Pt1420 Unit 1 Essay
If the equation is 7th degree then it has 7 roots.
Those roots can be complex or real.
Complex roots always come in pairs, so if it has one, then it has 2, the other one being the conjugate of the first one. This in other words, if one complex root is a + bi, then the other complex root is a – bi.
If at least one root were complex, then we would have a minimum of 2 complex roots with a maximum of 5 real roots.
The equation can have at most 6 complex roots (3 pairs) so the minimum number of real roots is equal to 1.
2. What is f(x) = x8 - 1 divided by x - 1?
I can solve this given equation through synthetic division or by long division since the given divisor is in the form of (x-a), synthetic division is the way to solve this. …show more content…
My Given equation x^8 -1 becomes: x^8 +0x^7+0x^6 +0x^5 +0x^4 +0x^3 +0x^2 +0x^1 -1
This result of either synthetic division or long division will get me a result of →
This is my answer →X^7 +x^6 +x^5 +x^4 +x^3 +x^2 +x -1
I can Prove this → if the solution correct by multiplying it by (x-1) and I will get x^8-1 back again.
For example my long division would like this :- x^8 +0x^7 +0x^6 +0x^5 +0x^4 +0x^3 + 0x^2 +0x^1 +0x^1 – 1 divided by (x-1) → it look like This 1 + 0 + 0+ 0 + 0 + 0 + 0 + 0 – 1 synthetically divided by 1.
3. What are the zeroes of f(x) = x3 + 5x2 - 7x +