Solve f(x)=3x^2-12x+15

$$f ( x ) = 3 x ^ { 2 } - 12 x + 15$$

Factor

$$3\left(x^{2}-4x+5\right)$$

Solution Steps

Factor out 3. Polynomial x^{2}-4x+5 is not factored since it does not have any rational roots.

$$3\left(x^{2}-4x+5\right)$$

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.

$$3x^{2}-12x+15=0$$

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.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 3\times 15}}{2\times 3}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 3\times 15}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-12\right)±\sqrt{144-12\times 15}}{2\times 3}$$

Multiply -12 times 15.

$$x=\frac{-\left(-12\right)±\sqrt{144-180}}{2\times 3}$$

Add 144 to -180.

$$x=\frac{-\left(-12\right)±\sqrt{-36}}{2\times 3}$$

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

$$3x^{2}-12x+15$$

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

$$x ^ 2 -4x +5 = 0$$

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

$$r + s = 4 rs = 5$$

Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 2. 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>

$$r = 2 - u s = 2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 5

$$(2 - u) (2 + u) = 5$$

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

$$4 - u^2 = 5$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = 5-4 = 1$$

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

$$u^2 = -1 u = \pm\sqrt{-1} = \pm i$$

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

$$r =2 - i s = 2 + i$$

Evaluate

$$3\left(x^{2}-4x+5\right)$$