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shaqadry

  • 2 years ago

Consider the polynomial P(x) = ax^4 - x^3 - 14x^2 + bx - 6 where a and b are constant. If one of the zeros of P(x) is 2 and the remainder is -36 when P(x) is devided by (x + 1). Find the values of a and b. How do I solve this?

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  1. RadEn
    • 2 years ago
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    it means P(2) = 0 and P(-1) = -36 now, aplly of them to P(x)

  2. shaqadry
    • 2 years ago
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    okay got the asnwers. thank you very much!

  3. RadEn
    • 2 years ago
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    a=6, b=-13, right ?

  4. shaqadry
    • 2 years ago
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    wait no i got a=2 b=19 :-/

  5. shaqadry
    • 2 years ago
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    can you show me your calculation? i want to see where i did wrong

  6. RadEn
    • 2 years ago
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    well, look this: P(x) = ax^4 - x^3 - 14x^2 + bx - 6 P(2) = 0, it means a(2)^4 - (2)^3 - 14(2)^2 + b(2) - 6 = 0 16a - 8 - 56 + 2b - 6 = 0 16a + 2b - 70 = 0 16a + 2b = 70 8a + b = 35 .......... (1st equation) then P(-1) = -36, means a(-1)^4 - (-1)^3 - 14(-1)^2 + b(-1) - 6 = 0 a + 1 - 14 - b - 6 = 0 a - b - 19 = 0 a - b = 19 ............ (2nd equation) solve for x and y, by using elimination and substitution method (1st) + (2nd), gives 9a = 54 (cancel out for b) a = 6 now, substitute a=6 to one of equation, let the (2nd eq) a-b=19 6 - b = 19 b = 6 - 19 = -13

  7. shaqadry
    • 2 years ago
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    why is the second equation is equal to 0 and not -36 since its the remainder?

  8. RadEn
    • 2 years ago
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    i used remainder theorem, if P(x) divided by (x+a) so its remainder is P(-a)

  9. shaqadry
    • 2 years ago
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    shouldnt it be a(-1)^4 - (-1)^3 - 14(-1)^2 + b(-1) - 6 = -36 ?

  10. RadEn
    • 2 years ago
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    opsss.. u are correct, sorry my bad because copy and paste from the1st equ... yeah, it should a(-1)^4 - (-1)^3 - 14(-1)^2 + b(-1) - 6 = -36 sorry, ur answer is right and mine was wrong

  11. RadEn
    • 2 years ago
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    dont angry to me, yea :)

  12. shaqadry
    • 2 years ago
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    but when i cant seem to factorise it completely :-/ there's a remainder. is it correct? or is my calculation wrong? bcs the 2nd questions ask me to factorise it completely,

  13. shaqadry
    • 2 years ago
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    not at all :)

  14. RadEn
    • 2 years ago
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    factor ? i think we didnt see and neednt factor there but elimination and substitution method

  15. shaqadry
    • 2 years ago
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    i dont understand

  16. RadEn
    • 2 years ago
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    oh sorry, miss comunication (because my english soo bad :(, sorry again ) if u have pounded the value of a and b, just plug of them to P(x), from here we can find the factors of P(x)

  17. shaqadry
    • 2 years ago
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    so what i should do is devide 2x^4 - x^3 - 14x^2 + 19x - 6 with x^2 - x -2 rite? i tried calculating it but i end up with remainders. what should i do?

  18. RadEn
    • 2 years ago
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    how come x^2-x-2 ? what is it

  19. shaqadry
    • 2 years ago
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    (x-2) and (x+1) is two of the four factors right? (x+1) is given. and x=2 is also given therefore x-2=0 correct? or am i wrong?

  20. RadEn
    • 2 years ago
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    if the problem told u one of the zeros or one of the factor, yea right but wrong for (x+1), it's not a root because it has remainder

  21. shaqadry
    • 2 years ago
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    oh right i see...so i should try to find the root by myself?

  22. RadEn
    • 2 years ago
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    yea, u might use trial and error or use synthetic division to find the other factors

  23. shaqadry
    • 2 years ago
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    okay i get. thank you so much, again! you've been a great help :)

  24. RadEn
    • 2 years ago
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    i have asked to wolfram, it said to me here : http://www.wolframalpha.com/input/?i=2x^4-x^3-14x^2%2B19x-6

  25. RadEn
    • 2 years ago
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    you're welcome :)

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