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anonymous

  • 5 years ago

f(x)=x^4-6x^3+10x^2+2x-15

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  1. anonymous
    • 5 years ago
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    What do you need to find?

  2. anonymous
    • 5 years ago
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    u need to solve for x

  3. amistre64
    • 5 years ago
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    1,3,5,15 -1,-3,-5,-15 lets try x=3 3 | 1 -6 10 2 -15 0 3 -9 3 15 ------------------ 1 -3 1 5 0 <-remainder is zero, this is a solution (x-3) ------------ lets try x = -1 -1 | 1 -3 1 5 0 -1 4 -5 ------------- 1 -4 5 0 <- remainder is zero, this is a solution. (x+1)(x-3) is what we got so far.. ------------------------------ the rest is a quadratic that can be figured out the normal way:

  4. amistre64
    • 5 years ago
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    x^2 -4x +5 = 0 16 - 4(5) = 16 - 20 = -4... there are no more real solutions the answer is: (x-3) (x+1) (x^2 -4x +5) since the quadratic is zeroless that leaves: x = 3 or x = -1 as our solutions.

  5. anonymous
    • 5 years ago
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    wow thanx that did help

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