Roots of a Rational Cubic

By Vieta's relations, $a+b+c = -a$ $ab + bc + ca = b$ $abc = -c$ ). Starting from the third equation, either $c=0$ (case 1) or $ab = -1$ (case 2)

Case 1: We proceed to the second equation. $ab = b$ )Thus $b=0$ or $a = 1$ .

Case 1.1: The first equation gives $a=0$ . This has solution set $a=0, b=0, c=0$ .

Case 1.2: The first equation gives $b = -2$ . This has solution set $a=1, b=-2, c=0$ .

Case 2: The third equation gives $b=-\frac{1}{a}$ while the first equation gives $c = \frac{1}{a} - 2a$ . Thus after substituting this into the second equation and simplifying, we have $(a-1)(1-2a^2 - 2a^3 ) = 0$ . The second bracket has no rational roots (by using rational root theorem) so we have that $a =1$ . This has solution set $a=1, b=-1, c=-1$ .

Consolidating the three cases (yielding 1 solution each) gives 3 triples.

By Vieta's formula,

-(a+b+c)=a-----------(1)

ab+ac+bc=b---------(2)

-abc=c----------------(3)

(i) When c=0

-(a+b)=a

2a=-b----------------(1a)

ab=b(from (2))------(2a)

If b=0,a=0 from (1a)

If b is not equal to 0,a=1 from 2a and b=-2 from (1a).

Therefore,we have two solutions in this case,namely (a,b,c) = (0,0,0) and (1,-2,0).

(ii) When c is not equal to 0

ab= -1

b= - $\frac {1}{a}$ -------(3a)

Substituting (3a) into (2),

-1- $\frac {c}{a}$ +ac = - $\frac {1}{a}$

From (1),

-b-c=2a

Substituting (3a) and rearranging,

c= $\frac {1-2a^2}{a}$ -------(1b)

Substituting (1b) into (2b),

$\frac {2a^2-1}{a^2}$ - 2 $a^2$ = - $\frac {1}{a}$

Multiplying both sides by $a^2$ and rearranging,

2 $a^4$ - 2 $a^2$ -a+1 = 0

Factorizing,

(a-1)(2 $a^3$ -1) = 0

a=1 or 2 $a^3$ - 1 = 0

2 $a^3$ - 1 = 0

Factorizing,

(2^(1/3)a -1)(2^(2/3) $a^2$ + 2^(1/3)a +1) = 0

a=1/(2^(1/3)) or 2^(2/3) $a^2$ + 2^(1/3)a +1=0

Consider the latter equation

The discriminant is 2^(2/3) - 4(2^(2/3)) < 0

Therefore,the two solutions are complex

a =1/(2^(1/3)) is irrational.

Thus,the only rational solution is a=1.

From (1b) and (3a), we get (a,b,c) = (1,-1,-1)

From (i) and (ii), we have 3 solutions

We can learn from the question that f(x)=(x-a)(x-b)(x-c). Because these are the three roots. And f(x)=x^3+ax^2+bx+c. So we factorization the f(x) and get: a+b+c=-a, ab+bc+ac=b, abc=-c. When c=0 there have ab=b, 2a+b=0. So there have two answers which are a=0,b=0 and a=1,b=-1/2. When c \neq 0 then ab=-1 and c=-2a-b and (a+b)c=b-ab. From these three equations we can get b^4+b^3-2b^2+2=(b+1)(b^3-2b+2)=0. (It can no longer factorization in rational numbers). So b=-1, a=1, c=-1 is the third answer. So there has three in total.

f(x) = x^3+ax^2+bx+c = (x-a)(x-b)(x-c), so by comparing the coefficients (Vieta's Theorem), we have three equations: a+b+c = -a ab+bc+ca = b abc = -c From the third equation, we have abc+c = 0 --> c(ab+1) = 0 --> c = 0 or ab = -1

In the first case, we have c = 0: Then ab = b --> b(a-1) = 0 so b = 0 or a =1. b = 0 gives a = -a --> a = 0. a = -1 gives -1+b = 1 --> b = 2.

In the second case, we have ab = -1: a+b+c = -a --> c = -2a-b ab+bc+ca = b --> c(a+b) = b-ab --> (-2a-b)(a+b) = b-ab --> -2a^2-3ab-b^2 = b-ab --> 2a^2+b^2+b+2ab = 0 Substituting ab=-1 and a=-1/b, we get: 2(-1/b)^2+b^2+b-2 = 0 --> b^4+b^3-2b^2+2 = 0. Using the rational root theorem, the only possibilities for rational roots are +/- 1,2 and only b = -1 works. Then a = 1 and c = -1.

Thus, we have three solutions: (0,0,0), (-1,2,0), and (1,-1,-1).

The only monic polynomial with roots a,b,c is (x−a)(x−b)(x−c), so we must have abc = −a − b − c ,

                     **b = ab + bc + ca**,

                     **c = − abc**

If c=0, this becomes , ab = −a−b , b = ab , thus, either b=0 and a=0 or a=1 and b=−2.

