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find the value of 'a' if ( x-a) is a factor of x^6-ax^5 x^4-x^3 3x-a 2. Whe

  1. profile image46
    sartaj76posted 7 years ago

    find the value of 'a' if ( x-a) is a factor of x^6-ax^5 x^4-x^3 3x-a 2.      When can i get...

    answer to this question plse ????

  2. scramble profile image63
    scrambleposted 7 years ago

    Did you miss some operators? Otherwise, are there supposed to be some parentheses around the exponents. I'm not sure what the expression is supposed to be.

  3. Uninvited Writer profile image82
    Uninvited Writerposted 7 years ago

    You might find it in the textbook you were probably supposed to study.

  4. Ben Evans profile image71
    Ben Evansposted 7 years ago

    Would this be x^6+ax^5+x^4-x^3+3x-a2

    Assuming this is a binomial expansion ie all terms are a, then
    (-a)^6=-2a or a=-(2)^1/5.

    However, if this is a binomial expansion. then this says 6a^5=3 and that is not true if a =-(2)^1/5

    So lets iterate and let a=1, then assume it is 1 of six possible roots.
    Then we have (-1)^6 -a(-1)^5+(-1)^4-(-1)^3+3(-1) -2a=1+a+1+1-3-2a=0


    -a=0 or -1=0 which means that -1 is not a root and a is therefore not 1.

    Lets test a=-1............so we have (1)^6-a(1)^5+(1)^4-(1)^3+3(1)-2a=
    1-a+1-1+3-2a=0 and 4+3 does not equal 0

    Lets try a=2

    We then have(-2)^6-a(-2)^5+(-2)^4-(-2)^3+3(-2)-2a=64+32a+16-8-6-2a which doesn't equal 0..

    Without going into specifics, letting a equal any integer larger than 2 will not cause a 0.

    I don't believe there is an integer solution for this.  I used to have a calculator that would solve this.  This is an iterative problem and is solved by converging on the answer.

    I don't have the time but would guess a is between 0 and 1.  Just an educated guess. big_smile

    Maybe I don't understand what equation you are using.
    This cant be a binomial expansion and it converges above 2 and less than -2.  This needs to be iterated until you converge on an answer.  Unless someone has a graphing calculator big_smile.