Synthetic division, quadratic formula and grouping












1












$begingroup$


I'm learning mathematics by myself. I've searched about this topic but I'm not able to get an answer. Also, my language is not English.



For example, let's factor this polynomial



$6x^4+13x^3+6x^2-3x-2$



I use synthetic division and I get $(x+1)^2$ as a factor.



So $6x^2+x-2$ is what's left.



Now I have two options: quadratic formula or grouping, if I'm right.



If I do grouping, the steps are



$6x^2+4x-3x-2$



$2x(3x+2)-1(3x+2)$



And I get $(2x-1)(3x+2)$, which it seems correct.



But if I decide to use quadratic formula I get $x_1=frac{1}{2},x_2=-frac{2}{3}$ as zeroes, and $(x-frac{1}{2})(x+frac{2}{3})$ as factors, which is wrong.



Obviously I'm failing to understand something but I can't find what. Is there any step I am missing? Why the results are so different? Shouldn't be the same?






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    The zeros don't determine the factors, since there are still leading coefficients to take care of. In particular, $(x - 1/2)(x + 2/3)$ is just off by a factor of $6$.
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 23:58










  • $begingroup$
    Yes, I've noticed that. But I though that the results shoud be just the same. What's the difference?
    $endgroup$
    – A.G.
    Dec 4 '18 at 0:06










  • $begingroup$
    Knowing the zeros doesn't determine the scaling: Both $x-1$ and $2(x - 1)$ are lines with roots at $x = 1$, right?
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 0:07










  • $begingroup$
    @A.G. dress few graphs with the same zeros, e.g. $(x-1)(x+2),;2(x-1)(x+2),;-3(x-1)(x+2)$ and observe.
    $endgroup$
    – user376343
    Dec 4 '18 at 10:55












  • $begingroup$
    @T.Bongers Now I get it. I didn't know about the leading coefficient. Thank you so much.
    $endgroup$
    – A.G.
    Dec 4 '18 at 16:06
















1












$begingroup$


I'm learning mathematics by myself. I've searched about this topic but I'm not able to get an answer. Also, my language is not English.



For example, let's factor this polynomial



$6x^4+13x^3+6x^2-3x-2$



I use synthetic division and I get $(x+1)^2$ as a factor.



So $6x^2+x-2$ is what's left.



Now I have two options: quadratic formula or grouping, if I'm right.



If I do grouping, the steps are



$6x^2+4x-3x-2$



$2x(3x+2)-1(3x+2)$



And I get $(2x-1)(3x+2)$, which it seems correct.



But if I decide to use quadratic formula I get $x_1=frac{1}{2},x_2=-frac{2}{3}$ as zeroes, and $(x-frac{1}{2})(x+frac{2}{3})$ as factors, which is wrong.



Obviously I'm failing to understand something but I can't find what. Is there any step I am missing? Why the results are so different? Shouldn't be the same?






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    The zeros don't determine the factors, since there are still leading coefficients to take care of. In particular, $(x - 1/2)(x + 2/3)$ is just off by a factor of $6$.
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 23:58










  • $begingroup$
    Yes, I've noticed that. But I though that the results shoud be just the same. What's the difference?
    $endgroup$
    – A.G.
    Dec 4 '18 at 0:06










  • $begingroup$
    Knowing the zeros doesn't determine the scaling: Both $x-1$ and $2(x - 1)$ are lines with roots at $x = 1$, right?
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 0:07










  • $begingroup$
    @A.G. dress few graphs with the same zeros, e.g. $(x-1)(x+2),;2(x-1)(x+2),;-3(x-1)(x+2)$ and observe.
    $endgroup$
    – user376343
    Dec 4 '18 at 10:55












  • $begingroup$
    @T.Bongers Now I get it. I didn't know about the leading coefficient. Thank you so much.
    $endgroup$
    – A.G.
    Dec 4 '18 at 16:06














1












1








1





$begingroup$


I'm learning mathematics by myself. I've searched about this topic but I'm not able to get an answer. Also, my language is not English.



For example, let's factor this polynomial



$6x^4+13x^3+6x^2-3x-2$



I use synthetic division and I get $(x+1)^2$ as a factor.



So $6x^2+x-2$ is what's left.



Now I have two options: quadratic formula or grouping, if I'm right.



If I do grouping, the steps are



$6x^2+4x-3x-2$



$2x(3x+2)-1(3x+2)$



And I get $(2x-1)(3x+2)$, which it seems correct.



But if I decide to use quadratic formula I get $x_1=frac{1}{2},x_2=-frac{2}{3}$ as zeroes, and $(x-frac{1}{2})(x+frac{2}{3})$ as factors, which is wrong.



Obviously I'm failing to understand something but I can't find what. Is there any step I am missing? Why the results are so different? Shouldn't be the same?






algebra-precalculus







share|cite|improve this question











$endgroup$




I'm learning mathematics by myself. I've searched about this topic but I'm not able to get an answer. Also, my language is not English.



For example, let's factor this polynomial



$6x^4+13x^3+6x^2-3x-2$



I use synthetic division and I get $(x+1)^2$ as a factor.



So $6x^2+x-2$ is what's left.



Now I have two options: quadratic formula or grouping, if I'm right.



If I do grouping, the steps are



$6x^2+4x-3x-2$



$2x(3x+2)-1(3x+2)$



And I get $(2x-1)(3x+2)$, which it seems correct.



But if I decide to use quadratic formula I get $x_1=frac{1}{2},x_2=-frac{2}{3}$ as zeroes, and $(x-frac{1}{2})(x+frac{2}{3})$ as factors, which is wrong.



