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  1. anonymous
    • one year ago
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    @pooja195

  2. anonymous
    • one year ago
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    @welshfella

  3. welshfella
    • one year ago
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    there are 3 zeros:- -1 duplicity 2 and 1 so in factor form it is (x+1)^2(x-1)

  4. anonymous
    • one year ago
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    I am confused.

  5. welshfella
    • one year ago
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    if you look at the graph you see that at x = -1 the graph touches the x-axis and the it pass through the point where x = 1 these are called the zeros of the function its a cubic function so it has 3 zeros;- 2 at x = -1 and 1 at x = 1 if we write (x + 1)^2(x - 1) = 0 we get x -1 = 0 or (x +1)^2 = 0 giving x = 1 and x = -1 (twice)

  6. welshfella
    • one year ago
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    so iy you expand (x - 1)(x + 1)(x + 1) you'll get the correct function

  7. welshfella
    • one year ago
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    do it by multiplying the last 2 factors first - giving x^2 + 2x + 1 so we have (x - 1)(x^2 + 2x+ 1) = ?

  8. anonymous
    • one year ago
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    \[x^3+x^2−x−1=?\]

  9. welshfella
    • one year ago
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    yep

  10. anonymous
    • one year ago
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    Thank you!

  11. welshfella
    • one year ago
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    welcome

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