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

More Algebra 2 problems.

__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

10b + 8c - 14 = 23

8b + 3c - 14 = 2

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

(8)(8b + 3c = 16)

-30b - 24c = -111

64b + 24c = 128

34b = 17

b = .5

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

4 + 3c - 14 = 2

3c = 12

c = 4

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

a + 4b + 6c = 23

-a + 6b + 2c = 14

10b + 8c = 37

2b + 5c = 21

10b + 8c = 37

(5)(2b + 5c = 21)

10b + 8c = 37

10b + 25c = 105

10b + 8c = 37

17c = 68

c = 4

2b + 5(4) = 21

2b + 20 = 21

2b = 1

b = 0.5

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) = x ^{4} + 2x^{3} + 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 x^{4} and -10. Label the rows x and +2

To get x^{4}, you have to multiply x by x^{3}, so put that on top of that column.

Multiply +2 by x^{3} and get 2x^{3}.

The next term in a(x) is 2x^{3}, and 2x^{3} - 2x^{3} = 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.

More Algebra 2 problems.