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  1. #1
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    Question factor theorem.

    Let f(x)=x^3-8x^2+17x-9. Use the factor theorem to find other solutions to f(x)-f(1)=0, besides x=1. My answer is 2,5 could that be right. thanks for looking.
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by kwtolley
    Let f(x)=x^3-8x^2+17x-9. Use the factor theorem to find other solutions to f(x)-f(1)=0, besides x=1. My answer is 2,5 could that be right. thanks for looking.
    f(x)=x^3-8x^2+17x-9
    f(x)=f(1)=1-8+17-9=1
    f(x)=x^3-8x^2+17x-9=1
    f(x)=x^3-8x^2+17x-10=0
    Yes, 2 and 5 are solutions of this equation besides 1.

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  3. #3
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    Hello, kwtolley!

    Another approach . . .

    Let f(x) \:=\:x^3 - 8x^2 + 17x - 9
    Use the factor theorem to find other solutions to f(x)-f(1)\,=\,0, besides x=1.
    My answers are: 2,\;5.
    We are given: . f(x) - f(1) \:= \:0

    ]. . . . . (x^3 - 8x^2 + 17x - 9) - (1^3 - 8\cdot1^2 + 17\cdot1 - 9) \;= \;0

    . . . . . . . . . (x^3 - 1^3) - 8(x^2 - 1^2) + 17(x - 1) - 9 + 9 \;= \;0

    . . . (x - 1)(x^2 + x + 1) - 8(x - 1)(x + 1) + 17(x - 1) \;= \;0

    . . . . . . . . . . . . . . . (x - 1)(x^2 + x + 1 - 8[x+1] + 17) \;= \;0

    . . . . . . . . . . . . . . . . . . . . . . . . . (x - 1)(x^2 - 7x + 10) \;= \;0

    . . . . . . . . . . . . . . . . . . . . . . . . . (x - 1)(x - 2)(x - 5) \;= \;0


    Therefore, the solutions are: . x \:= \:1,\;2,\;5
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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Soroban
    Hello, kwtolley!

    Another approach . . .


    We are given: . f(x) - f(1) \:= \:0

    ]. . . . . (x^3 - 8x^2 + 17x - 9) - (1^3 - 8\cdot1^2 + 17\cdot1 - 9) \;= \;0

    . . . . . . . . . (x^3 - 1^3) - 8(x^2 - 1^2) + 17(x - 1) - 9 + 9 \;= \;0

    . . . (x - 1)(x^2 + x + 1) - 8(x - 1)(x + 1) + 17(x - 1) \;= \;0

    . . . . . . . . . . . . . . . (x - 1)(x^2 + x + 1 - 8[x+1] + 17) \;= \;0

    . . . . . . . . . . . . . . . . . . . . . . . . . (x - 1)(x^2 - 7x + 10) \;= \;0

    . . . . . . . . . . . . . . . . . . . . . . . . . (x - 1)(x - 2)(x - 5) \;= \;0


    Therefore, the solutions are: . x \:= \:1,\;2,\;5
    Great.
    Alternative approach are always interesting.

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  5. #5
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    Factor theorem

    Thanks to everyone for looking it over with me.
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