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Math Help - Synthetic Substitution

  1. #1
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    Question Synthetic Substitution

    I missed alot of school so i dont know how to do this

    use synthetic substitution to find f(-2). show all work

    f(x) = -9x^5 + 5x^3 - 2x + 1

    f(x) = 3x^3 - 4x + 6

    given a polynomial and one of its factors, find the remaining factors of the polynomial. some factors may not be binomials.

    2x^3 + 15x^2 - 14x -48 ; x - 2

    x^3 + 6x^2 - x - 30 ; x + 5

    Find values of k so that each remainder is 5

    (2x^2 - 8x + k) / (x - 7)

    (x^4 + kx^3 - 7x^2 + 8x + 25) / (x - 2)
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  2. #2
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    Hello thisishard

    Welcome to Math Help Forum!
    Quote Originally Posted by thisishard View Post
    I missed alot of school so i dont know how to do this

    use synthetic substitution to find f(-2). show all work

    f(x) = -9x^5 + 5x^3 - 2x + 1

    f(x) = 3x^3 - 4x + 6
    Have you tried Googling 'synthetic substitution'? I found this very clear explanation very easily: http://hanlonmath.com/pdfFiles/29.Sy...bstitution.pdf

    given a polynomial and one of its factors, find the remaining factors of the polynomial. some factors may not be binomials.

    2x^3 + 15x^2 - 14x -48 ; x - 2

    x^3 + 6x^2 - x - 30 ; x + 5
    And I Googled 'synthetic division' and came up with: Synthetic Division
    Find values of k so that each remainder is 5

    (2x^2 - 8x + k) / (x - 7)

    (x^4 + kx^3 - 7x^2 + 8x + 25) / (x - 2)
    Use the remainder theorem. In the first of these f(7) = 5, and in the second f(2) = 5. So work out what each of these expressions is in terms of k, and then solve each equation to find k.

    Grandad
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  3. #3
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    I still don't understand!!!!
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  4. #4
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    Synthetic Substitution and Division

    Hello again thisishard!

    I can't explain synthetic substitution and division any more clearly than in the articles that I directed you to.

    It is difficult to set the working out using LaTeX so I've done it by hand, and attached image files.

    use synthetic substitution to find f(-2). show all work

    f(x) = -9x^5 + 5x^3 - 2x + 1
    The working for this is shown in the attachment I've called SyntheticSubstitution.jpg. The answer: f(-2) = 253.
    f(x) = 3x^3 - 4x + 6
    I hope you can do this one now.
    given a polynomial and one of its factors, find the remaining factors of the polynomial. some factors may not be binomials.

    2x^3 + 15x^2 - 14x -48 ; x - 2
    I've done the division using ordinary long division, and also synthetic long division. The working is shown in Division.jpg. The other factor, therefore, is 2x^2+19x+24, which factorises further into (2x+3)(x+8).
    x^3 + 6x^2 - x - 30 ; x + 5
    I'll leave this one for you to try.
    Find values of k so that each remainder is 5

    (2x^2 - 8x + k) / (x - 7)
    When x = 7,\; 2x^2-8x+k = 154+k. So, using the Remainder Theorem, if the remainder is 5:
    154+k=5

    \Rightarrow k = -149
    (x^4 + kx^3 - 7x^2 + 8x + 25) / (x - 2)
    Again, I'll leave you to have a go at this.

    If you want us to confirm that you have the remaining answers right, get back to us.

    Grandad
    Attached Thumbnails Attached Thumbnails Synthetic Substitution-syntheticsubstitution.jpg   Synthetic Substitution-division.jpg  
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  5. #5
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    given a polynomial and one of its factors, find the remaining factors of the polynomial. some factors may not be binomials.

    x^3 + 6x^2 - x - 30 ; x + 5

    this would be
    -5_| 1 6 -1 -30
    -5 -5 30
    ------------------
    1 1 -6 0

    would that be the whole answer because it doesn't seem like it
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  6. #6
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    Hello thisishard
    Quote Originally Posted by thisishard View Post
    given a polynomial and one of its factors, find the remaining factors of the polynomial. some factors may not be binomials.

    x^3 + 6x^2 - x - 30 ; x + 5

    this would be
    -5_| 1 6 -1 -30
    -5 -5 30
    ------------------
    1 1 -6 0

    would that be the whole answer because it doesn't seem like it
    Your working shows that when x^3+6x^2-x-30 is divided by (x+5) the quotient is x^2+x-6 (that's the meaning of the 1\; 1\; {-6}) and the remainder is 0.

    So you have shown that
    x^3+6x^2-x-30=(x+5)(x^2+x-6)
    Now you can factorise the second term further into (x+3)(x-2). So (x+3) and (x-2) are the remaining factors of the original polynomial.

    Grandad
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  7. #7
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    Huh??? second term?
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  8. #8
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    I learn best if i see some one do a problem completly so will you please finish it its the only part im stuck on

    I already did the rest thanks to you
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  9. #9
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    Hello thisishard

    I don't really see where the problem is. We have shown that
    x^3+6x^2-x-30=(x+5)(x^2+x-6)
    The second term on the right-hand-side - that is, (x^2 +x-6) - can be factorised:
    x^2+x-6 = (x+3)(x-2)
    So (x+3) and (x-2) are the remaining factors of x^3+6x^2-x-30.

    Grandad
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    where did the right side come from?
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  11. #11
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    Hello thisishard
    Quote Originally Posted by thisishard View Post
    where did the right side come from?
    The synthetic division you did, that gave the result 1 1 -6 0, means that when x^3+6x^2-x-30 is divided by (x+5) the answer (the quotient) is \color{red}(1)\color{black}x^2+\color{red}(1)\colo  r{black}x\color{red}-6 and the remainder is \color{red}0. In other words:
    \frac{x^3+6x^2-x-30}{x+5} =x^2+x-6
    So:
    x^3+6x^2-x-30=(x+5)(x^2+x-6)
    OK?

    Grandad
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