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Alex73 [517]
3 years ago
11

HELP HELP HELP HELP HELP HELP​

Mathematics
2 answers:
Zina [86]3 years ago
7 0

Answer:

Just use photo

math and you'll get the answer within seconds!!

Step-by-step explanation:

trust me it really works!!!

NikAS [45]3 years ago
4 0

Problem 1

I'm assuming your teacher means "list all the possible rational roots".

If so, then we use the rational root theorem. This is the idea where we'll divide the factors of the last term over the factors of the first coefficient.

  • Factors of 5:  1, 5
  • Factors of 3:  1, 3

If we divide said factors in the order mentioned, then we get this list of possible positive rational roots:

1/1, 1/3, 5/1, 5/3

That list simplifies to

1, 1/3, 5, 5/3

That takes care of the positive possible rational roots. We double the list to incorporate the negative possible rational roots as well

-1, -1/3, -5, -5/3

<h3>Answer:   1, -1, 1/3, -1/3, 5, -5, 5/3, -5/3</h3>

Note: We can rewrite that list in bold into \pm 1, \pm\frac{1}{3}, \pm 5, \pm \frac{5}{3} to shorten it.

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

Problem 2

We'll follow the same idea as the previous problem

Factors of 36:  1, 2, 3, 4, 6, 9, 12, 18, 36

Factors of 4:  1, 2, 4

To keep track of everything, I made a spreadsheet table. Refer to the diagram below. The factors of 36 are laid across the top while the factors of 4 are along the left side. Each inner cell represents dividing the headers.

For example, 36/4 = 9 is in the bottom right corner. The first table is before the fractions are reduced, and the second table after the fractions are reduced.

As the second table shows, we have a few repeats. The unique items here are:

1,2,3,4,6,9,12,18,36,1/2,3/2,9/2,1/4,3/4,9/4

We'll then mirror this set to include negative values as well to write the full list of all possible rational roots.

<h3>Answer: </h3><h3>1,2,3,4,6,9,12,18,36,1/2,3/2,9/2,1/4,3/4,9/4</h3><h3>-1,-2,-3,-4,-6,-9,-12,-18,-36,-1/2,-3/2,-9/2,-1/4,-3/4,-9/4 </h3>

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

Problem 3

  • Factors of 15: 1, 3, 5, 15
  • Factors of 5:  1, 5

Divide the factors of 15 over the factors of 5 to get

1/1, 1/5, 3/1, 3/5, 5/1, 5/5, 15/1, 15/5

That list reduces to

1, 1/5, 3, 3/5, 5, 1, 15, 3

we can then toss out the duplicates to get

1, 3, 5, 15, 1/5, 3/5

Don't forget to include the negative values as well

<h3>Answer: </h3><h3>1, 3, 5, 15, 1/5, 3/5</h3><h3>-1, -3, -5, -15, -1/5, -3/5</h3>

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

Problem 4

The steps will be effectively the same as before. You can write out the factors like done in problem 3, or make a table like done in problem 2. I'll skip the steps shown here unless you need me to go over them.

<h3>Answer: </h3><h3>1, 3, 9, 1/2, 3/2, 9/2</h3><h3>-1, -3, -9, -1/2, -3/2, -9/2</h3>

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

Problem 5

There's not really much new here, except for the fact that the first term has coefficient of 1. This is a special case where all we have to do is list the factors of the last term -40 to generate all the roots we need. This is because each time we're dividing over +1 or -1 which doesn't really change the overall result (except for the sign part).

Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40

Don't forget the negative values as well

<h3>Answer:</h3><h3>1, 2, 4, 5, 8, 10, 20, 40</h3><h3>-1, -2, -4, -5, -8, -10, -20, -40</h3>

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