a. What is the maximum # of zeroes g(x) could have?>>>> 4 (all the x's right)?
b. What is the maximum # of turning points g(x) could have?>>>> 10 turning points (is it adding up the exponents?)
c. list all of the potential rational zeroes of g(x)>>>> (Rational zeros: P= X^4 +1 Q= 6= +1, +2, +3, +6 the answer)
d. Is (x – 3) a factor of g(x)? >>> Explain. x3/ x^44x^3+2x^2x+6 someone explain to me how to do long division i kind of forgot.
e. Write g(x) in its fully factored form.>>>> don't know ?
f. Find all zeroes (real and complex) of g(x)>>> i think it might be (x+2)
b. What is the maximum # of turning points g(x) could have?>>>> 10 turning points (is it adding up the exponents?)
c. list all of the potential rational zeroes of g(x)>>>> (Rational zeros: P= X^4 +1 Q= 6= +1, +2, +3, +6 the answer)
d. Is (x – 3) a factor of g(x)? >>> Explain. x3/ x^44x^3+2x^2x+6 someone explain to me how to do long division i kind of forgot.
e. Write g(x) in its fully factored form.>>>> don't know ?
f. Find all zeroes (real and complex) of g(x)>>> i think it might be (x+2)

a. Four because (you're right) the max. power of x.
b. Three, because when you equate g ' (x) = 0, you have at most three roots.
c. Those are the possible factors of +6, (you got that right too) .Because the coefficient of x^4 is 1.
d. Applying the Remainder/Factor theorems, you can get the first factor of the polynomial.i.e. if you get g(a) = 0, then (xa) is a factor.
So (x+3) is a factor of g(x)
Notice that (x+2) is also a factor of g(x)  sub. x=2 into the polynomial.
e. Now you just do a long division of polynomial. Divide g(x) by (x+3)(x+2) or (x^2 + 5x + 6)
You do this in the same way as long division of number , only be careful in handling the powers of
x.
If you do it corectly, the quotient polynomial will be (x^2  x +1)
So g(x) = (x+3)(x+2) (x^2  x + 1)
f. The real zeros of g(x) are 2 and 3
The complex zeros are 0.5 + 0.866 i and 0.5  0.866 i.
Regards
Edem
b. Three, because when you equate g ' (x) = 0, you have at most three roots.
c. Those are the possible factors of +6, (you got that right too) .Because the coefficient of x^4 is 1.
d. Applying the Remainder/Factor theorems, you can get the first factor of the polynomial.i.e. if you get g(a) = 0, then (xa) is a factor.
So (x+3) is a factor of g(x)
Notice that (x+2) is also a factor of g(x)  sub. x=2 into the polynomial.
e. Now you just do a long division of polynomial. Divide g(x) by (x+3)(x+2) or (x^2 + 5x + 6)
You do this in the same way as long division of number , only be careful in handling the powers of
x.
If you do it corectly, the quotient polynomial will be (x^2  x +1)
So g(x) = (x+3)(x+2) (x^2  x + 1)
f. The real zeros of g(x) are 2 and 3
The complex zeros are 0.5 + 0.866 i and 0.5  0.866 i.
Regards
Edem

a. Yes, the highest power of x.
b. No, if the function went up and down 10 times, then the right constant term could make it cross the xaxis 11 times. In a you said this was impossible. For this reason, the answer is 1 less than the max # of zeroes: 3.
c. Yes, apply the rational root theorem.
d. No need to do long division, compute g(3). If g(3) = 0, then (x  3) is a factor, and vice versa.
e. Like in the last problem, to find roots is to find factors. Try your possible rational roots, and see which if any work. Hopefully you remember synthetic division: if it gives you 0 at the end, your number is a root and the rest of the bottom line is the list of coefficients for the polynomial you get for factoring out (x  your root). Don't be afraid to try other methods of factoring such as by grouping.
f. Hopefully, you managed to do this in the last step. Otherwise, you were left with a 2nd degree irreducible polynomial. Use the quadratic formula to get its irrational/complex roots.
Good luck!
b. No, if the function went up and down 10 times, then the right constant term could make it cross the xaxis 11 times. In a you said this was impossible. For this reason, the answer is 1 less than the max # of zeroes: 3.
c. Yes, apply the rational root theorem.
d. No need to do long division, compute g(3). If g(3) = 0, then (x  3) is a factor, and vice versa.
e. Like in the last problem, to find roots is to find factors. Try your possible rational roots, and see which if any work. Hopefully you remember synthetic division: if it gives you 0 at the end, your number is a root and the rest of the bottom line is the list of coefficients for the polynomial you get for factoring out (x  your root). Don't be afraid to try other methods of factoring such as by grouping.
f. Hopefully, you managed to do this in the last step. Otherwise, you were left with a 2nd degree irreducible polynomial. Use the quadratic formula to get its irrational/complex roots.
Good luck!