Question 218142: what are the zeros for f(x)=x^4-x^3+3x^2-9x-54
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! f(x)= x^4 -x^3 +3x^2 -9x -54
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zeros =p/q= (factors of constant)/ (factors of leading coeff)= +/- 54/1
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or zeroes are factors of 54,,,+/- 1,54,2,27,3,18,6,9
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If we plot using a graphing calc, we find zero's at x=-2,,,x=+3
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using synthetic division to confirm,, and find imaginary zeros
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.try x=3
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.,,,,,3|,,,1,,,-1,,,3,,,-9,,,-54
.,,,,,,,,,,,,,,3,,,,6,,,,27,,+54
.,,,,,,,------------------------
.,,,,,,,,,,1,,,2,,,,9,,,18,,,,0,,,ok,,,,now try x=-2
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,,,,,,-2|,,1,,,2,,,,9,,,18
.,,,,,,,,,,,,,,-2,,,0,,,-18
,,,,,,,,---------------------
,,,,,,,,,,,1,,,0,,,,9,,,,,0,,,,,,,,ok
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Remainder is ,,1x^2 +0x +9
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.This factors to (x+3i)(x-3i),,,,,imaginary zeros,,,(use quadratic formula if you don't remember)
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Set of zeros is,,,,x=3,,,x=-2,,,x=+3i,,,,x=-3i
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this can be confirmed by multiplying above to get original
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check ,,,,(x^2 +9) (x+2)(x-3) = (x^3 +2x^2 +9x +18)(x-3) = x^4 -x^3 +3x^2 =9x -54,,,,,ok
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