More Algebra 2 problems.

__August 2017, Part I__

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.

*13. Which expression is equivalent to (4x ^{3} + 9x − 5) / (2x − 1) , where x ≠ 1/2 ?
Which equation approximates the amount of Iridium-192 present after t days?
1) 2x^{2} + x + 5
2) 2x^{2} - x + 5
3) 2x^{2} + 11/2 + 1 / (2(2x-1))
4) 2x^{2} - x + 4 + 1 / (2x - 1)
*

**Answer: 1) 2x ^{2} + x + 5**

First of all, "x ≠ 1/2" is specified because the expression would be

*undefined*because the denominator of the fraction would be zero. It isn't needed for anything else.

You could multiply choices 1 and 3 by *2x - 1* to see if you got *4x ^{3} + 9x − 5*. The other two choices would be messier, but you would have your answer before you got to them.

Or you can do the polynomial division:

*14. The inverse of the function f(x) = (x + 1) / (x − 2) is
1) f*

^{-1}= (x + 1)/(x + 2) 2) f

^{-1}= (2x + 1)/(x - 1) 3) f

^{-1}= (x + 1)/(x - 2) 4) f

^{-1}= (x - 1)/(x + 1)

**Answer: 2) f ^{-1} = (2x + 1)/(x - 1)**

Write the function at y = (x + 1) / (x - 2)

Switch the x's and y's: x = (y + 1) / (y - 2)

Solve for

*y*by getting all the y variables on the left side.

Any term without a y in it goes to the right side.

Finally, factor the y from the expression on the left, and then divide.

xy - 2x = y + 1

xy - y - 2x = 1

xy - y = 2x + 1

y(x - 1) = 2x + 1

y = (2x + 1)(x - 1)

*15. Which expression has been rewritten correctly to form a true statement?
1) (x + 2)*

^{2}+ 2(x + 2) − 8 = (x + 6)x 2) x

^{4}+ 4x

^{2}+ 9x

^{2}y

^{2}− 36y

^{2}= (x + 3y)

^{2}(x − 2)

^{2}3) x

^{3}+ 3x

^{2}− 4xy

^{2}− 12y

^{2}= (x - 2y)(x + 3)

^{2}4) (x

^{2}− 4)

^{2}− 5(x

^{2}− 4) − 6 = (x

^{2}- 7)(x

^{2}- 6)

**Answer:4) B and D**

You can multiply these, and in this case, it won't take so long because the first one is correct. Or you can put each side of each equation into the graphing calculator and compare the Tables of Values.

You can also simplify the multiplication by using substitutions.

Suppose you let y = (x + 2), then choice 1 becomes *y ^{2} + 2y - 8 = (y + 4)(y - 2)*

If you factor the polynomial, the factors of -8 that add to +2 are +4 and -2, so this works.

You could have done something for choice 4, letting w = x^{2} - 4, so that the equation could be be rewritten as:

^{2}- 5w - 6 = (w - 3)(w - 2)

However, if you multiply those binomials, you'll get + 6, not - 6.

You can see that in Choice 2, the left side of the equation ends with *- 36y ^{2}*. However, on the right side, 3y

^{2}is 9y

^{2}and (-2)

^{2}is 4, both

*positive*numbers, which is a contradiction.

In Choice 3, you can see on the right side that the

*y*term is

*not*squared, but y

^{2}appears in two terms. This isn't possible.

Comments and questions welcome.

More Algebra 2 problems.

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