Aaron S. answered • 05/29/13

With Aaron's help, math is as easy as pi

**1. **I'm not sure what you meant by that first ^. I'll assume that it was the 9th root of 16, i.e. 16^(1/9) or ^{9}√(16). In that case, we have (notice when the exponents multiply because we are raising numbers to powers and when they add because we are multiplying numbers with the same base):

( 16^(1/9) ) ^5 = 16^(5/9) = ( 2^4 ) ^ (5/9) = 2^(20/9) = 2^(18/9)*2^(2/9) = 2^2*2^(2/9) = 4*(^{9}√(2^{2}))

=4*(^{9}√4)

**2. **10^(3/7) = ^{7}√(10^{3}) = ^{7}√(1000)

**3. **Here, we use the idea that exponents can distribute to terms (being multiplied) in parentheses and that raising a number to a negative exponent is the same as the reciprocal of that number to the power of the negative of that exponent, i.e. that:

x^{-m} = 1/x^{m}

So:

(8x^{3}y^{2})^{-3} = 1/( 8x^{3}y^{2} )^{3} = 1/( 83(x3)3(y2)3 )

= 1/ ( 8^{3}x^{9}y^{6} )

**4**. I assume the problem wanted you to list all possible **rational** roots. Otherwise you couldn't list all possible roots (since there are an infinite amount). Anyway, to do this, you find all factors of the constant term (36) and divide all such factors by all possible factors of the coefficient of the highest degree term (so the coefficient of x^{3}, which is 1 in this case). Then you take plus or minus any of these results. To summarize:

+/- { any factor of 36}/{any factor of 1} = +/- {1,2,3,4,6,9,12,24,36}/{1} = { +/- 1, +/- 2, +/-3, +/-4, +/-6, +/- 9, +/- 12, +/- 24, +/-36 }.

Theoretically, you could test all of these using your favorite method to determine which one(s) are zeros, but given that there are 18 of them, this seems unpleasant. But thankfully I see that we can factor this by grouping, which we always try to do when we have 4 terms. So:

f(x)=x^{3}+4x^{2}+9x+36 = x^{2}*(x+4) + 9*(x+4) = (x+4)*(x^{2}+9)

So x=-4 is the only **real** zero, and I'm not sure whether you know about complex numbers, but x = +/- 3**i** are the remaining **complex** zeros.

Hope I helped.