If c≠0 , the third equation becomes ab = −1 ,

substituting b= −1/a into the first equation yields,

c = − 2a + 1/a , and then the second equation becomes

−1/a = 1 − 1/(a^2) − 2 (a^2)+1 . Multiplying through by a^2 yields

2 a^4 − 2 a^2 − a + 1 = 0.

The solution a = 1 is readily observable (on factorization) , 2 a^2 (a^2 - 1) - (a-1) = 0 which gives, (a-1) [2 a^2 (a+1) - 1] = 0 So, dividing through by (a−1) yields

2 a^3+ 2 a^2 −1=0 ,

which has one irrational and two complex roots (by computation). The solution a=1 leads to b = c = −1 .

Thus the only ordered triples are (0,0,0) , (1,−2,0) and (1,−1,−1) , with corresponding polynomials x^3, x^3 + x^2− 2 x * and * x^3 + x^2 −x−1 , respectively.

Note that by Vieta's Formulas we have the system of equations

$\begin{cases}a+b+c&=-a,\\ab+ac+bc&=b,\\abc&=-c.\end{cases}$

From the third equation we either have $c=0$ or $ab=-1$ . If the former holds, then our original two equations become $a+b=-a$ and $ab=b$ . Once again, from this second equation we either have $a=1$ or $b=0$ , and these give $b=-2$ and $a=0$ respectively. Therefore, two possible ordered triples $(a,b,c)$ are $(0,0,0)$ and $(1,-2,0)$ . Both of these check.

If, on the other hand, we have $ab=-1$ , then the algebra gets a bit messier. We wish to try and isolate one of the variables in a single equation that we can possibly work with to extract roots. Note that the first equation becomes $2a+b+c=0\implies -c=2a+b$ . Plugging this and $ab=-1$ into the second equation gives

$\begin{aligned}ab+c(a+b)&=b\\1+(2a+b)(a+b)&=-b\\1+2a^2+3ab+b^2&=-b\\2a^2+b^2+b-2&=0.\end{aligned}$

Now, we substitute $b=-\frac1a$ into this equation (in the hopes of continuing our isolation) to get

$2a^2+\dfrac1{a^2}-\dfrac1a+1=0\implies (a-1)(2a^3-2a^2-1)=0.$

It is clear that by the Rational Root Theorem the cubic has no rational solutions, so the only rational solution to this equation is $a=1$ . Substitution gives the third and final ordered triple $(a,b,c)=(1,-1,-1)$ . This checks, as $x^3+x^2-x-1$ factors into $(x-1)(x+1)^2$ . This gives $\boxed{3}$ possible ordered triplets that work.

$(x-a)(x-b)(x-c)=x^3+ax^2+bx+c$

From Vieta's formulae, we have...

$a=-(a+b+c)$

$b=ab+bc+ca$

$c=-abc$

Solving the system, we get...

$(a,b,c)=(0,0,0),(1,-2,0),(1,-1,-1)$

Total integral solutions: $\fbox{3}$

Using Vieta's Theorem, we have three equations:

$1) a + b + c = -a$

$2) ab + ac + bc = b$

$3) abc = - c$

From looking at equation 3, the motivation is to divide the case into two:

CASE I

If $c = 0$ , we get:

From equation 1:

$4) 2a + b = 0$

From equation 2:

$5) ab - b = 0 \Rightarrow b(a-1) = 0$

From $5$ , either $b = 0$ , or $a-1 = 0$ .

If $b = 0$ , then from $4)$ we get $a = 0.$

If $a-1=0$ , or $a = 1$ , then from $4)$ we get $b = -2$

From this case $(c = 0)$ , we get 2 solutions $(0, 0, 0), (1, -2, 0)$ , which are rational solutions.

CASE II

If $c \neq 0$ , then

From equation 1:

$4) 2a + b + c = 0 \Rightarrow c = -(2a+b)$

From equation 2:

$5) ac + bc - b = 1 \Rightarrow c(a+b) = b + 1$

From equation 3:

$6) ab = -1 \Rightarrow a = -1/b$

Substituting $c$ from equation $4)$ to $5)$ we get:

$-(2a +b)(a+b) = b + 1$

Substitung $a = -1/b$ to above equation we get:

$-(\frac{-2}{b}+b)(\frac{-1}{b}+b) = b + 1 \Rightarrow -(\frac{b^2-2}{b}) (\frac{b^2-1}{b}) = b + 1$

$-(b^2-2) (b^2-1) = (b + 1)b^2 \Rightarrow b^2(b+1) + (b^2-2)(b^2-1) = 0$

$b^2(b+1) + (b^2-2)(b-1)(b+1) = 0 \Rightarrow (b+1) (b^2 + (b^2-2)(b-1)) = 0 \Rightarrow (b+1) (b^3 -2b+2) = 0$

From above, we get $b = -1$ or $b^3-2b+2 = 0$

If $b = -1$ then $a = 1, c = -1$ . We get 1 more rational solution here: $(1, -1, -1)$

If $b^3 -2b+2 = 0$ , then we don't have any rational solutions for that $b$ simply by checking rational root test for $b = \pm1, \pm2$

Hence, there are $\boxed{3}$ ordered triple rational numbers of $(a,b,c)$

Let (x - a) (x- b) (x - c) = x^3 + ax^2 + bx + c. Then, we have

x^3 - (a + b + c)x^2 + (ab +bc +ac)x -abc = x^3 + ax^2 + bx + c implying

a = - (a + b + c) , b = (ab + bc + ac) and c = -abc.