Obviously I'm failing to understand something but I can't find what. Is there any step I am missing? Why the results are so different? Shouldn't be the same?






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 0:00







A.G.

















asked Dec 3 '18 at 23:55









A.G.A.G.

63




63












  • $begingroup$
    The zeros don't determine the factors, since there are still leading coefficients to take care of. In particular, $(x - 1/2)(x + 2/3)$ is just off by a factor of $6$.
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 23:58










  • $begingroup$
    Yes, I've noticed that. But I though that the results shoud be just the same. What's the difference?
    $endgroup$
    – A.G.
    Dec 4 '18 at 0:06










  • $begingroup$
    Knowing the zeros doesn't determine the scaling: Both $x-1$ and $2(x - 1)$ are lines with roots at $x = 1$, right?
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 0:07










  • $begingroup$
    @A.G. dress few graphs with the same zeros, e.g. $(x-1)(x+2),;2(x-1)(x+2),;-3(x-1)(x+2)$ and observe.
    $endgroup$
    – user376343
    Dec 4 '18 at 10:55












  • $begingroup$
    @T.Bongers Now I get it. I didn't know about the leading coefficient. Thank you so much.
    $endgroup$
    – A.G.
    Dec 4 '18 at 16:06


















  • $begingroup$
    The zeros don't determine the factors, since there are still leading coefficients to take care of. In particular, $(x - 1/2)(x + 2/3)$ is just off by a factor of $6$.
    $endgroup$
    – T. Bongers
    Dec 3 '18 at 23:58










  • $begingroup$
    Yes, I've noticed that. But I though that the results shoud be just the same. What's the difference?
    $endgroup$
    – A.G.
    Dec 4 '18 at 0:06










  • $begingroup$
    Knowing the zeros doesn't determine the scaling: Both $x-1$ and $2(x - 1)$ are lines with roots at $x = 1$, right?
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 0:07










  • $begingroup$
    @A.G. dress few graphs with the same zeros, e.g. $(x-1)(x+2),;2(x-1)(x+2),;-3(x-1)(x+2)$ and observe.
    $endgroup$
    – user376343
    Dec 4 '18 at 10:55












  • $begingroup$
    @T.Bongers Now I get it. I didn't know about the leading coefficient. Thank you so much.
    $endgroup$
    – A.G.
    Dec 4 '18 at 16:06
















$begingroup$
The zeros don't determine the factors, since there are still leading coefficients to take care of. In particular, $(x - 1/2)(x + 2/3)$ is just off by a factor of $6$.
$endgroup$
– T. Bongers
Dec 3 '18 at 23:58




$begingroup$
The zeros don't determine the factors, since there are still leading coefficients to take care of. In particular, $(x - 1/2)(x + 2/3)$ is just off by a factor of $6$.
$endgroup$
– T. Bongers
Dec 3 '18 at 23:58












$begingroup$
Yes, I've noticed that. But I though that the results shoud be just the same. What's the difference?
$endgroup$
– A.G.
Dec 4 '18 at 0:06




$begingroup$
Yes, I've noticed that. But I though that the results shoud be just the same. What's the difference?
$endgroup$
– A.G.
Dec 4 '18 at 0:06












$begingroup$
Knowing the zeros doesn't determine the scaling: Both $x-1$ and $2(x - 1)$ are lines with roots at $x = 1$, right?
$endgroup$
– T. Bongers
Dec 4 '18 at 0:07




$begingroup$
Knowing the zeros doesn't determine the scaling: Both $x-1$ and $2(x - 1)$ are lines with roots at $x = 1$, right?
$endgroup$
– T. Bongers
Dec 4 '18 at 0:07












$begingroup$
@A.G. dress few graphs with the same zeros, e.g. $(x-1)(x+2),;2(x-1)(x+2),;-3(x-1)(x+2)$ and observe.
$endgroup$
– user376343
Dec 4 '18 at 10:55






$begingroup$
@A.G. dress few graphs with the same zeros, e.g. $(x-1)(x+2),;2(x-1)(x+2),;-3(x-1)(x+2)$ and observe.
$endgroup$
– user376343
Dec 4 '18 at 10:55














$begingroup$
@T.Bongers Now I get it. I didn't know about the leading coefficient. Thank you so much.
$endgroup$
– A.G.
Dec 4 '18 at 16:06




$begingroup$
@T.Bongers Now I get it. I didn't know about the leading coefficient. Thank you so much.
$endgroup$
– A.G.
Dec 4 '18 at 16:06










1 Answer
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$begingroup$

Note that $(2x-1) = 2(x-frac{1}{2})$ and $(3x+2)=3(x+frac{2}{3})$ so if either of them are zero then you can cancel out the leading coefficient to solve for $x$.






share|cite|improve this answer









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    $begingroup$

    Note that $(2x-1) = 2(x-frac{1}{2})$ and $(3x+2)=3(x+frac{2}{3})$ so if either of them are zero then you can cancel out the leading coefficient to solve for $x$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Note that $(2x-1) = 2(x-frac{1}{2})$ and $(3x+2)=3(x+frac{2}{3})$ so if either of them are zero then you can cancel out the leading coefficient to solve for $x$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Note that $(2x-1) = 2(x-frac{1}{2})$ and $(3x+2)=3(x+frac{2}{3})$ so if either of them are zero then you can cancel out the leading coefficient to solve for $x$.






        share|cite|improve this answer









        $endgroup$



        Note that $(2x-1) = 2(x-frac{1}{2})$ and $(3x+2)=3(x+frac{2}{3})$ so if either of them are zero then you can cancel out the leading coefficient to solve for $x$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 23:58









        CyclotomicFieldCyclotomicField

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