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Math Help - Factor theorem!

  1. #1
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    Factor theorem!

    ok, im stuck with this question:

    Find the values of p and q if 4x^2-4x-3 is a factor of the expression 8x^4 + px^3 + qx^2 + x + 3. Hence, factorise the expression completely

    It is based on factor theorem...if u've ever heard of it...

    The answer is p= -12, q= -6 and the factors are (x-1)(2x-3)(2x+1)^2
    I got the answers from the answer key behind the book

    Its due tomorrow so can someone please help meee?
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  2. #2
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    I am getting -14 and -4.

    8x^{4}-12x^{3}-6x^{2}+x+3 has 2 real solutions and two non-real. Are you sure you typed it correctly. One sign is wrong then it's all wrong.

    If you factor 4x^{2}-4x-3=(2x-3)(2x+1)

    As you can see, two soluitons are 3/2 and -1/2

    Plug these into the quartic and you get:

    27p+18q+360=0

    -p+2q+24=0

    Solve these two for p and q and you get q=-14 and p=-4

    Sub those in and you get:

    8x^{4}-4x^{3}-14x^{2}+x+3=(x+1)(2x-3)(2x-1)(2x+1)

    The factorization you said was in the book when expanded gives:

    8x^{4}-12x^{3}-6x^{2}+\boxed{7}x+3
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  3. #3
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    wow...I get you. Thanks!!! I checked the book and it said 8x^4+px^3+qx^2+x+3
    so i guess they have printing error...

    but i dont get how you expand it though...i got different answers...

    I got 8x^4-4x^3-10x^2-x-3 when i expanded  (2x^2-x-3)(4x^2-1^2) since (x+1)(2x-3) = 2x^2-x-3 and (2x-1)(2x+1)= (4x^2-1^2)
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  4. #4
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    It's not -10x^2, it's -14x^2. Be careful. One number, one sign and it's all kaput.

    Also, you have negatives where plus should be. You are using different expressions now. Which is which?.

    I thought it was 4x^{2}-4x-3. You now have 2x^{2}-x-3
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  5. #5
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    hmmm...for that, i got  (4x^2-1^2)(2x^2-x-3) = 8x^4-4x^3-12x^2+2x^2-x-3 = 8x^4-4x^3-10x^2-x-3

    and for (2x^2+x-1)(4x^2-4x-3), i got = 8x^4-4x^3-14x^2+x+3 O_O
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