Today in class we worked with synthetic substituation and the rational root therom. To start out we went over the worksheet.

Both long division and the synthetic substitution methods can be used to find points on the graph. In this example we are trying to find what f(2) is.

Long Division:

3(x^3)+4(x^2)+11x-10 divided by (x-2)

-3(x^3)+6(x^2)

                   10(x^2)+11x

                   -10(x^2)+20x

                                        31x-10

                                       -31x+62

                                                  Remainder 52

The remainder is the y coordinate, so f(2) = 52 so on the graph (2,52)

Synthetic Substitution: You simply multiply add, multiply add. So 2 x 0 is 0 and 3+0 is 3. Then 2 x 3 is 6 and 4+6 is 10. Just continue this until you’ve used all the numbers.

2]       3             4         11        -10

                          6          20       62

   —————————————-

           3             10       31        [52]

The number in brackets is the y coordinate so (2,52).

Next we are told to find all the zeros. They give us 2/3 and when you use synthetic substitution you can simplify to  a quadratic formula.

2/3]          3               4             11            -10

                                     2               4              10

———————————————–

                  3               6                15             [0]

Then the coefficients at the end can be used to write a quadratic equation. So 3(x^2)+6x+15. Unfortunately, the other two zeros (we know there are 3 zeros because it is a 3rd degree polynomial) are not real numbers…they are IMAGINARY! so to find them we use the quadratic formula

-b  plus/minus  sq rt (b^2-4ac)/2a

so just plug in the numbers from out quadratic equation.

-6  plus/minus  sqrt (6^2-4(3)(15))/2(3)

-6  plus/minus  sqrt (-144)/6

Now since there is a negative under the sqrt sign we would normally stop but instead we are using imaginary numbers. In imaginary land the sqrt(-1)=i. So in out equation we can change the sqrt of (-144_ to the sqrt (144) x sqrt(-1). That simplifies to 12 times i. Then the 6′s cancel out to -1 and we are left with out two other zeros.

-1  plus/minus 12i

So our zeros are (2/3), (-1+12i), and (-1-12i). Keep in mind that imaginary roots always come in pairs (because of the plus/minus).

Next we moved on to the Rational Root Therom. We factor, the numbers we are mainly concerned with is the leading coefficient and the constant term so the first and the last numbers. To find all the rational zeros in a function you can use this to narrow the search.

plus/minus {factors of constant term/factors of leading coefficient}

Lets consider (x^3)-(8x^2)+17x-10

the factors of the constant (10) are 1, 2, 5, 10

the factor of the leading (10) is 1

so plus/minus {1,2,5,10/1} so the possible zeros are +1,-1,+2,-2,+5,-5,+10,-10

If there was more than one factor on the bottom you would have to write down all possible combinations, including fractions. Then to find which are actually zeros you can use the synthetic substitution method to find one and the quadratic formula to find the others

5]      1    -8     17     -10

                 5     -15     10

—————————

        1     -3        2      [0]

Then use the quadratic formula on (x^2)-3x+2

The Zeros are 1, 2, 5

We just tried a couple more examples and that was it. Tomorrow hopefully we will learn a short cut to the rational root therom. The next scribe is whoever hasn’t gone yet.

Today we started the class with a brief review of what we learned on monday.  After this we looked at the problem we were assigned for home work which was to find the zeros of the equation x^3-13x^2+59x-87 knowing that one of the zeros was x=3.  To do this we found our coeffients for the equation which were 1,-13,59,and -87.  Then by using the format we learned monday we were able to solve this problem. 

3/  1  -13   59   -87

      _     3    -30    87

_________________

      1    -10    29      0

Knowing the coefficents for the equation for finding the zeros we can make the equation x^2-10x+29=0.  Then using the quadratic formula( )  we find that x= 10+-squareroot(-16)/2.  Since we cant take the square root of negative our calculators gave us an error when we plugged the coefficients into our quadractic programs.  We all immediatly assumed that the answer to the problem was no solution but then Mr. B told us the story of John and Betty.

http://mathforum.org/johnandbetty/

This story explained the usefulness of the varible i. i=squareroot(-1).  In our problem because we have the squareroot of -16 we can make this 4i.  This is because the square root of -16 = the square root of -1 or i times the square root of 16.  Then since the square root of 16 is 4 we end up with x=5+2i, x=5-2i and x=3 as our zeros.  So in conclusion the most important thing we learned today was in other countries instead of playing like kids here, kids like to create new math problems.  To solve them they simply ask “Can we do that?” and answer with “Why not?”

today in mr. bieniek’s 3rd hour class, se learned some wonderful things (as always). for example, we talked more about synthetic division.