Case i: c = 0 implies 2a +b =0 and b = ab.

If b = 0 then a = 0 so that (0,0,0) is a solution

If b is not equal to 0 then a =1 and b = -2 making (1,-2,0) a solution.

Case ii: c is not equal to 0. this implies that 1 + ab = 0.

We have 2a + b + c = 0 and b = ab + bc +ac. Multiplying by a in the 2nd equation we get,

ab =a^2 b + abc + a^2 c implies -1 = -a -c +a^2 c implies a^2 c - a - c + 1 = 0 implies

c (a^2 - 1) - (a - 1) = 0 implies (a - 1) (c(a + 1) - 1 ) = 0

Case i: if a = 1, c = -1 and b = -1 making (1,-1,-1) a solution.

Case ii: c( a + 1) = 1 implies c =1/ (a +1). Putting c = 1/ (a + 1) in a^2 c - a - c + 1 = 0 we get that

2a^3 + 2a^2 = 1. We claim that this has no rational solutions. If not, let a = p/q be a rational solution with gcd (p,q) = 1.

Then, we have 2 p^2 (p + q) = q^3.

Cae i: p is even and q is odd. In this case lhs is even and rhs is odd which is a contradiction.

Case ii: p is odd and q is even. In this case, 8 divides rhs whereas 8 does not divide lhs, since the only factor of 2 in lhs is just 2 itself, p^2 and p + q being odd. This is again a contradiction.

Case iii: p is odd and q is also odd. In this case lhs is even where as rhs is odd, this is again a contradiction.

Hence 2a^3 +2a^2 = 1 has no rational solutions.

So, the no of ordered rational triples with the required property is just 3.

We get $a+b+c = -a$ , $ab+bc+ac = b$ , $abc = -c$ . If $c = 0$ , we get $ab = b$ and $2a= -b$ , which leads to $(0,0,0)$ and $(1,-2,0)$ . Otherwise we have $ab = -1$ , and $(a+b)c = b+1$ , and $c = -(2a+b)$ , so $ab = -1$ and $(a+b)(2a+b) +b+1 = 0$ . Plugging in $a = -1/b$ and multiplying through by $b^2$ , we get $b^4+b^3-2b^2+2 = 0$ . Note that $b = -1$ is a solution, which corresponds to $(1,-1,-1)$ . Dividing out by $b+1$ gives $b^3 -2b+2 = 0$ , and this has no rational roots (by the Rational Root Theorem; it suffices to check $\pm 1$ and $\pm 2$ ).

So there are a total of $\fbox{3}$ solutions.

The only monic polynomial with roots a,b,c is (x−a)(x−b)(x−c), so we must have

abc=−a−b−c,=ab+bc+ca,=−abc. If c=0, this becomes

ab=−a−b,=ab, and thus either b=0 and a=0 or a=1 and b=−2.

If c≠0, the third equation becomes ab=−1; substituting b=−1/a into the first equation yields

c=−2a+1a, and then the second equation becomes

−1a=−1+2−1a2−2a2+1. Multiplying through by a2 yields

2a4−2a2−a+1=0. The solution a=1 is readily guessed, and dividing through by a−1 yields

2a3+2a2−1=0, which has one irrational and two complex roots (computation). The solution a=1 leads to b=c=−1.

Thus the only ordered triples are (0,0,0), (1,−2,0) and (1,−1,−1), with corresponding polynomials x3, x3+x2−2x and x3+x2−x−1, respectively.

if f has roots a,b and c then x=a and x = b and x = c when f(x) = 0. Therefore we can then equate f(x) = (x-a)(x-b)(x-c). Multiplying the RHS and then equating the coefficients of both sides will yield 3 simultaneous equations witch through manipulation will yield the correct set (a,b,c) satisfying f.

Suppose $(a, b, c)$ are the rational roots of $f(x) = x^3 + ax^2 + bx = c$ . It's obvious that $(0, 0, 0)$ are such triplet. Now suppose $a$ , $b$ , $c$ $\neq 0$ . And by the rational roots theorem, $a$ , $b$ , $c$ are integers. Also by Vieta's formulas,

(1) $a + b + c = -a$

(2) $abc = -c$

From (2), we get $ab = -1$ or $(a, b) =$ $(1, -1)$ , $(-1, 1)$ . Plugging in (1), we get $(a, b, c)$ = $(1, -1, -1)$ , $(-1, 1, 1)$ . Thus, our answer is 3.

By Vieta's Rule, we have $a+b+c=-a, ab+bc+ca=b, abc=-c$ . Solving we have two trivial solutions $(0,0,0)$ and $(1,-1,-1)$ . At last if we eliminate $a,c$ , we have the equation $b^3-2b+2=0$ which has only 1 real root. [this can be seen using its discriminant]. So, in all we have three ordered pair of $a,b,c$