Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

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January 2019, Part III

All Questions in Part I are worth 4 credits. Partial credit can be earned.


33.Solve the following system of equations algebraically for all values of a, b, and c.

a + 4b + 6c = 23 a + 2b + c = 2 6b + 2c = a + 14

Answer:
Substitution:
Rewrite the last equation as a = 6b + 2c - 14
Rewrite the first two equations:

6b + 2c - 14 + 4b + 6c = 23
6b + 2c - 14 + 2b + c = 2

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10b + 8c - 14 = 23
8b + 3c - 14 = 2

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(-3)(10b + 8c = 37)
(8)(8b + 3c = 16)

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-30b - 24c = -111
64b + 24c = 128
34b = 17
b = .5

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8(.5) + 3c - 14 = 2
4 + 3c - 14 = 2
3c = 12
c = 4

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a + 2(.5) + (4) = 2
a + 1 + 4 = 2
a = -3

a = -3, b = 0.5, c = 4

Elimination:
Rewrite the last equation as -a + 6b + 2c = 14

a + 4b + 6c = 23
a + 2b + c = 2
2b + 5c = 21

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a + 4b + 6c = 23
-a + 6b + 2c = 14
10b + 8c = 37

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2b + 5c = 21
10b + 8c = 37

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(5)(2b + 5c = 21)
10b + 8c = 37

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10b + 25c = 105
10b + 8c = 37
17c = 68
c = 4

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2b + 5(4) = 21
2b + 20 = 21
2b = 1
b = 0.5

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a + 2(.5) + (4) = 2
a + 1 + 4 = 2
a = -3

a = -3, b = 0.5, c = 4

Checking the work:
a + 4b + 6c = -3 + 4(.5) + 6(4) = -3 + 2 + 24 = 23 (check)
a + 2b + c = 2 = -3 + 2(.5) + 4 = -3 + 1 + 4 = 2 (check)
6b + 2c = a + 14
6(.5) + 2(4) = (-3) + 14
3 + 8 = 11 (check)

34. Given a(x) = x⁴ + 2x³ + 4x - 10 and b(x) = x + 2, determine a(x)/b(x) in the form q(x) + r(x)/b(x).
Is b(x) a factor of a(x)? Explain

Answer:
Divide the polynomial and leave the remainder as a fraction over (x + 2).
I used the Reverse Area Model, which is something I've only recently started doing.
Normally, I would only draw one table, but I expanded it here for clarity.
Some students who understood preferred it to long division. Others prefer to stick with long division.

Start by filling in x⁴ and -10. Label the rows x and +2
To get x⁴, you have to multiply x by x³, so put that on top of that column.
Multiply +2 by x³ and get 2x³.
The next term in a(x) is 2x³, and 2x³ - 2x³ = 0, so right 0 in the top row, second column. 0 divided by x is 0, so 0 gets written on top. And 0 times +2 is 0, so 0 goes on the bottom.
The next term is 4x, and 4x - 0 = 4x, so write 4x in the next column. 4x / x = 4, so write 4 on top. Then 4 times 2 = 8, so put 8 on the bottom.
We didn't want 8. We needed -10. That means that there is a remainder. Subtract -10 - 8 = -18. That remainder is put in a fraction over (x + 2).

b(x) is NOT a factor of a(x) because there is a remainder.
If something is a factor then there can be no remainder.

If people want to see the long division version of this, I can write it on scratch paper and scan it in.



Comments and questions welcome.

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