Question

Use the Fundamental Theorem of Algebra to find the number of complex roots of the polynomial, including any repeated roots.

x3+3x+17

Answer 1

@wolfe8

Answer 2

Theres no multiple choice?? lmao

Answer 3

I don't have multiple choices

Answer 4

Oh ill help you on the next one i know the formula i just dont know how to do the problem lol

Answer 5

Ok I'm not home and I used a software for this so this is what I have for now until I get home and do it myself:

Change coordinates by substituting x = lambda/y+y, where lambda is a constant value that will be determined later.
If x = lambda/y+y then y = 1/2 (sqrt(x^2-4 lambda)+x) which will be used during back substitution:
17+3 (y+lambda/y)+(y+lambda/y)^3 = 0
Transform the rational equation into a polynomial equation.
Multiply both sides by y^3 and collect in terms of y:
17 y^3+y^6+lambda^3+y^4 (3+3 lambda)+y^2 (3 lambda+3 lambda^2) = 0
Find an appropriate value for lambda in order to make the coefficients of y^2 and y^4 both zero.
Substitute lambda = -1 and then z = y^3, yielding a quadratic equation in the variable z:
-1+17 z+z^2 = 0
Solve for z.
Find the positive solution to the quadratic equation:
z = 1/2 (-17+sqrt(293))
Perform back substitution on z = 1/2 (sqrt(293)-17).
Substitute back for z = y^3:
y^3 = 1/2 (-17+sqrt(293))
Take the cube root of both sides.
Taking cube roots gives (1/2 (sqrt(293)-17))^(1/3) times the third roots of unity:
y = (1/2 (-17+sqrt(293)))^(1/3) or y = -(1/2 (17-sqrt(293)))^(1/3) or y = (-1)^(2/3) (1/2 (-17+sqrt(293)))^(1/3)
Look at the first equation: Perform back substitution on y = (1/2 (sqrt(293)-17))^(1/3).
Substitute back for y = sqrt(x^2+4)/2+x/2:
x/2+sqrt(4+x^2)/2 = (1/2 (-17+sqrt(293)))^(1/3) or y = -(1/2 (17-sqrt(293)))^(1/3) or y = (-1)^(2/3) (1/2 (-17+sqrt(293)))^(1/3)
Solve for x.
Solve the radical equation in x:
x = -(2/(-17+sqrt(293)))^(1/3)+(1/2 (-17+sqrt(293)))^(1/3) or y = -(1/2 (17-sqrt(293)))^(1/3) or y = (-1)^(2/3) (1/2 (-17+sqrt(293)))^(1/3)
Look at the second equation: Perform back substitution on y = -(1/2 (17-sqrt(293)))^(1/3).
Substitute back for y = sqrt(x^2+4)/2+x/2:
x = -(2/(-17+sqrt(293)))^(1/3)+(1/2 (-17+sqrt(293)))^(1/3) or x/2+sqrt(4+x^2)/2 = -(1/2 (17-sqrt(293)))^(1/3) or y = (-1)^(2/3) (1/2 (-17+sqrt(293)))^(1/3)
Solve for x.
Solve the radical equation in x:
x = -(2/(-17+sqrt(293)))^(1/3)+(1/2 (-17+sqrt(293)))^(1/3) or x = -(1/2 (17-sqrt(293)))^(1/3)-(-1)^(2/3) (2/(-17+sqrt(293)))^(1/3) or y = (-1)^(2/3) (1/2 (-17+sqrt(293)))^(1/3)
Look at the third equation: Perform back substitution on y = (-1)^(2/3) (1/2 (sqrt(293)-17))^(1/3).
Substitute back for y = sqrt(x^2+4)/2+x/2:
x = -(2/(-17+sqrt(293)))^(1/3)+(1/2 (-17+sqrt(293)))^(1/3) or x = -(1/2 (17-sqrt(293)))^(1/3)-(-1)^(2/3) (2/(-17+sqrt(293)))^(1/3) or x/2+sqrt(4+x^2)/2 = (-1)^(2/3) (1/2 (-17+sqrt(293)))^(1/3)
Solve for x.
Solve the radical equation in x:
Answer: |
| x = -(2/(-17+sqrt(293)))^(1/3)+(1/2 (-17+sqrt(293)))^(1/3) or x = -(1/2 (17-sqrt(293)))^(1/3)-(-1)^(2/3) (2/(-17+sqrt(293)))^(1/3) or x = (-2/(-17+sqrt(293)))^(1/3)+(-1)^(2/3) (1/2 (-17+sqrt(293)))^(1/3)

Answer 6

Great! Thank you so much!

Answer 7

This is insane lol

Answer 8

Hahaha

Answer 9

The problem asks you to find the NUMBER of complex roots, does it not?

Answer 10

The equation \[x^3+3x+17=0\] has how many roots in total?

Answer 11

One of the corollaries of the FToA is that every polynomial in a single variable x with real coefficients (no \(i\) or \(\sqrt{-1}\) anywhere) can be written as a product of a constant, factors such as \((x+a)\) where \(a\) is real, and factors such as \((x^2+ax + b)\) where \(a,b\) are real and \(a^2-4b <0\), which means \((x^2+ax+b)\) has only complex roots.

Answer 12

If you can figure out how many roots the equation has (which you can, just by looking at it and observing the highest power of \(x\)) and recalling that a polynomial in one variable with only real coefficients must have an even number of complex roots (because they come in conjugate pairs), can you conclude anything about this problem?