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
8 answers
1, -3,3
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 :)
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 :)
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
enough already, you try. Note #7 looks like #1
3. should be 3 + 2 + 2
#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?
#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?
6. So 1?
7. 8
7. 8
I dont understand the answer to two. Can you explain?