600\right)$$