My polynomial: $2x^3 +7x^2+12x+9$.

Now, I've tried both of the techniques given in this Wikihow page, but neither of them worked for this problem. Synthetic division is something which I think would work, but it seems like a lot of work to first use the Rational Zeros Theorem to check all the possibilities using synthetic division and then come up with a quadratic equation.

The solution given is as follows: solution in Slader.

Obviously, one cannot use synthetic division to factor our $(2x +3)$ for the division works only for factors of the form $(x-c)$. So my question is, how do you factor a polynomial like this?


The polynom $2x^3 +7x^2+12x+9$ is a polynomial with coefficients in $\mathbb{Q}$, there is a result saying that the roots living in $\mathbb{Q}$ are of the form $\frac{a}{b}$ where $a$ divides thecoefficient $a_0$ and $b$ divides the dominant coefficient of the polynomial.

So in our case : $$a\in \{ -9,-3,-1,1,3,9\}$$ $$b\in \{-2,-1,1,2\}$$ which can be reduced to $$b\in \{1,2\}$$ because otherwise each fraction appears twice.

Since all the coefficient are positive, a positive root will automatically give a quantity $> 0$. So :$$a\in \{ -9,-3,-1\}$$ $$b\in \{2,1\}$$ From here you can try the $6$ possible rationnal roots.

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  • $\begingroup$ I got all of the above explanation, except for ${-2, -1, 1, 2}$ getting reduced to ${1, 2}$. What reasoning/theorem did we use to reduce it? 'Otherwise each fraction appears twice'? Meaning? $\endgroup$ – MathEnthusiast May 29 '16 at 8:04
  • $\begingroup$ If you take $x \in a$, $y\in b$ then $-x \in a$ and $-y \in b$ but $a/b=-a/-b$ and $-a/b =a/-b$, so all the fractions (positive and negative) can be reached with using only the positive divisors in $b$. $\endgroup$ – Bérénice May 29 '16 at 8:09
  • $\begingroup$ Got it, thanks! $\endgroup$ – MathEnthusiast May 29 '16 at 8:10

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