Synthetic Division:

f(x) = x³-13x²+59x+87

we use the coefficients and set it up like this:

3          1          -13           59          -87         

             ?              3             -30         87                      

             1            -10        29           0     

 the one goes down, which is then multiplied by the 3 and put under the - 13. the 3 is added to the -13 to get -10, which is put under the 3. the 10  is  multiplied by 3 to get -30, which goes under the 59. the 59 and -30 added together to get 29, which goes under the -30. 29 is multiplied by 3 to  get  87, which is added to -87 to get 0. this makes x=3, a zero in this equation. this is now a quadratic equation, so we can use the quadratic formula to find the other zeros. we know that there are 2 other zeros, because the highest degree of the polynomial is three. pretty easy to remember, right?

x=-b±?b²-4ac

             2a

now we plug in the values and simplify

x=10±?100-116

            2

x=10±?-16

           2

from what we know of fractions, ?-16 = 4i. so we plug that in instead.

x=(10?4i)/2

simplified, x=5?2i. so the 3 zeros for this equation are x=3, x=5+2i, and x=5-2i.

today in math we learned a couple more things like how apparently math is all imaginary!!! so remember, whenever you are frustrated with math, just think “it’s all pretend. it’s not real!”

So today we recapped a process many of us may not have used since grade school, long division.  But, as was discovered early on in the class this long division was not the simple process we remember.  Instead it was used to simplify complex polynomials into quadratic equations so we could then factor them out to find the remaining “zeros.”  I use the term zeros lightly because we were not only trying to find the zeros but we were finding the f(x)=9 or -9 or 5 or -2.  Virtually we were drawing a horizontal line on the graph of a polynomial and trying to find what the coordinates of the points that it intersected on the graph.

An example of a polynomial would be something like:

f(x) = 1x³ – 9x² – x + 105

-with 1 being the leading coefficient and 3 being the degree

or the equation can even be simplified to  f(x) = x((1x – 9) x – 1) + 105 by factoring out the x’s so that you have an easier equation to use without a calculator.

Next we would be asked to simplify these equations down to the “zeros” based on one of them being given to us.  In this problem we’ll say that x=6 and we’ll plug it into the problems.

f(6) = 1(6³) – 9(6²) – 6 + 105

simplified: f(6) = -9

or

f(6) = 6 ((6 – 9) 6 – 1) + 105

simplified: f(6) = -9

This -9 is the horizontal line that crosses specific points on the graph one of them being 6.  So now what we can say we are finding is actually not the “zeros”  but the “negative 9′s.”

But, 6 is only one of the 3 points f(x) = -9 crosses and now we must find the other two using the first equation.  This is when we utilize the more complicated but still easy long division.

———-–—---x² – 3x  – 19

(x-6)  / x³  - 9x²- x +  105  

————x³ - 6x²

—————–-3x² -x

—————-– 3x²+18x

———————-—-19x+105

————————--19x +114

————————————--9

Sorry if the math is a little confusing but it is basically a long division problem with the answer at the top and the remainder at the bottom.  Well look at that! -9 is our remainder and we now have a quadratic equation as our answer which can be easily factored for us to find the remaining “zeros.”  -9 being our remainder is not a coincidence because:

* If f(x) is a polynomial, then f(c) equals the remainder when f(x) is divided by (x-c).

and it is possible to not have a remainder meaning that one of he “zeros” would be zero!

So the next step would be to factor the quadratic.

x² – 3x – 19

(x – 6.10977…) ( x + 3.10977…)

remaining “zeros” : 6.10977… , -3.10977…

So there.  Hopefully that was understandable.  But, there has to be an easier way to get the quadratic equation….well there is.  It is called Synthetic Substitution/Division.

6 /  1   -9    -1      105

—- ~    6    -18    -114

—–1   -3    -19     -9

So time to use your imagination…there are many arrows in this diagramthat represent multiplication:

6 / -> 1(bottom left corner) -> 6 (underneath -9):         6 x 1 = 6

6 / -> -3 -> -18:        6 x -3 =18

6 / -> -19 -> -114:        6 x -19 = -114

and the columns of numbers are exactly what they look like, subtraction.

1 – 0 = 1

-9 – 6 = -3

-1 – (-18) = -19

105 – (-114) = -9

Simplified:

6 /  1   -9    -1      105

—- ~    6    -18    -114

—–1   -3    -19     -9

So what this table basically gives us are the coefficients for the quadratic    (1 & 3 & -19) and our f(x)= -9.

So that wraps up the lesson on using long and synthetic division/substitution to find the factors of a polynomial function.  Hopefully this helps anyone who may not have gotten it the first time.

<3 Jenny

Check us out at bienprecal. Nice! On your mobile device you will see a new Twitter link in the header at http://prec.alcul.us