find all the zeros of x^4+4x^3-2x^2-12x+9. if you know that two of its zeroes are -1 and -1?
Answers
Answer:
(x - 4 + 5i)(x - 4 - 5i) =
x^2 - 4x - 5ix - 4x + 16 + 20i + 5ix - 2i + 25
x^2 - 8x + 41
Dividing the original polynomial by that we get
. . . . . . . . . . . . . . . . . . . . . . x^2 . - 4x - 45
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x^2 - 8x + 41 | x^4 - 12x^3 + 28x^2 + 196x - 1845
. . . . . . . . . . . x^4 - 8x^3 . + 41x^2
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. . . . . . . . . . . . . . . -4x^3 - 13x^2 + 196x
. . . . . . . . . . . . . . . -4x^3 + 32x^2 - 164x
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. . . . . . . . . . . . . . . . . . . . . -45x^2 + 360x - 1845
. . . . . . . . . . . . . . . . . . . . . -45x^2 + 360x - 1845
x^2 - 4x - 45 factors to (x - 9)(x + 5), which leaves 9 and -5 as our other zeros.