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Math Help - Factorising quadratic polynomials

  1. #1
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    Factorising quadratic polynomials

    Hi, im doing a Maths quiz, and am only stuck now on a couple more questions where i was told my anwer is wrong....if anyone could show me the light, it would be appreciated.


    1) Given:

    x^3+1 = ( x+a ) ( x^2+bx+c )

    Find the values of a, b and c. Please enter your answer as a list [in brackets], in the form: [a, b, c]

    2) Factorise the following polynomial completely over the integers.

    3 x^4+18 x^2+15
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  2. #2
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    Quote Originally Posted by Maccabhoy View Post
    Hi, im doing a Maths quiz, and am only stuck now on a couple more questions where i was told my anwer is wrong....if anyone could show me the light, it would be appreciated.


    1) Given:

    x^3+1 = ( x+a ) ( x^2+bx+c )

    Find the values of a, b and c. Please enter your answer as a list [in brackets], in the form: [a, b, c]

    2) Factorise the following polynomial completely over the integers.

    3 x^4+18 x^2+15
    1. Use the factor theorem.

    If P(x) = x^3 + 1 and (x + a) is a factor, then P(-a) = 0.

    So (-a)^3 + 1 = 0

    -a^3 + 1 = 0

    1 = a^3

    a = 1 (I assume you're factorising over the reals...)


    So we have x^3 + 1 = (x + 1)(x^2 + bx + c).

    Now use long division of \frac{x^3 + 1}{x + 1} or the sum of two cubes rule to find what b and c are.
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  3. #3
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    Quote Originally Posted by Maccabhoy View Post
    Hi, im doing a Maths quiz, and am only stuck now on a couple more questions where i was told my anwer is wrong....if anyone could show me the light, it would be appreciated.


    1) Given:

    x^3+1 = ( x+a ) ( x^2+bx+c )

    Find the values of a, b and c. Please enter your answer as a list [in brackets], in the form: [a, b, c]

    2) Factorise the following polynomial completely over the integers.

    3 x^4+18 x^2+15
    3x^4 + 18x^2 + 15

    This is a quadratic if written with X = x^2

     = 3X^2 + 18X + 15

     = 3(X^2 + 6X + 5)

     = 3(X + 5)(X + 1)

     = 3(x^2 + 5)(x^2 + 1)


    These can now only be factorised over Reals, so this is as far as you need to go.
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    Quote Originally Posted by Prove It View Post
    1. Use the factor theorem.

    If P(x) = x^3 + 1 and (x + a) is a factor, then P(-a) = 0.

    So (-a)^3 + 1 = 0

    -a^3 + 1 = 0

    1 = a^3

    a = 1 (I assume you're factorising over the reals...)


    So we have x^3 + 1 = (x + 1)(x^2 + bx + c).

    Now use long division of \frac{x^3 + 1}{x + 1} or the sum of two cubes rule to find what b and c are.
    Thanks for the info. Much appreciated. But sorry for being a bit slow, but not sure what to divide to get b and c
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    Could I also get a hint for this....im actually not just sure, if its my syntax for the quiz program giving me the wrong answer...

    Find the constants a and b such that
    x^2+3 x+a = ( x+3 ) ( x+b )
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  6. #6
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    Quote Originally Posted by Maccabhoy View Post
    Could I also get a hint for this....im actually not just sure, if its my syntax for the quiz program giving me the wrong answer...

    Find the constants a and b such that
    x^2+3 x+a = ( x+3 ) ( x+b )
    Expand the right hand side.

    You should get x^2 + 3x + a = x^2 + (3 + b)x + 3b

    You should be able to find a and b now.


    Also for the question x^3 + 1 = (x + a)(x^2 + bx + c), it might be easiest to use the sum of two squares rule.


    \alpha ^3 + \beta ^3 = (\alpha + \beta)(\alpha ^2 - \alpha \beta + \beta^2)


    In your case, \alpha = x, \beta = 1.

    Have a go.
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