How to tell whether the roots are the only rational roots for a given polynomial Find all the rational roots of the polynomial $p(x)=2x^4-5x^3+7x^2-25x-15$. I only found $x=3, -\frac{1}{2}$. I am not sure whether there is any other rational roots. 
Is there a way to tell whether these are the ONLY rational roots other than doing the long polynomial division?
Thanks.
 A: You can use Vieta's relations: the sum of the roots (in $\mathbf C$) is $\frac 52$. Now the sum of the found roots is already $\frac 52$. So if  there were other rational roots, they would be opposite. Now the equation can be written as 
$$2x^4+7x^2+15=5x(x^2+5),$$
and having opposite roots would imply they're roots of $x^2+5$, which has no real root. 
A: You can use the Rational Roots Test:
Lemma Let $P(X)=a_nX^n+...+a_1x+a_0$ be a polynomial with integer coefficient and $m,n$ integers. If $x =\frac{m}{k}$ is a root of $P(X)$ then 
$$m|a_0 \,;\, k|a_n$$
The proof is a pretty simple divisibility problem.
For your exercise, the Lemma tells you that $m \in \{ \pm 1, \pm 3, \pm 5 \pm 15 \}$ and $k \in \{1, 2\}$.
Therefore, the only potential rational roots are 
$$\{ \pm 1, \pm 3, \pm 5 ,\pm 15, \pm \frac12 , \pm \frac32, \pm \frac52, \pm \frac{15}2 \}$$
The only thing you need is now to test all of them. ANd remember, you don't need to necessarily calculate them, you only need to see if you get 0 or not [For example $P(-\frac{15}{2})$ is clearly much larger than 0].
A: $$p(x)=2x^4-5x^3+7x^2-25x-15$$
$$p'(x)=8x^3-15x^2+14x-25$$
$$p''(x)=24x^2-30x+14$$
The discriminant of $p''(x)$ is,
$$D=30^2-4\cdot14\cdot24=900-1344\lt0$$
Thus, $p''(x)\gt0$ for all $x\in\mathbb R$.
Thus, $p'(x)$ is a strictly increasing function.
Thus, the $p(x)$ is concave up everywhere and can cut the $x$-axis at most twice. Hence, $p(x)$ has at most two real, and hence at most two rational roots.
A: Alternate method: Exhaust search of potential rational roots
Imagine the polynomial can be written as four linear factors with integer coefficients:
$$p(x)=(a_1x-b_1)(a_2x-b_2)(a_3x-r_3)(a_4x-b_4)$$
If you expand this out you can see that $a_1a_2a_3a_4=2$ and $r_1r_2r_3r_4=-15$.
So the possible values for the $a_i$'s are $\pm1$ and $\pm2$. Also the possible values for the $b_i$'s is $\pm1$, $\pm3$, $\pm5$, $\pm15$.
The roots of the above factorized polynomial will be $x=\frac{b_i}{a_i}$
So the only possible rational roots are: $\pm1$,$\pm3$,$\pm5$,$\pm15$,$\pm\frac{1}{2}$,$\pm\frac{3}{2}$,$\pm\frac{5}{2}$,$\pm\frac{15}{2}$
You can then exhaustively test these 16 potential roots to find that the two you found are the only ones.
$p(1)=-36$
$p(-1)=24$
$p(3)=0$
$p(-3)=420$
$p(5)=660$
$p(-5)=2160$
$p(15)=85560$
$p(-15)=120060$
$p(\frac{1}{2})=-\frac{105}{4}$
$p(-\frac{1}{2})=0$
$p(\frac{3}{2})=-\frac{87}{2}$
$p(-\frac{3}{2})=\frac{261}{4}$
$p(\frac{5}{2})=-\frac{135}{4}$
$p(-\frac{5}{2})=\frac{495}{2}$
$p(\frac{15}{2})=4410$
$p(-\frac{15}{2})=\frac{36015}{4}$
A: Alternate method: Polynomial Short Division
For $p(x)=2x^4−5x^3+7x^2−25x−15$ you have found two factors: $(x-3)$ and $(2x+1)$.
Like you would do for long division short division needs us to multiple the two factors together:
$$(x-3)(2x+1)=2x-5x-3$$
Next start writing $p(x)$ as a multiple of this new quadratic expression. Compare the first terms: $2x^4$ and $2x^2$ the factor there is $x^2$. So:
$$2x^4−5x^3+7x^2−25x−15=x^2(2x^2-5x-3)+10x^2−25x−15$$
Adjust the bit at the end as needed for equality. Then repeat this step on the next largest term: $10x^2$ compared to $2x^2$ gives a factor of 5. So:
$$2x^4−5x^3+7x^2−25x−15=x^2(2x^2-5x-3)+5(2x^2−5x−3)+0$$
As you've gotten to $+0$ at the end you can stop. Do one last factorize of the quadratic expression and:
$$2x^4−5x^3+7x^2−25x−15=(x^2+5)(2x^2-5x-3)$$
You can then look at the new bracket and consider if it has any rational factors. It doesn't factorize so you know there are only two.
A: Common sense and the rational roots theorem.  You have $3, -1/2$ as roots so you know
$p(x)=2x^4-5x^3+7x^2-25x-15 = (x - 3)(x + 1/2)(2x^2 + bx + 10)=(x-3)(2x + 1)(x^2 + b'x + 5)$.
So by rational roots the remaining rational roots, if any are $\pm(1,5)$.  Very short list to check.  
Although with very little extra work you can figure what $b'$ is and use the quadratic equation.
... or at least that be is not $\pm 6$ so $(x^2 + b'x + 5)$ is not factorable and has no rational roots.
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The thing people tend to forget about the ration roots test is that although with the rational roots test you get a long list of potential solutions: $\pm 1, \pm 3, \pm 5 \pm 15, \pm 1/2, \pm 3/2, \pm 5/2, \pm 15/2$, the product of the rational roots must multiply to end coefficient divided by the leading coefficint (-15/2) or a factor if there are not all rational roots. 
So as you have 3 and -1/2.  There are either 2 rational roots and they must multiply to -15/2(3x-1/2)=5, or there are none.  So that's either 1 and 5 or -1 and -5 or no more rational roots.
A: Develop $(x-3)(x+\frac{1}{2})(ax^2+bx+c)$. Find $a,b,c$ by identification. Then, you can calculate the other roots.
