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Math Help - Delta and factorization of polynomials?

  1. #1
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    Delta and factorization of polynomials?

    I am wondering if Delta (b square minus 4ac) has any relation or can determine if a quadratic polynomial (with a degree of 2) can be factorized?

    To put it simply, let say, a quadratic expression 30x^2 + 37x + 84 , it can either be factorized into (x-a)(x-b) where a and b are nice integers by cross method or cannot be factorized into integral a and b. So besides trial and error test with cross method, is there any quick way or using delta can determine whether it can be factorized into integral a and b in form of (x-a)(x-b)?

    thanks all
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    Quote Originally Posted by kenny1999 View Post
    I am wondering if Delta (b square minus 4ac) has any relation or can determine if a quadratic polynomial (with a degree of 2) can be factorized?

    To put it simply, let say, a quadratic expression 30x^2 + 37x + 84 , it can either be factorized into (x-a)(x-b) where a and b are nice integers by cross method or cannot be factorized into integral a and b. So besides trial and error test with cross method, is there any quick way or using delta can determine whether it can be factorized into integral a and b in form of (x-a)(x-b)?

    thanks all
    Recall that if a Quadratic can be factorised, then it will have solutions.

    Therefore, if there are not any solutions, it can not be factorised.


    i.e. If \Delta < 0 the function can not be factorised.


    Also, recall that you only get one solution if the factorised form of the Quadratic is a Perfect Square.

    Therefore if \Delta = 0 the function will be factorised into a perfect square.
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    If \Delta\geq 0 then you can find the roots:

    x_1=\frac{-b-\sqrt{\Delta}}{2a}, \ x_2=\frac{-b+\sqrt{\Delta}}{2a}

    Then you can factorize the quadratic using the roots in the following identity:

    ax^2+bx+c=a(x-x_1)(x-x_2)

    (The identity holds even the roots are complex numbers.)
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    Quote Originally Posted by red_dog View Post
    If \Delta\geq 0 then you can find the roots:

    x_1=\frac{-b-\sqrt{\Delta}}{2a}, \ x_2=\frac{-b+\sqrt{\Delta}}{2a}

    Then you can factorize the quadratic using the roots in the following identity:

    ax^2+bx+c=a(x-x_1)(x-x_2)

    (The identity holds even the roots are complex numbers.)
    True. I was assuming that the OP was factorising over the Real Number field.
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    yes, thank all for the answer

    but I want to know.....

    Usually by using the general cross method to factorize a quadratic polynomials x^2+bx+c, we will either be able to get (x-p)(x-q) where
    p and q are integral values, or unable to factorize. But whether it can be factorized is highly determined by trial-and-error in cross method.

    My question is (was) , is there any other way, rather than trial and error in the cross-method to determine whether
    ax^2 + bx + c can be factorized into the form of (x-p)(x-q), where p and q are INTEGERS

    I understand delta, if delta is less than 0, then the surd will be undefined in real region, so there is no value of x

    and I also understand that if delta = 0 or bigger than 0 there will be real value of x, but my question is how to know if the roots of a
    quadratic equation is an integer or not (not integers). Thanks all in advance
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    Quote Originally Posted by kenny1999 View Post
    yes, thank all for the answer

    but I want to know.....

    Usually by using the general cross method to factorize a quadratic polynomials x^2+bx+c, we will either be able to get (x-p)(x-q) where
    p and q are integral values, or unable to factorize. But whether it can be factorized is highly determined by trial-and-error in cross method.

    My question is (was) , is there any other way, rather than trial and error in the cross-method to determine whether
    ax^2 + bx + c can be factorized into the form of (x-p)(x-q), where p and q are INTEGERS

    I understand delta, if delta is less than 0, then the surd will be undefined in real region, so there is no value of x

    and I also understand that if delta = 0 or bigger than 0 there will be real value of x, but my question is how to know if the roots of a
    quadratic equation is an integer or not (not integers). Thanks all in advance
    If \Delta is a perfect square, then you will be able to factorise so that p and q are integers.
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    Assume ax^2 + bx + c = a(x-p)(x-q)
    = ax^2 -a(p+q) x + apq

    Equate the coefficients
    p + q = -b/a
    pq = c/a
    check if b/a is integer or not and c/a is integer or not..

    if they are not.. then p and q are not integer..

    if they are integer..see if you can solve them for integers...

    probably this is what you were looking for.
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    Quote Originally Posted by Prove It View Post
    If \Delta is a perfect square, then you will be able to factorise so that p and q are integers.
    why you can conclude that?

    delta is only PART of the story

    it has denominator 2a

    an integral value divides an integral value 2a not necessarily an integers.

