Polynomials and Triangles

f(x) = 4x³ - 4x² - 11x + 6

Here is our polynomial. A polynomial is an expression with at least two different algebraic terms.

Y-Intercept
When graphing a polynomial, you need to know the y-intercept. This is the point at which a graph crosses the y-axis (the vertical axis). To do this, you look for the constant. In our equation, 6 is the constant term. This means the y-intercept is 6, or (0,6). We need to write it as an ordered pair because it is a point on a graph.

Descartes
The next step in graphing if finding the number of possible real and imaginary solutions of the polynomial. The solutions are where the graph crosses the x-axis (the horizontal axis). We can already tell how many solutions the polynomial will have due to the largest exponent (the degree). The degree of our polynomial is 3. Therefore, we will have three solutions.

To find the possible positive solutions of the expression, we take the signs of all of the terms and see how many times the signs change.
The sign of 4 is +, the sign of -4 is -, the sign of -11 is -, and the sign of 6 is +.
+ - - +
The + changes to - and the - changes to +. This means there are either two or zero positive solutions. There might be zero positive solutions because there could be two imaginary solutions. Imaginary solutions only come in pairs. Therefore, when there are two or more solutions, you can take pairs out and say they could be imaginary.

To find the possible negative solutions of the expression, we take the expression and plug -x in for x. This might change some of the signs. Then we follow the same procedure as the one for positive solutions and count the number of changes.
f(x) = 4x³ - 4x² - 11x + 6 --> f(x) = -4x³ - 4x² + 11x + 6
The sign of -4 is -, the sign of -4 is -, the sign of 11 is +, and the sign of 6 is +.
- - + +
Only the one - changes to a +. This means there is only one possible negative solution.
Here is a chart showing the possible combinations of solutions:
+ | - | i 
2 | 1 | 0
0 | 1 | 2

Rational Root Test
The next step to graphing is to find all of the possible rational roots. In order to do this, we use the rational root test. To do this, we put all of the factors of the constant term (6 in our polynomial) over all of the factors of the leading coefficient (4 in our polynomial). This looks like this:
±1, 2, 3, 6
±1, 2 ,4
Then we divide and get all of the possible rational roots.
±1, 2, 3, 6,¹/₂, ³/₂, ¹/₄, ³/₄
These are our possible rational roots. There is a ± in front of them to show that our solutions could be positive or negative. This is because not all factors are positive.

Synthetic Division
Now, we need to find out which possible solutions are the right ones. We will do this through the use of synthetic division. This involves writing out all of the coefficients of the polynomial in a row, including the ones that don't show up. If there is a missing term, you put zero in for it. This is because a missing term could be 0x or 0. Then you put the possible rational root you want to test to the left of the coefficients. You then drop the leading coefficient down and multiply it by the possible root. Then you add the product to the next coefficient and drop down the sum. You take this sum and multiply it by the possible root and add the product to the next coefficient. You repeat this process until you get to the final coefficient. When you add the last product to the last coefficient, you should get zero. If the sum is zero, the possible root is one of the roots. If the sum isn't zero, then the possible root isn't a solution.
2 | 4 -4 -11 6
         8   8 -6  
     4  4  -3  0

¹/₂| 4  4  -3
          2    3 
      4  6    0

^-3/₂| 4  6
          -6   
       4  0
We now know all three of our solutions. x = 2, ¹/₂, and ^-3/₂
Our factors are (x - 2)(2x - 1)(2x + 3).

This solutions fit both Descartes and the rational root test because there are two positive roots and one negative root. All of the roots are also possible solutions according to the rational root test.

Behavior of the Graph
We know the beginning and ending behavior of the graph will be similar to that of a line because the degree of the polynomial is odd. If the degree was even, it would start and end like a parabola.
The graph will cross the x-axis at every root as opposed to bouncing off of it because each root has an odd multiplicity of one. If they had even multiplicities, the graph would hit the axis and bounce off.

Graph

By looking at the graph, we can see that all of our roots are there and it begins and ends like a line. The y-intercept is also 6. You can check this by plugging the equation into a graphing calculator and looking for the solutions in the table. You will find the roots in the x-value column and zero in the y-value column next to the roots.

Triangles
45-45-90 Triangle
The three angles of this triangle are 45°, 45°, and 90°.
The side lengths that corresponds to the two 45° angles is √2.
The side length that corresponds to the 90° angle (the hypotenuse) is 2.
The sin(45°) = ^√2/₂
The cos(45°) = ^√2/₂
The tan(45°) = 1

30-60-90 Triangle
The three angles of this triangle are 30°, 60°, and 90°.
The side length that corresponds to the 30° angle is 1.
The side length that corresponds to the 60° angle is √3.
The side length that corresponds to the 90° angle is 2.
The sin(30°) = ¹/₂
The cos(30°) = ^√3/₂
The tan(30°) = ^√3/₃

60-30-90 Triangle
The three angles of this triangle are 60°, 30°, and 90°.
The side length that corresponds to the 60° angle is √3.
The side length that corresponds to the 30° angle is 1.
The side length that corresponds to the 90° angle is 2.
The sin(60°) = ^√3/₂
The cos(60°) = ¹/₂
The tan(60°) = √3