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My doubt is from a PRMO model paper I had today. The image of the question is given below ( I had to have my lunch, so thought of a quicker way to put my question and ended up with this) : My problem

What I tried :

I felt that $f(x) - x^3$ can give me the value of the quadratic part of the polynomial. As a result, taking the quadratic part to be of the form $ax^2 +bx + c$, the differences I get are :

  1. $f(1) - 1^3 = a + b + c = 0$
  2. $f(2) - 2^3 = 4a + 2b + c = -4$
  3. $f(3) - 3^3 = 9a + 3b + c = -18$

I am not an expert at solving 3 linear equations in 3 variables , but I tried and ended up pulling out my hair (trying to be a bit literary; hope you won't mind the wordings , but rather concentrate on the question). I tried to take 2 equations at a time, and ended getting multiple values for the same variables.

I will be grateful to anyone who is willing to help me out.

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  • $\begingroup$ Note : the question didn't have options, it was a numerical answer-type question. $\endgroup$ – Spectre Sep 27 '20 at 8:08
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    $\begingroup$ It’s $-18$, not $-9$. Note that $3a+b$ is $-4$ and $8a+2b$ is $-18$ thus $4a+b$ is $-9$. $\endgroup$ – Mindlack Sep 27 '20 at 8:17
  • $\begingroup$ Thanks for the correction $\endgroup$ – Spectre Sep 27 '20 at 9:00
  • $\begingroup$ I am, at times, a bit careless ... $\endgroup$ – Spectre Sep 27 '20 at 9:04
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  1. $f(1) - 1^3 = a + b + c = 0$
  2. $f(2) - 2^3 = 4a + 2b + c = -4$
  3. $f(3) - 3^3 = 9a + 3b + c = -18$

The third equation is wrong

Subtracting eq(1) from eq(2)

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

$3a +b = -4$------------------------------ eq(3)

Subtracting eq(1) from eq(3)

$9a + 3b + c -(4a + 3b + c) = -18-(-4)$

$5a = -22$ $a = \frac{-2}{5}$ Substitute this in all equations and you'll get the answers

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Note that $f(x) = x^2$ satisfy the three given equations. But we want of degree $3$ so we add $(x-1)(x-2)(x-3)$ and get $$f(x) = (x-1)(x-2)(x-3) + x^2$$ with $f(4) = 6+16 = 22$.

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  • $\begingroup$ Just what user suggested, @cgss $\endgroup$ – Spectre Sep 27 '20 at 9:12
  • $\begingroup$ I am sorry, I don't understand what you are saying. $\endgroup$ – cgss Sep 27 '20 at 10:23
  • $\begingroup$ Refer the solution by @user $\endgroup$ – Spectre Sep 27 '20 at 10:30
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    $\begingroup$ I got you now. It's not the same though. User used lagrange polynomials with a slight change to take advantage of the fact that $a_3 = 1$. I used polynomial division on $f(x) - x^2$. $\endgroup$ – cgss Sep 27 '20 at 10:35
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Hint: Consider $f(x)-x^2$ instead. It's so much easier to work with polynomials that are 0 at given points.

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  • $\begingroup$ But the deduction technique and the use of linear equations in two variables (those that TimCrosby and user suggested) are exactly what I wanted... I like to take the hard route at times ... :) $\endgroup$ – Spectre Sep 27 '20 at 9:09
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While the solutions involving $x^2$ are likely the ones intended by the question-setter, it is also easy to quickly get the value of $f(4)$ without extracting the quadratic by using finite differences. Let $g(n) = f(n) - n^3$, as in the OP, be quadratic. The first-order differences are:

$$\Delta g(1) = g(2)-g(1) = -4 \\ \Delta g(2) = g(3) - g(2) = -14.$$

So the second-order difference is $\Delta^2 g(1) = \Delta g(2) - \Delta g(1) = -10$, which for any quadratic (or lower) polynomial is constant. Hence $\Delta^2 g(2) = -10$, so $\Delta g(3) = \Delta g(2) -10 = -24$, and $g(4) = g(3) -24 = -42$.

So $f(4) = 4^3 -42 = 22$.

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We have $$a+b+c=0,$$ $$4a+2b+c=-4$$ and $$9a+3b+c=-18,$$ which gives $$(a,b,c)=(-5,11,-6)$$ and $$f(x)=x^3-5x^2+11x-6.$$ Thus, $$f(4)=22.$$

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We don't need to determine explicitly the coefficients for the polynomial, indeed we have that by uniqueness

$$f(x)=(x-1)(x-2)(x-3)+\frac12(x-2)(x-3)-4(x-1)(x-3)+\frac92(x-1)(x-2)$$

which satisfies by construction the given conditions with $f(1)=1$, $f(2)=4$ and $f(3)=9$, then

$$f(4)=(3)(2)(1)+\frac12(2)(1)-4(3)(1)+\frac92(3)(2)=6+1-12+27=22$$

As noticed by other answers, more trickly we have that

$$f(x)=(x-1)(x-2)(x-3)+x^2$$

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  • $\begingroup$ But such kind of deduction was a bit hard for me though ... by the way, thanks for that answer ... I'll someday learn to deduce better ... $\endgroup$ – Spectre Sep 27 '20 at 9:06
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    $\begingroup$ @Spectre Yes I understand that! Anyway try to think to similar problem also for simpler cases to get confident with it. Bye $\endgroup$ – user Sep 27 '20 at 9:09
  • $\begingroup$ Thanks for that advice too, @user ! I like it ! $\endgroup$ – Spectre Sep 27 '20 at 9:10
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    $\begingroup$ Unfortunately, I didn't accept your answer as I can't accept many answers at a time, and also because I needed to learn how to solve linear equations in 3 variables. $\endgroup$ – Spectre Sep 27 '20 at 9:11
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    $\begingroup$ @Spectre That's fine! You are doing a great work! Bye $\endgroup$ – user Sep 27 '20 at 9:21
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Given $f(x)$ , a monic cubic polynomial.

$f(1) = 1$, $f(2) = 4$, $f(3) = 9$

By factor theorem we can show that $(x-1)$,$(x-2)$ and $(x-3)$ are factors of $f(x) - x^2$

We have three linear factors thus we can write $f(x)$ as,

$f(x) - x^2 = k(x-1)(x-2)(x-3)$, where k is some constant

But $f(x)$ is a monic polynomial, therefore k = 1

thus our polynomial becomes, $f(x) - x^2 = (x-1)(x-2)(x-3)$

Putting x = 4,

$f(4) = 6 + 16 = 22$

This is common technique to solve competition math problems, You can read more about it here https://brilliant.org/wiki/polynomial-interpolation/

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