*I'll be reviewing a*

**New York State Regents Exam**Question every day from now until the Regents exams begin next month. At least, that is the plan.### June 2014, Questions 30 and 31

**30.** *The function f has a domain of {1, 3, 5, 7} and a range of {2, 4, 6}. Could f be represented by {(1, 2), (3, 4), (5, 6), (7, 2)}? Justify your answer.*

I'm still not happy about the wording of this one. I've been told that I'm "splitting hairs", but, in my opinion, the question is asked backward.

The question is asking if the set given *could be* function *f*. In other words, is it a function and does it have the correct domain and range?

Let's answer that question.

The set of ordered pairs is a function because no x value is mapped to more than one y value. The domain (the x values) are {1, 3, 5, 7} as in *f*. The range (the y values) are {2, 4, 6} which matches the range of *f*. So, yes, it could represent f.

Remember that "yes" or "no" without an explanation is worth ZERO points.

**31.** *Factor the expression x ^{4} + 6x^{2} - 7 completely.*

Usually, a “factor completely” question has a Greatest Common Factor (GCF) component to it. This one doesn't, but it does have an extra step that isn't obvious at first.

If you are thrown off by the exponents being 4 and 2, instead of 2 and 1, that's okay. They will act pretty much the same way.

x^{4} + 6x^{2} - 7 factors into **(x ^{2} + 7) (x^{2} - 1)**. You now have two binomials, each with an x

^{2}term. If you look closely, you can see that we aren't finished. We can factor further.

Using the **Difference of Squares Rule** (x^{2} - 1) can be factored into **(x – 1)(x + 1)**.

This makes the final answer:

**(x**

^{2}+ 7) (x – 1)(x + 1)Note that (x

^{2}+ 7) has no real roots and cannot be factored further.

Any questions?

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

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