*The following problems were taken from the*

**ALGEBRA II (Common Core)**Regents Exam given on Friday, January 27, 2017.Previous problems can be found here

### Part 1

**3. When factored completely, m^{5} + m^{3} - 6^{m} is equivalent to
**

(4) m(m^{2} + 3)(m^{2} - 2)

Start with: m^{5} + m^{3} - 6^{m}

Each term has *m*, factor it: (m)(m^{4} + m^{2} - 6)

Factor into two binomials: What are two factors of -6 that add up to +1: +3 and -2

So it factors into __m(m ^{2} + 3)(m^{2} - 2)__.

**4. If sin^{2}(32°) + cos^{2}(M) = 1, then M equals
**

(1) 32°.

The rule is: sin^{2}x + cos^{2}x = 1.

In this case x = 32°, so M must be 32° as well.

Continue to the next problems

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