Solve 4x^2+24x+3

$$4 x ^ { 2 } + 24 x + 3$$

Factor

$$4\left(x-\left(-\frac{\sqrt{33}}{2}-3\right)\right)\left(x-\left(\frac{\sqrt{33}}{2}-3\right)\right)$$

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.

$$4x^{2}+24x+3=0$$

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.

$$x=\frac{-24±\sqrt{24^{2}-4\times 4\times 3}}{2\times 4}$$

Square 24.

$$x=\frac{-24±\sqrt{576-4\times 4\times 3}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-24±\sqrt{576-16\times 3}}{2\times 4}$$

Multiply -16 times 3.

$$x=\frac{-24±\sqrt{576-48}}{2\times 4}$$

Add 576 to -48.

$$x=\frac{-24±\sqrt{528}}{2\times 4}$$

Take the square root of 528.

$$x=\frac{-24±4\sqrt{33}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-24±4\sqrt{33}}{8}$$

Now solve the equation x=\frac{-24±4\sqrt{33}}{8} when ± is plus. Add -24 to 4\sqrt{33}.

$$x=\frac{4\sqrt{33}-24}{8}$$

Divide -24+4\sqrt{33} by 8.

$$x=\frac{\sqrt{33}}{2}-3$$

Now solve the equation x=\frac{-24±4\sqrt{33}}{8} when ± is minus. Subtract 4\sqrt{33} from -24.

$$x=\frac{-4\sqrt{33}-24}{8}$$

Divide -24-4\sqrt{33} by 8.

$$x=-\frac{\sqrt{33}}{2}-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3+\frac{\sqrt{33}}{2} for x_{1} and -3-\frac{\sqrt{33}}{2} for x_{2}.

$$4x^{2}+24x+3=4\left(x-\left(\frac{\sqrt{33}}{2}-3\right)\right)\left(x-\left(-\frac{\sqrt{33}}{2}-3\right)\right)$$

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 4

$$x ^ 2 +6x +\frac{3}{4} = 0$$

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

$$r + s = -6 rs = \frac{3}{4}$$

Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -3. 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>

$$r = -3 - u s = -3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{4}

$$(-3 - u) (-3 + u) = \frac{3}{4}$$

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

$$9 - u^2 = \frac{3}{4}$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = \frac{3}{4}-9 = -\frac{33}{4}$$

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

$$u^2 = \frac{33}{4} u = \pm\sqrt{\frac{33}{4}} = \pm \frac{\sqrt{33}}{2}$$

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

$$r =-3 - \frac{\sqrt{33}}{2} = -5.872 s = -3 + \frac{\sqrt{33}}{2} = -0.128$$

Evaluate

$$4x^{2}+24x+3$$