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EleoNora [17]
3 years ago
5

Create a quadratic polynomial function f(x) and a linear binomial in the form (x − a).

Mathematics
2 answers:
Dima020 [189]3 years ago
7 0
Part 1. In the long division, you find the greatest factor that could divide the dividend. You do this one at a time per term. Then, you find the product of the factor and the divisor, then subtract it from the dividend. The cycle goes on until all the terms are divided:
               x  +  6
            ----------------------------
    x - 9 |  x² - 3x + 6
             - x² - 9x
             -------------------
                       6x + 6
                     - 6x - 54
                     --------------
                           60

There quotient is (x+6) with a quotient of 60.


Part 2.  The solution is already given. Set the binomial to 0, such that x-a = 0 is equal to x=a. Using this a to substitute the x terms in the given function:
f(9) = (9)² - 3(9) + 6
f(9) = 60


Part 3. The steps shown are from the concept of Factor and Remainder Theorem. When you substitute x=a to the function, the answer could determine if x=a is a factor or not. If the answer is zero, then x=a is a factor. If not, the answer represents the remainder.

Therefore, x = 9 is not a factor of the given function. It yields a remainder of 60 which coincides with Part 1.
bazaltina [42]3 years ago
5 0

Answer:

Step-by-step explanation:

Part 1. Show all work using long division to divide your polynomial by the binomial.

                      X + 8

          ----------------------------

X-10 |   x^2 - 4x + 8

           (-x^2 + 10x)

             6x + 8

             (-6x + 8)

              68

X + 8 + 68 / x - 10

Part 2. Show all work to evaluate f(a) using the function you created.

f(x) = x^2 - 4x + 8

f(10) = (10)^2 - 4(10) + 8

f(10) = 68

Part 3. Use complete sentences to explain how the remainder theorem is used to determine whether your linear binomial is a factor of your polynomial function.

When you substitute f(x) to f(a) you would be able to use this to determine if f(x) to f(a) is a factor or not. If the answer is zero then it is not a factor, if the answer is not zero then it will have a remainder.  

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  • Choice B) Billy misidentified the first quartile value
  • Choice C) Billy misidentified the median
  • Choice D) Billy misidentified the third quartile value

The correct box plot is shown below.

=====================================================

Explanation:

The first step when constructing a box plot is to sort the data from smallest to largest. That has already been done for us, so we can move to the next step.

For step 2, we need to find the median. This is the middle-most value. Write out all of the values on one horizontal line. Then cross off the first and last values. You'll end up with a smaller set, with two values tossed out. Repeat this process (cross off the first and last items) and keep doing this until you work your way toward the very middle position. There are 19 numbers in this list, which is an odd number, this means that the middle value is in slot 10 (note how 19/2 = 9.5 which rounds to 10). We have 9 values below the median and 9 values above the median. So 9+1+9 = 19 items total.

You should find that the value 55 is at slot 9, and at the very middle. So 55 is the median. Billy marked on the box plot that the median is at 50, so this is one mistake he made. Choice C is one of the answers

Note: the median is the vertical bar inside the box of the box-and-whisker-plot.

-----------------------------

Once you have the median, break the list into two halves like so:

L = {25,25,30,30,35,35,40,45,50}

U = {60,60,60,70,85,90,95,95,105}

L is the lower set of values smaller than the median, U is the upper set of values larger than the median. Set L and set U have 9 items each. Note the median of 55 is not part of set L nor is it part of set U either.

Repeat the process to find the median, but you'll do so on sets L and U separately. The median of set L is 35, so this is the value of Q1 or the first quartile.

The left edge of the box is the visual location of Q1. We can see that Billy marked Q1 at 30, but it should be at 35. So this is another mistake he made. Choice B is one of the answers

The median of set U is 85. We have four values below 85, and four values above 85 in set U. This makes Q3 to be 85. However, Billy marked the third quartile at 80 instead of 85. This is another mistake. Choice D is one of the answers.

--------------------------------

To summarize his 3 mistakes, they are:

  • He misidentified Q1  (Choice B)
  • He misidentified the median (Choice C)
  • He misidentified Q3 (choice D)

The good news is that he got the minimum correct because 25 is the smallest value. The min value is the left-most tip of the left whisker. The max is also in the correct location since 105 is the largest item. As you can probably guess, the max is located at the right-most tip of the right whisker.

See the attached image below to see how Billy should have drawn the box plot.

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