# Synthetic division, quadratic formula and grouping

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?

• 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 $6$. – user296602 Dec 3 '18 at 23:58
• Yes, I've noticed that. But I though that the results shoud be just the same. What's the difference? – A.G. Dec 4 '18 at 0:06
• Knowing the zeros doesn't determine the scaling: Both $x-1$ and $2(x - 1)$ are lines with roots at $x = 1$, right? – user296602 Dec 4 '18 at 0:07
• @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. – user376343 Dec 4 '18 at 10:55
• @T.Bongers Now I get it. I didn't know about the leading coefficient. Thank you so much. – 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 $$x$$.