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.
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.
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].
$$p(x)=2x^4-5x^3+7x^2-25x-15$$ $$p'(x)=8x^3-15x^2+14x-25$$ $$p''(x)=24x^2-30x+14$$ 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.
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.
$p(1)=-36$
$p(-1)=24$
$p(3)=0$
$p(-3)=420$
$p(5)=660$
$p(-5)=2160$
$p(15)=85560$
$p(-15)=120060$
$p(\frac{1}{2})=-\frac{105}{4}$
$p(-\frac{1}{2})=0$
$p(\frac{3}{2})=-\frac{87}{2}$
$p(-\frac{3}{2})=\frac{261}{4}$
$p(\frac{5}{2})=-\frac{135}{4}$
$p(-\frac{5}{2})=\frac{495}{2}$
$p(\frac{15}{2})=4410$
$p(-\frac{15}{2})=\frac{36015}{4}$
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.
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.
Develop $(x-3)(x+\frac{1}{2})(ax^2+bx+c)$. Find $a,b,c$ by identification. Then, you can calculate the other roots.
