1. How many real roots does y=x^3+x^2+9x+9 have?

2. If you know that a 6th degree polynomial has at least 5 complex roots, how many complex roots does it have in total?

3. A polynomial function has 3 real roots and 4 complex roots. What is the
degree of the polynomial?

4. What is the maximum number of complex roots that a polynomial of degree 5 can have?

5. How many roots does a polynomial of degree 9 have

6. How many roots does a linear equation with positive slope have?

7. How many complex roots does y=x^3-2x^2+4x-8 have?

I have done the majority of the assignment, these are just what im stuck on. Showing your work and an explanation would be appreciated. If you can only give an answer that is fine too. Thank you

1. How many real roots does y=x^3+x^2+9x+9 have?

0 = x(x^2+9) + 1(x^2+9)
0 = (x+1)((x^2+9)
0 = (x+1)(x -3i)(x+3i)
well, x = -1 works :)

#4. four, since they come in pairs

#5. nine- see the Fundamental Theorem of Algebra
#6. a linear equation is degree 1, so ...
#7. x^3-2x^2+4x-8 = (x^2+4)(x-2) So, what do you think?

1, -3,3

2. complex roots come in pairs, if you have five then you better have six

3. 3 + 2

enough already, you try. Note #7 looks like #1

3. should be 3 + 2 + 2

6. So 1?

7. 8

I dont understand the answer to two. Can you explain?

1. To determine the number of real roots of the polynomial y = x^3 + x^2 + 9x + 9, you can use the Rule of Signs. It states that the number of positive real roots is equal to the number of sign changes in the coefficients of the polynomial or less by an even number (0, 2, 4, etc.).

In this case, there are two sign changes, from positive to negative between the second and third term and from negative to positive between the third and fourth term. Since this number is less than an even number, the polynomial has exactly 2 or 0 positive real roots.

Next, we check the number of negative real roots by substituting -x for x in the polynomial. So, we evaluate y = (-x)^3 + (-x)^2 + 9(-x) + 9, which simplifies to y = -x^3 + x^2 - 9x + 9. Here, we have one sign change from negative to positive between the first and second term.

Therefore, the polynomial has exactly 1 negative real root.

In conclusion, the polynomial y = x^3 + x^2 + 9x + 9 has exactly 2 positive real roots and 1 negative real root.

2. If a 6th-degree polynomial has at least 5 complex roots, it means that it can have at most 6 - 5 = 1 real root. Since complex roots occur in pairs (conjugate pairs), the remaining 1 root must be a real root.

Therefore, the 6th-degree polynomial has exactly 1 real root and 5 complex roots.

3. The total number of roots of a polynomial is equal to its degree. Since a polynomial with 3 real roots and 4 complex roots has a total of 3 + 4 = 7 roots, the degree of the polynomial is 7.

4. A polynomial of degree 5 can have a maximum of 5 complex roots. This is because the Fundamental Theorem of Algebra states that a polynomial of degree n has exactly n complex roots, counting multiplicities. Since complex roots occur in conjugate pairs, there can be at most n/2 complex roots, which in this case is 5/2 = 2.5. However, since we can only have whole numbers of roots, the maximum number of complex roots for a polynomial of degree 5 is 5.

5. A polynomial of degree 9 can have a maximum of 9 roots, counting multiplicities. This is because the degree of the polynomial represents the highest power of x in the polynomial, and each power corresponds to a possible root. Furthermore, the Fundamental Theorem of Algebra states that a polynomial of degree n has exactly n complex roots, so all 9 roots can potentially be complex.

6. A linear equation in the form y = mx + b, where m is the slope and b is the y-intercept, has exactly one root (one solution). This is because it represents a straight line, and a line intersects the x-axis at only one point unless it is parallel to the x-axis.

7. To determine the number of complex roots of the polynomial y = x^3 - 2x^2 + 4x - 8, we can use the Rule of Signs.

First, we check the number of positive real roots. There are no sign changes between the coefficients of the polynomial, so the polynomial has either 0 or an even number of positive real roots.

Next, we substitute -x for x in the polynomial to check the number of negative real roots. Evaluating y = (-x)^3 - 2(-x)^2 + 4(-x) - 8, we get y = -x^3 - 2x^2 - 4x - 8. Here, we have two sign changes from negative to positive between the first and second term and from negative to positive between the second and third term. So, the polynomial has 2 or 0 negative real roots.

Since the total number of complex roots is equal to the degree of the polynomial minus the number of real roots, the polynomial y = x^3 - 2x^2 + 4x - 8 has 3 - 2 = 1 complex root.