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Please, give me a hint to find a polynomial of 4th order. I only know 2 roots: 1-square...
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We'll recall the fact that complex roots and irrational roots come in pairs. Therefore, if x1 = 1 - sqrt2 is a root of the polynomial, then x2 = 1 + sqrt 2 is also a root of the polynomial. Also, if x3 = 2 - i is another root of polynomial, then it's conjugate, x4 = 2 + i, is a root of the polynomial, too.
We'll write the 4th order polynomial as a product of linear factors:
P(x) = (x - x1)(x - x2)(x - x3)(x - x4)
P(x) = (x - 1 + sqrt2)(x - 1 - sqrt2)(x - 2 + i)(x - 2 - i)
P(x) = [(x-1)^2 - 2]*[(x-2)^2 - i^2]
But i^2 = -1. Expanding the binomials inside brackets, we'll get:
P(x) = (x^2 - 2x + 1 - 2)*(x^2 - 4x + 4 + 1)
Combining like terms inside brackets, we'll get:
P(x) = (x^2 - 2x - 1)*(x^2 - 4x + 5)
We'll remove the brackets:
P(x) = x^4 - 4x^3 + 5x^2 - 2x^3 + 8x^2 - 10x - x^2 + 4x - 5
Combining like terms, we'll get:
P(x) = x^4 - 6x^3 + 12x^2 - 6x - 5
The 4th order requested polynomial is: P(x) = x^4 - 6x^3 + 12x^2 - 6x - 5.
Posted by giorgiana1976 on May 11, 2011 at 2:07 PM (Answer #1)
The complex roots of a polynomial are always found as conjugate pairs. For the root 2 - i, the complex conjugate is 2 + i.
The polynomial has the roots 2 - i and 2 + i
Irrational roots are also found as pairs. For the root 1 - sqrt 2, the root that forms the pair is 1 + sqrt 2
The required polynomial has the roots: 2 - i and 2 + i and 1 - sqrt 2 and 1 + sqrt 2. As you only require the method to solve for the polynomial and not the polynomial itself , I am only giving you a hint.
The 4th order polynomial that has to be found is:
(x - (2 - i))(x - (2 + i))(x - (1- sqrt 2))(x - (1 + sqrt 2))
Posted by justaguide on May 11, 2011 at 2:46 PM (Answer #2)
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