Anyone how to do question 17 ii?

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  • ?
    Lv 7
    5 months ago
    Favorite Answer

    4x³ - 6x² + 1 = (x - 2)Q(x) + ax - 3

     

    Subtract ax - 3 from both sides

    4x³ - 6x² - ax + 1 =  (x - 2)Q(x)

     

    Divide both sides by (x - 2) but the final term must be -2 since -2 * -2 = 4

    4x² + 2x - 2 remainder(-a + 6x) = Q(x)

     

    Subtract the remainder from both sides

    4x² + 2x - 2 = Q(x) + (a - 6)x

     

    since the coefficient of x in the original term is zero

    a - 6 = 0

    a = 6 <–––––

     

    Q(x) = 4x² + 2x - 2 <––––––

    --------------------

     

    Check

    (x - 2)Q(x) + ax - 3 = 4x³ - 6x² + 1

     

    (x - 2)(4x² + 2x - 2) + 6x - 3 = 4x³ - 6x² + 1

     

    (4x³ + 2x² - 2x) + (- 8x² -4x + 4) + 6x - 3 = 4x³ - 6x² + 1

     

    4x³ + (2x² - 8x²) + (-2x - 4x + 6x) + (4 - 3) = 4x³ - 6x² + 1

     

    4x³ - 6x² + 0x + 1 = 4x³ - 6x² + 1 ✔︎

  • 5 months ago

    Thats the whole question. The value of A is 6 and thats all I have. So I dont really know how to move on to II frok this

  • Alan
    Lv 7
    5 months ago

    I would want to see the 

    rest of the question before answering that. 

    What you have by itself doesn't make any sense.  

    P(x) - 1 is divided by 3x^2 + 5      

    then what ? 

    with a remainder of what?

    with no remainder?

     

    Explain whether 3x^2 + 5  is a factor  

    P(x)-1  ?

    No one can based on your information alone 

    [P(x)- 1]/  [3x^2 + 5]  =  (who knows ? with a remainder?  

    without a remainder? on  a Tuesday?

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