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Given that $P(x) =ax^5+8x^4+bx^3-10x^2+9x+c$, $P(x) $ is exactly divisible by $2x^3+5x^2-x-6$. Then how many among the following statements is/are true?

I) $a=4$   II) $b=5$   III) $c=-18$   IV) $a+b-c=-19$   V) $a^2+b^2<c^2$

  1. $0$   2) $1$   3) $2$   4) $3$

This question is supposed to be done in 1 to 2 minutes. I have no idea how to solve it. I factorised $2x^3+5x^2-x-6=(x-1)(x+2)(2x+3)$. After that came three equations with three variables, which are impossible to solve in such little time. Could anyone please help me?

SarGe
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Popular Power
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    Typeset mathematical terms using MathJax instead of posting images. Here's the MathJax tutorial. – SarGe Jul 08 '20 at 09:27
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    -@Popular Power, at least I don't see any other method other than solving three equations. – SarGe Jul 08 '20 at 09:36
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    Try using "long division" to divide your $P(x)$ by each of those factors in turn. Or by the entire $2x^3 ...$ expression. You will find there are conditions that have to be satisfied by $a$, $b$ and $c$ in order for the process to come out without a remainder. – Prime Mover Jul 08 '20 at 09:43
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    I agree with @SarGe - I don't see any shortcuts. Evaluating $P(x)$ at $x=1$ tells us quickly that at least two statements are false, but with the answer choices given that does not help. – gandalf61 Jul 08 '20 at 09:44
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    Using a computer, I get that $a = \frac{636}{143}$, $b = -\frac{121}{13}$ and $c = -\frac{306}{143}$, so the correct answer must be option $A$ or $0$. This would be very hard to do by hand, so there must be some other way. Maybe the rational root theorem or Vieta's formulas could help? – Toby Mak Jul 08 '20 at 10:01
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    $1$ to $2$ minutes for such an exercise is really not much. You were on the right track. The problem is : If one wants to apply an easier approach, one must find it out immediately under such a time pressure. – Peter Jul 08 '20 at 10:26
  • If you use your factorisation to write three equations in terms of $a,b,c$, you can assume $a=4$, solve for $b$ and $c$, then check if it is divisible by $2x^3+5x^2-x-6$. The problem is that although you can get the correct answer this way, this is a very time-consuming process. – Toby Mak Jul 08 '20 at 10:45
  • @gandalf61 Could you please tell how you conclude that two statements are false with just $P(1)$? – Popular Power Jul 08 '20 at 15:57
  • @PrimeMover the equations are extremely horrible. – Popular Power Jul 08 '20 at 15:58
  • @PopularPower I said at least two statements are false. We know that $P(1)=0$, so $a+b+c=-7$. This means that statement IV is false. And if statements I, II and III were all true then $a+b+c=-9$. So at least one of I, II or III is also false. – gandalf61 Jul 08 '20 at 20:01

1 Answers1

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Quick (partial) answer: Work modulo $5$ to eliminate the first four possible answers. Try some estimations for the last elimination.

A long answer follows, shows that i also tried the helpless path first. Please extract only the part of calculus in $\Bbb F_5$ for the "quick answer".

Didactically, this is a bad problem. Instead of attracting people to do it, it depicts maths as a guessing contest. People that never answered such problems at their time find it also useful to present mathematics as a race against time, without computers in the era of computers, best using only virtual paper and virtual pen. Well, let's see. My answer in such conditions during an exam would be now, after years, to compute quickly in 20" something like $(2x^2-x+1)(2x^3 +5x^2-x-6)$ with my hidden choice of the quotient, and offer the examiner the same question with replaced polynomial $ax^5 + 8x^4 +bx^3 - 6x^2 + 5x +c$ so that we can both start the race under democratic conditions and finally compare. (It turns out however that my choice was too friendly for this examiner, and that she or he has indeed a chance.)

