To find a polynomial

Question

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) :

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.

Answer 1

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

Answer 2

Hint: Consider $f(x)-x^2$ instead. It's so much easier to work with polynomials that are 0 at given points.

Answer 3

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

Answer 4

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

Answer 5

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, $$f(4)=22.$$

Answer 6

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

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

Answer 7

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/