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


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



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?

  • $\begingroup$ 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 $6$. $\endgroup$ – user296602 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? $\endgroup$ – user296602 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

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 $x$.

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