    This is my question.
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    Quote Originally Posted by kenny1999 View Post
    why you can conclude that?

    delta is only PART of the story

    it has denominator 2a

    an integral value divides an integral value 2a not necessarily an integers.

    This is my question.
    I should have said rational numbers.

    Having said that...

    If \Delta is a perfect square, then you have rational roots of the form x = \frac{a}{b}.

    Therefore bx - a is a factor.

    So you can still have an integer value of your p and q. You will also have an integer multiple of x.
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    Quote Originally Posted by tutor.mathitutor View Post
    Assume ax^2 + bx + c = a(x-p)(x-q)
    = ax^2 -a(p+q) x + apq

    Equate the coefficients
    p + q = -b/a
    pq = c/a
    check if b/a is integer or not and c/a is integer or not..

    if they are not.. then p and q are not integer..

    if they are integer..see if you can solve them for integers...

    probably this is what you were looking for.
    i think you understand what i am asking for, thanks

    but how can you conclude that if p+q is an integer
    and p x q is an integer, then p and q itself must also be an integer?
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  11. #11
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    Quote Originally Posted by Prove It View Post
    I should have said rational numbers.

    Having said that...

    If \Delta is a perfect square, then you have rational roots of the form x = \frac{a}{b}.

    Therefore bx - a is a factor.

    So you can still have an integer value of your [tex]p[\math] and q. You will also have an integer multiple of x.
    can i conclude that

    if delta < 0

    the quadratic expression cannot be factorised

    while if delta = 0

    it can be factorised into (x-p)^2 where p is rational

    and while if delta > 0 and is a perfect square

    then it can be factorised into (x-p)(x-q) where p and q are rational

    and while if delta > 0 and not a perfect square

    then it can also be factorised into (x-p)(x-q) but p and q are irrational

    ??
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  12. #12
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    Quote Originally Posted by kenny1999 View Post
    can i conclude that

    if delta < 0

    the quadratic expression cannot be factorised

    while if delta = 0

    it can be factorised into (x-p)^2 where p is rational

    and while if delta > 0 and is a perfect square

    then it can be factorised into (x-p)(x-q) where p and q are rational

    and while if delta > 0 and not a perfect square

    then it can also be factorised into (x-p)(x-q) but p and q are irrational

    ??
    It would actually be

    If \Delta > 0 and a perfect square, then it can be factorised into (mx - p)(nx - q) where m, n, p, q are rational numbers.

    If \Delta > 0 and not a perfect square, then it can be factorised into (mx - p)(nx - q) where m, n, p, q are irrational numbers.
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    Quote Originally Posted by Prove It View Post
    I should have said rational numbers.

    Having said that...

    If \Delta is a perfect square, then you have rational roots of the form x = \frac{a}{b}.

    Therefore bx - a is a factor.

    So you can still have an integer value of your p and q. You will also have an integer multiple of x.
    you are right, but I have a problem

    I am talking about factorizing a quadratic "expression", i.e. ax^2+bx+c

    but not quadratic eqauation i.e. ax^2+bx+c=0

    equation can multiply the denominator to both sides
    but expression cannot.
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  14. #14
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    Quote Originally Posted by Prove It View Post
    It would actually be

    If \Delta > 0 and a perfect square, then it can be factorised into (mx - p)(nx - q) where m, n, p, q are rational numbers.

    If \Delta > 0 and not a perfect square, then it can be factorised into (mx - p)(nx - q) where m, n, p, q are irrational numbers.
    ok
    but why?
    can you prove to me? thanks!
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