Find the value of the polynomial

Algebra Level 5

Let $Q(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$ be a polynomial with integer coefficients, and $0 \le a_i < 3$ for all $0 \le i \le n$ .

Given that $Q\big(\sqrt 3\big) = 20 + 17\sqrt 3$ , what is the value of $Q(2)$ ?

The answer is 86.

3 solutions

Ajinkya Shivashankar
Mar 4, 2017

Lets begin with writing 20 and 17 in ternary.

$20=2*3^2+0*3+2$ and also $17=3^2+2*3+2$

Thus $Q(\sqrt{3})=20+17\sqrt{3}=(\sqrt{3})^5+2*(\sqrt{3})^4+2*(\sqrt{3})^3+0*(\sqrt{3})^2+2*\sqrt{3}+2$

So $Q(x)=x^5+2*x^4+2x^3+2x+2$

Thus $Q(2)=86$

This representation is unique because there is only 1 way a number can be written in ternary form.(i.e the coefficients must be less than 3 and non-negative) For example 3^4 can never be written in some combination of 3^3, 3^2 and 3 and a constant if their coefficients are less than 3. So if a term of 3^4 appears, it has to come from 3^4.

Aakash Khandelwal
Mar 3, 2017

$Q(x) = x^{5} + 2x^{4} + 2x^{3}+2x+2$

Why must $Q(x) = x^{5} + 2x^{4} + 2x^{3}+2x+2$ be true?

Pi Han Goh - 4 years, 3 months ago

It is true...I got the same after spending 10 minutes. You would like to know how?

Skanda Prasad - 3 years, 11 months ago

Yes sure. $\quad$

Pi Han Goh - 3 years, 11 months ago

@Pi Han Goh – $Q(\sqrt{3})=a_{0}+\sqrt{3}a_{1}+3a_{2}+3\sqrt{3}a_{3}+9a_{4}+9\sqrt{3}a_{5}+...$ I spent time substituting values $0,1,2$ for $a_{i}$ to the terms containing $\sqrt{3}$ to make the sum equal to $17\sqrt{3}$ and to the terms without $\sqrt{3}$ to make the sum equal to $20$ .

I got the following combination.

$a_{0}=2,$ $a_{2}=0$ and $a_{4}=2$ which will give me 20 when added.

$a_{1}=2$ , $a_{3}=2$ and $a_{5}=1$ which will give $17\sqrt{3}$ when added.

And note that substituting $\sqrt{3}$ for further terms will make numbers too big like $81$ , $81\sqrt{3}$ etc which we don't want. So those terms can be regarded as $0$ .

Now we know the values of $a_{i}$ s and we can substitute them in $Q(x)$ .

Hence, we get that polynomial.

Hope I am clear.

Skanda Prasad - 3 years, 11 months ago

K T
Jun 13, 2019

$Q(\sqrt{3})=20+17\sqrt{3}=a_0+ \sqrt{3}a_1+ 3a_2+ 3\sqrt{3}a_3+...= a_0+ 3a_2+9a_4+...+\sqrt{3}(a_1+3a_3+9a_5+...)$

So we have $a_0+ 3a_2+9a_4+...=20$ and $a_1+3a_3+9a_5+...=17$ .

In effect, we are looking for the trinary representations of

$20_{10}=202_3$ so that $a_4=2, a_2=0, a_0=2$

and $17_{10}=122_3$ so that $a_5=1, a_3=2, a_1=2$ .

$Q(2)=2×2^0+2×2^1+0×2^2+2×2^3+2×2^4+1×2^5=\boxed{86}$

Find the value of the polynomial

The answer is 86.

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