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zheka24 [161]
3 years ago
5

Find all roots given that x = 3 + i is a solution. x^4 - 10x^3 + 29x^2 - 10x - 50 = 0

Mathematics
1 answer:
Basile [38]3 years ago
5 0

Answer:

x = 3 + i, x = 3 - i , x = 5 , and x = -1

Step-by-step explanation:

We know that given the polynomial with real coefficients, if x = 3 + i is a root, then x = 3 - i must be another root. These two roots can combine multiplicative factors as shown:

(x - 3 - i) * (x - 3 + i) = (x-3)^2 + 1 = x^2 - 6 x + 9 + 1 = x^2 - 6 x + 10

Then we divide the given polynomial: x^4 - 10x^3 + 29x^2 - 10x - 50  by the quadratic one we just found, and get a perfect division:

{x^4 - 10x^3 + 29x^2 - 10x - 50} / {x^2 - 6 x + 10} = x^2 - 4 x - 5

which can be factored out by grouping as shown below:

x^2 - 4 x - 5 = x^2 - 5 x + x - 5 = x (x - 5) + x - 5 = (x - 5 ) * (x + 1)

then the full factor form of the quartic polynomial given is:

(x - 3 - i) * (x - 3 + i) * (x - 5 ) * (x + 1)

and therefore the associated roots are:

x = 3 + i, x = 3 - i , x = 5 , and x = -1

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