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Math Help - how to factor four terms

  1. #1
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    how to factor four terms

    Hi all;
    How do you factor 2x^3 + 5x^2 -11x + 4? not quite sure where to start I can see that the last three terms make an ordinary quadratic and I know its in the right format with highest power first and thats about it.

    Thanks.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Look at 2x^3 + 5x^2 -11x + 4=0, and try to guess one solution, it is x=1, now divide 2x^3 + 5x^2 -11x + 4 with (x-1), you will get: 2 x^2+7 x-4
    which is parabola and her solutions are x=-4, x=1/2

    Hence: 2x^3 + 5x^2 -11x + 4=(x-1)(x+4)(2x-1)
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  3. #3
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    The simplest approach is to start by looking for rational "zeroes" of the polynomial- rational number values of x that make 2x^3+ 5x^2- 11x+ 4= 0. The "rational root theorem" says that if \frac{m}{n} is a zero of the polynomial, then the numerator, m, must divide the constant term, 4, and the denominator, n, must divide the leading coefficient, 2. The only factors of 4 are \pm 1, \pm 2, and \pm 4 while the only factors of 2 are \pm 1 and \pm 2.

    That means that the only possible rational roots of the polynomial are \pm 1, pm 2, \pm 4 and \pm \frac{1}{2}. We have to try each of those in the polynomial to see if one makes the polynomial equal to 0. Fortunately, we are successful immediately: [tex]2(1^3)+ 5(1^2)- 11(1)+ 4= 7- 11+ 4= 4- 4= 0.

    That tells us that x- 1 is a factor. Now we can find another factor by long division, or its variant, synthetic division, or by "comparing coefficients": the other factor must be quadratic, of the form ax^2+ bx+ c. Multiplying (x- 1)(ax^2+ bx+ c) we get ax^3+ bx^2+ cx- ax^2- bx- c= ax^3+ (b- a)x^2+ (c- b)x- c= 2x^3+ 5x^2- 11x+ 4 so we must have a= 2 and -c= 4. Then b-a= b- 2= 5 so b= 7. As a check c- b= -4- 7= -11.

    Now we have [tex]2x^3+ 5x^2- 11x+ 4= (x- 1)(2x^2+ 7x- 4). We could continue to look at the possible rational roots \pm 1, \pm 2, \pm 4 and \pm \frac{1}{2} to find factors for 2x^2+ 7x- 4 but since it is quadratic we can use the quadratic formula: if 2x^2+ 7x- 4= 0 then x= \frac{-7\pm\sqrt{7^2- 4(2)(-4)}}{2(2)} = \frac{-7\pm\sqrt{40+ 32}}{4}= \frac{-7\pm\sqrt{72}}{4}= \frac{-7\pm 6\sqrt{2}}{4}.

    Those roots are not rational so if, by "factor", we mean "factor with integer coefficients", we cannot factor it any more: 2x^3+ 5x^2- 11x+ 4= (x- 1)(2x^2+ 7x- 4).

    But in a more general sense, 2x^3+ 5x^2- 11x+ 4= 16(x- 1)\left(x- \frac{7}{4}+\frac{3\sqrt{2}}{2}\right)\left(x-\frac{7}{4}-\frac{3\sqrt{2}}{2}\right)
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    7^2=49
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  5. #5
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    Shortcut method (when you're lucky)

    If the coefficients (including 6) add up to 0, this means that 1 is one of the roots.

    By factoring out (x - 1), then you're just left with solving a quadratic equation.

    Question: wasn't this problem posted before on another thread?
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