Note: I am trying to find a way to display the synthetic division in this, but I want to post these notes already. I will update when I’ve found a way to do that.
Topics:
• review of homework problems on roots and factors
• Review of two methods for solving a quadratic equation (other than factoring and using the zero product property, which was mentioned last time), with a quick reminder that we define 
The methods:
If the quadratic equation does not have a 1st degree (linear) term, it can be solved using the square root property:
Examples:
use the square root property:
use the square root property:
Any quadratic equation can be solved by putting it into the form
and using the quadratic formula
Even though this could always be used to solve a quadratic equation, it is usually better to use factoring or the square root property when you can do so. But when all else fails you can use the quadratic formula, and you need to memorize it – something I don’t say very often!
• Important theorems about roots and factors (see the notes below)
• Using all of this to find all of the roots of a polynomial, and/or a complete factorization of the polynomial over the complex numbers.
The important theorems about roots and factors: (don’t forget we had some other important theorems about roots, factors, and remainders last time)
The Fundamental Theorem of Algebra:
If
is a polynomial of degree
, then
In this theorem, it is important to remember that the complex numbers include the real numbers. The theorem is saying that every non-constant polynomial has at least one root, but you have to allow for the possibility that the root is not a real number. See the second example above where we used the square root property and found two imaginary roots.
A consequence of the Fundamental Theorem of Algebra:
The polynomial
will have exactly
, if we also count multiple roots according to their multiplicities.
Remember that a multiple root is a root whose factor appears more than once in the factorization of the polynomial.
If the polynomial
has roots
, then
where
is the leading coefficient of
See the examples below.
There is one more theorem, which I will mention after the examples.
Examples:
Find all the roots of the polynomial, and use them to find a compete factorization of the polynomial.
[10.7 (a)]
Looking at this polynomial, we see that its degree is 3 so we are expecting it to have 3 roots. Also, there is a common factor of 2 in all the terms, so we will factor that out first:
Now we can try to guess the roots by looking at the graph of
[I’ll fill this in later]
So we get the complete factorization (don’t forget the 2 which we factored out at the start!):
[10.7(b)]
We see that this polynomial has degree 4, so we are expecting 4 roots. Also the leading coefficient is 1, so there will not be an additional numerical factor as there was in the previous example.
By looking at the graph, it appears that there are roots at -5, -2, 4, and 6. We can verify this by synthetic division:
[I will fill this in later]
So we have a complete factorization:
[10.7(c)]
We see that this polynomial has degree 3, so we are expecting 3 roots. Also, the leading coefficient is 1.
By looking at the graph (or just guessing), it appears that there is a root at -1, but no other real number roots. We can use synthetic division to verify this and find out what to do next:
[I will fill this in later]
The quotient after dividing by x+1 is 
Notice that these are complex conjugates – they are the same except that the imaginary parts have opposite signs. This is an instance of the last theorem for today, which I will give below.
So we have a complete factorization
![p(x) = (x+1)\left[x-\left(\frac{1}{2} \pm \frac{\sqrt{3}}{2}i\right)\right]\left[x-\left(\frac{1}{2} \pm \frac{\sqrt{3}}{2}i\right)\right]](https://s0.wp.com/latex.php?latex=p%28x%29+%3D+%28x%2B1%29%5Cleft%5Bx-%5Cleft%28%5Cfrac%7B1%7D%7B2%7D+%5Cpm+%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7Di%5Cright%29%5Cright%5D%5Cleft%5Bx-%5Cleft%28%5Cfrac%7B1%7D%7B2%7D+%5Cpm+%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7Di%5Cright%29%5Cright%5D&bg=ffffff&fg=000000&s=0 "p(x) = (x+1)\left[x-\left(\frac{1}{2} \pm \frac{\sqrt{3}}{2}i\right)\right]\left[x-\left(\frac{1}{2} \pm \frac{\sqrt{3}}{2}i\right)\right]")
Here is the final theorem of the day:
If the polynomial
Homework: please note that the WeBWorK does not cover everything! There are other problems from the textbook you need to do.
• Review the examples and theorems discussed in class. Make especially sure that you understand the following:
The relationship between a root of a polynomial and the factor of the polynomial that the root corresponds to
How to verify a root of a polynomial (by substitution, or by division)
How to use the square root property correctly, and why it’s wrong to “square root both sides”
How to use the quadratic formula and simplify your results
• Do the following problems from the textbook:
In Session 10, Exercises 10.3(a-c) and 10.4(a-c and f-h). We will discuss the graphs in 10.4 next time. Students may put their solutions to these problems on the board at the start of class next time: please make sure that you show us how you verified your roots! (It’s not enough to just say what you read off the graph! You must check that they actually are roots, either by substitution as the textbook does, or by synthetic or long division.)
• Do the WeBWorK: due by Sunday evening, but do not wait until the last minute! And don’t forget to make use of Piazza to discuss!
• Information and resources for next time will be posted soon. Please look for this and take a look at the resources before we meet.
• There will be a quiz next time: the topic will be using one root to find a complete factorization of a polynomial
Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!
and a root 
, from previously using factoring and the zero product property to solve a quadratic equation. For example, to solve 
, we can factor the left-hand side:
or 
or 
corresponds to the root 5, 
corresponds to the root -2.
.
by 
, what would the remainder be? It is not practical to do this division (there are 99 “missing terms” that you would have to replace with 0’s), but we can use the remainder theorem with 
, and in this case 
, so the remainder would be
?
and 
.]
and 
.]
.
, in other words, if the value of the function is the same at x as it is at -x. Then the graph will have mirror symmetry around the y-axis. [Think of the graph of 
]
, in other words, if the value of the function is the opposite when we change the sign of x. Then the graph will be symmetric by rotation of 180 degrees around the origin. [Think of the graph of 
]
, we make a new function 
. My Desmos graph uses the basic quadratic function 
, so we are considering the graphs of functions of the form 
. By using desmos to play around with different values for c, we found that this moves the graph upward or downward by the amount c, depending on whether c is positive or negative. 
. By using desmos to play around with different values for c, we found that this moves the graph to the left or to the right by the amount c, depending on whether c is positive or negative: it moves to the left if c is positive, and to the right if c is negative. 
, and we will consider the graph of 
for various values of a which are positive. Notice that the original function has a y-intercept at (o,1) and also that its graph passes through the points (-1, 0) and (1, 2). Pay attention to what happens to those points when we make changes. By using desmos to play around with different values for a which were positive, we found that this transformation stretches or compresses the graph toward the x-axis, depending on whether a is greater than 1 or a is between 0 and 1 . 
for 
in the formula for the function. Again we are starting with the graph of 
, and we will consider the graph of 
for various values of a which are positive. Notice again that the original function has a y-intercept at (o,1) and also that its graph passes through the points (-1, 0) and (1, 2). Pay attention to what happens to those points when we make changes. By using desmos to play around with different values for a which were positive, we found that this transformation stretches or compresses the graph toward the y-axis, depending on whether a is greater than 1 or a is between 0 and 1 . 
of the equation of a parabola can be explained using these transformations on the basic parabola 
.
by transforming the basic graph 
. Then we looked at the graph to answer the question posed by the inequality: for which values of x is 
? From the graph we can see why we only needed to test one x-value in each interval cut off by the solutions to the corresponding equation 
, because in each interval the graph always lies either over the line y=2, or under the line y=2: the graph does not jump from over to under in the middle of those intervals! This only happens because the graph of our function is continuous (no breaks or jumps or gaps). We will return to this next time and in the future.