It is a bad problem and we are in time trouble, and the examiner may address an "exactly divisibility" either in the ring $\Bbb Z[x]$, i.e. $a,b,c\in\Bbb Z$ and the quotient is also in $\Bbb Z[x]$, or in the bigger ring $\Bbb Q[x]$. So after five seconds we have one more problem and decide to work in the bigger ring. I hate denominators.

We start with the equality of the shape $$ \begin{aligned} (\dagger)\qquad& \\ P(x) &=ax^5+8x^4+bx^3-10x^2+9x+c \\ &=(sx^2+tx+u)(2x^3+5x^2-x-6) \\ &=(sx^2+tx+u)(x-1)(x+2)(2x+3) \end{aligned} $$ for some unknowns $s,t,u$ that we would immediately buy in the exam for a lot of money. So $P(1)=0$ gives $a+b+c=-7$. So $a+b=-c-7$ and (IV) gets simpler, $c=6$. Some 20" were gone.

The other "simple" equation we can get, from $P(-2)=0$, is $32a+8b-c=70$. (The third one goes outside the two minutes.) We add and get $33a+9b=63$, so $11a+3b=21$. (Ugly.) We try to use what may hold (or not).

  • Assume $a=4$. Then $b$ has some denominator $\ne 1$, so $c$ also has it, this excludes the integer values for $b,c$ in the next answers. We deal with the last one. Well $b=(21-44)/3$ is roughly $-7$, so $a+c\approx 0$. At any rate, $c$ is smaller in modulus than $b$. So under the given assumption, the "how many answer" is 0 or 1. We borrow a coin from the examiner now, two minutes are over.

  • While she or he is searching for the coin, we consider $b=5$. That $11a=21-15$ has no chance to make $a$ integer, so $c$ is also not an integer, and $a\approx 0$. So $c\approx -7-5$. Bad luck, the last inequality seems to work. The "how many answer" is 1 or 2. If the coin is already there, we would flip the coin and in both cases take the 1 for probabilistic reasons. And fail.

  • Well, we need some points in the artistic note, and need to explain the decision to flip the coin. We then ask, "... may i please write down..." (since it is easier to explain), the examiner understands, and moreover prefers this. Self-confident we write down the relation $(\dagger)$ above and that inserting the roots $1,-2$ we get $a+b+c=-7$, then $33a+9b=63$, so $11a+3b=21$. And we exclaim: "How beautiful! You see, if there are denominators, they are not involving the primes two and/or five". This is useful, since taking $(\dagger)$ modulo five we get: $$ \tag{$\dagger_{\Bbb F_5}$} (\dots)(x-1)(x+2)2(x-1) \\ = ax^5 +3x^4+bx^3-x+c\ . $$ Let us solve it in the field with five elements $\Bbb F_5$! The equation obtained after taking $x=1$ is $a+b+c=-7=-2=3$, well known already. Derive and plug in $x=1$ again. We get $0=12+3b-1$. So $b=3$, excluding (II). Then $a=11a=21-3b=2$, excluding (I). We have $a+c=0$, so $c=-2=3$, excluding (III) and (IV). You smile and give the examiner the two minutes to understand the (new) reason for the coin.

  • OK, we need the information at the archimedean place for (V). For this i only see the way which writes down also the third equations, adjust "slightly" the coefficients to get a numerical approximation of the solution. This would be enough to eliminate (V), since (in modulus) $c$ is very small compared to both $a,b$.

Not as an answer, but as a check to see that one needs luck to chose the one path that may separate quickly the answers, sage:

sage: var('a,b,c,x');
sage: P(x) = a*x^5 + 8*x^4 + b*x^3 - 10*x^2 + 9*x + c
sage: dic = solve([P(1)==0, P(-2)==0, P(-3/2)==0], [a,b,c], solution_dict=1)[0]
sage: dic
{a: 636/143, b: -121/13, c: -306/143}
sage: A, B, C = dic[a], dic[b], dic[c]
sage: F = GF(5)
sage: F(A), F(B), F(C)
(2, 3, 3)
sage: A.n(), B.n(), C.n()
(4.44755244755245, -9.30769230769231, -2.13986013986014)
dan_fulea
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