# Practice Paper 2 Question 8

A polynomial \(f(x) = x^n + a_{n-1}x^{n-1} \cdots + a_1x + a_0,\) with integer coefficients \(a_i,\) has roots at \(1, 2, 4, \ldots, 2^{n-1}.\) What possible values can \(f(0)\) take?

## Related topics

## Hints

- Hint 1How else can you write a polynomial when you know all of its roots?
- Hint 2... specifically, a polynomial that has a root at \(x_0\) is divisible by the binomial \(x-x_0.\) How does this help you answer the first hint?
- Hint 3... more specifically, \(x-x_0\) is a factor of \(f(x),\) i.e. \(f(x)=(x-x_0)g(x)\) where \(g(x)\) is a polynomial of degree \(n-1.\) How does this help you answer the first hint?
- Hint 4Finding \(f(0)\) is the same as evaluating this new expression at \(0.\) What do you obtain?
- Hint 5What manipulations can you perform on exponents when the bases are equal?
- Hint 6Specifically, try writing the product as a single number raised to a power, where the latter is an expression.
- Hint 7Looking at the expression (sum) in the power, do you notice a familiar relationship between its terms?

## Solution

Seeing as the coefficient of \(x^n\) is 1, we can write \(f(x)=\prod_{i=0}^{n-1}(x - 2^i).\) Thus, \[ \begin{align} f(0)&=\prod_{i=0}^{n-1}-2^i \\&= \prod_{i=0}^{n-1}(-1)\cdot2^i \\&= (-1)^n\;2^{\sum_0^{n-1}i} \\&= (-1)^n\;2^{n(n-1)/2}. \end{align} \] There are no duplicate roots since \(f\) is of degree \(n\) and we were told it has \(n\) distinct roots.

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