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Math Help - Please Help!!Alegbra

  1. #1
    cathystntn123
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    Please Help!!Alegbra

    Can anyone help me with these three sums!!


    (x+5)(x-4)(x-a)=xto the cube+bx to be square-23x+c=0
    Find the values of a,b, and c


    x to be squard+6x-k is afactor of x to be cube+kx to be squared-x-k!!Find the value of k!!Hence solve the following equation x to cubed+kx to be squared-x-k=0


    (x to be squared+ax+b) is a factor of 2x to cube+ax to be squared+13x+b

    Find the values of a and b

    Thanks alot!!
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  2. #2
    Senior Member DivideBy0's Avatar
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    Here are my ideas for the questions...

    In question 1, expanding we get

    (x^2+x-20)(x-a)=x^3+bx^2-23x+c=0

    x^3-ax^2+x^2-ax-20x+20a=x^3+bx^2-23x+c=0

    x^3+(1-a)x^2-(a+20)x+20a=x^3+bx^2-23x+c=0

    So we have

    1-a=b ...[1]

    a+20=23 ...[2]

    20a=c ...[3]

    From [2],
    a=3
    Then from [1]
    b=-2
    And from [2]
    c=60

    For question 2, x^2+6x-k | x^3+kx^2-x-k

    If you factor x^2+6x-k you get
    x^2+6x-9+9-k=(x+3)^2-(9+k)=(x+3+\sqrt{9+k})(x+3-\sqrt{9+k})
    Then by the factor theorem...

    (-3-\sqrt{9+k})^3+k(-3-\sqrt{9+k})^2-(-3-\sqrt{9+k})-k=0

    and

    (-3+\sqrt{9+k})^3+k(-3+\sqrt{9+k})^2-(-3+\sqrt{9+k})-k=0

    You should be able to solve either for k...

    Haha! My calculator got k = 7 after a very... long time. And I had to resort to iteration. Surely there is a more efficient way to do this.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by cathystntn123 View Post
    x to be squard+6x-k is afactor of x to be cube+kx to be squared-x-k!!Find the value of k!!Hence solve the following equation x to cubed+kx to be squared-x-k=0
    We know that x^2 + 6x - k is a factor of x^3 + kx^2 - x - k.

    Since the quadratic is a factor of the cubic we must have that
    (x^2 + 6x - k)(x - a) = x^3 + kx^2 - x - k

    x^3 + (6 - a)x^2 - (6a + k)x + ak = x^3 + kx^2 - x - k

    Matching coefficients we know that
    6 - a = k
    and
    -6a - k = -1
    and
    ak = -k

    Rewriting the first equation gives:
    k = 6 - a

    So
    -6a - (6 - a) = -1
    and
    a(6 - a) = -(6 - a)

    The first of these says:
    -6a - 6 + a = -1 \implies -5a = 5 \implies a = -1

    The second of these says:
    6a - a^2 = 6a + a \implies a^2 + a = 0 \implies a = -1, 0

    The only solution common to both of these is that a = -1. Thus
    k = 6 - (-1) = 7.

    Thus
    (x^2 + 6x - 7)(x + 1) = x^3 + 7x^2 - x - 7
    which you can verify.

    So now we solve
    x^3 + 7x^2 - x - 7 = 0

    (x + 1)(x^2 + 6x - 7) = 0

    (x + 1)(x + 7)(x - 1) = 0

    So x = -7, -1, 1

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by cathystntn123 View Post
    (x to be squared+ax+b) is a factor of 2x to cube+ax to be squared+13x+b

    Find the values of a and b
    Use the same process as I used before:
    (x^2 + ax + b)(2x + c) = 2x^3 + ax^2 + 13x + b
    (Why do we need the "2" on the linear factor? Because we know the coefficient of the leading term of the cubic is a 2 and this is the only place this factor can come from.)

    This will give you three equations in three unknowns. What I would recommend is that you solve for c first, then put b in terms of a. The result will be a cubic equation for a that factors:
    a^3 + a^2 + 13a + 13 = (a + 1)(a^2 + 13) = 0

    So I get that a = -1, b = 7, and c = 1.

    (Make sure you can fill in the intermediate steps.)

    -Dan
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  5. #5
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by cathystntn123 View Post
    x to be squard+6x-k is afactor of x to be cube+kx to be squared-x-k!!Find the value of k!!Hence solve the following equation x to cubed+kx to be squared-x-k=0

    We have x^3+kx-x-k=(x^2+6x-k)(x+l)
    Expand the right side: x^3+kx-x-k=x^3+(l+6)x^2+(6l-k)x-kl
    Then \left\{\begin{array}{lll}l+6=k\\6l-k=-1\\-kl=-k\end{array}\right.

    From the first two equations we get k=7, \ l=1 wich also satisfy the third equation.

    Now, we have to solve the equation x^3+7x^2=x-7=0
    x^2(x+7)-(x+7)=0\Leftrightarrow (x+7)(x^2-1)=0\Rightarrow x_1=-7,x_2=-1,x_3=1
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  6. #6
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    Hello, Cathy!

    I approached them differently . . . Here's #3.


    3)\;\;x^2 +ax+b .is a factor of . 2x^3+ax^2+13x+b
    Find the values of a and b
    I used long division . . .
    Code:
                                      2x  -  a
                      ---------------------------
          x + ax + b ) 2x +  ax + 13x  +  b
                        2x + 2ax + 2bx
                       -----------------------
                              -ax + (13-2b)x + b
                              -ax  -  ax  -  ab
                              --------------------
                                  (13-2b+a)x + (b+ab)

    Since the division "comes out even", the remainder is zero.

    Hence: . \begin{array}{cccccccc}13-2b+a^2& = & 0& \Rightarrow & a^2+13 & = & 2b & {\color{blue}[1]}\\b+ab & = & 0 & \Rightarrow & b(a+1) & = & 0 & {\color{blue}[2]}\end{array}


    From [2], we have: . b = 0 .or . a = -1

    If b=0, [1] becomes: . a^2+13\:=\:0 . . .which has no real roots

    If a=-1, [1] becomes: . 1+13 = 2b\quad\Rightarrow\quad b = 7


    The problem is: . (2x^3 - x^2 + 13x + 7) \div (x^2-x+7) \:=\:2x + 1

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