Solve 3k^2-12k+9

$$3 k ^ { 2 } - 12 k + 9$$

Factor

$$3\left(k-3\right)\left(k-1\right)$$

Solution Steps

Factor out 3.

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

Consider k^{2}-4k+3. Factor the expression by grouping. First, the expression needs to be rewritten as k^{2}+ak+bk+3. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

Rewrite k^{2}-4k+3 as \left(k^{2}-3k\right)+\left(-k+3\right).

$$\left(k^{2}-3k\right)+\left(-k+3\right)$$

Factor out k in the first and -1 in the second group.

$$k\left(k-3\right)-\left(k-3\right)$$

Factor out common term k-3 by using distributive property.

$$\left(k-3\right)\left(k-1\right)$$

Rewrite the complete factored expression.

$$3\left(k-3\right)\left(k-1\right)$$

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.

$$3k^{2}-12k+9=0$$

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.

$$k=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 3\times 9}}{2\times 3}$$

Square -12.

$$k=\frac{-\left(-12\right)±\sqrt{144-4\times 3\times 9}}{2\times 3}$$

Multiply -4 times 3.

$$k=\frac{-\left(-12\right)±\sqrt{144-12\times 9}}{2\times 3}$$

Multiply -12 times 9.

$$k=\frac{-\left(-12\right)±\sqrt{144-108}}{2\times 3}$$

Add 144 to -108.

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

Take the square root of 36.

$$k=\frac{-\left(-12\right)±6}{2\times 3}$$

The opposite of -12 is 12.

$$k=\frac{12±6}{2\times 3}$$

Multiply 2 times 3.

$$k=\frac{12±6}{6}$$

Now solve the equation k=\frac{12±6}{6} when ± is plus. Add 12 to 6.

$$k=\frac{18}{6}$$

Divide 18 by 6.

$$k=3$$

Now solve the equation k=\frac{12±6}{6} when ± is minus. Subtract 6 from 12.

$$k=\frac{6}{6}$$

Divide 6 by 6.

$$k=1$$

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

$$3k^{2}-12k+9=3\left(k-3\right)\left(k-1\right)$$

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

$$x ^ 2 -4x +3 = 0$$

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

$$r + s = 4 rs = 3$$

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 = 3

$$(2 - u) (2 + u) = 3$$

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

$$4 - u^2 = 3$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = 3-4 = -1$$

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

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

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

$$r =2 - 1 = 1 s = 2 + 1 = 3$$

Evaluate

$$3\left(k-3\right)\left(k-1\right)$$