Solve 4a^2-24a+4

$$4 a ^ { 2 } - 24 a + 4$$

Factor

$$4\left(a-\left(3-2\sqrt{2}\right)\right)\left(a-\left(2\sqrt{2}+3\right)\right)$$

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.

$$4a^{2}-24a+4=0$$

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.

$$a=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 4\times 4}}{2\times 4}$$

Square -24.

$$a=\frac{-\left(-24\right)±\sqrt{576-4\times 4\times 4}}{2\times 4}$$

Multiply -4 times 4.

$$a=\frac{-\left(-24\right)±\sqrt{576-16\times 4}}{2\times 4}$$

Multiply -16 times 4.

$$a=\frac{-\left(-24\right)±\sqrt{576-64}}{2\times 4}$$

Add 576 to -64.

$$a=\frac{-\left(-24\right)±\sqrt{512}}{2\times 4}$$

Take the square root of 512.

$$a=\frac{-\left(-24\right)±16\sqrt{2}}{2\times 4}$$

The opposite of -24 is 24.

$$a=\frac{24±16\sqrt{2}}{2\times 4}$$

Multiply 2 times 4.

$$a=\frac{24±16\sqrt{2}}{8}$$

Now solve the equation a=\frac{24±16\sqrt{2}}{8} when ± is plus. Add 24 to 16\sqrt{2}.

$$a=\frac{16\sqrt{2}+24}{8}$$

Divide 24+16\sqrt{2} by 8.

$$a=2\sqrt{2}+3$$

Now solve the equation a=\frac{24±16\sqrt{2}}{8} when ± is minus. Subtract 16\sqrt{2} from 24.

$$a=\frac{24-16\sqrt{2}}{8}$$

Divide 24-16\sqrt{2} by 8.

$$a=3-2\sqrt{2}$$

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

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

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

$$x ^ 2 -6x +1 = 0$$

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

$$r + s = 6 rs = 1$$

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 = 1

$$(3 - u) (3 + u) = 1$$

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

$$9 - u^2 = 1$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = 1-9 = -8$$

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

$$u^2 = 8 u = \pm\sqrt{8} = \pm \sqrt{8}$$

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

$$r =3 - \sqrt{8} = 0.172 s = 3 + \sqrt{8} = 5.828$$

Evaluate

$$4\left(a^{2}-6a+1\right)$$