The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field.

# 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?

The above links are provided as is. They are not affiliated with the Climb Foundation unless otherwise specified.

## Hints

• Hint 1
How 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 4
Finding $$f(0)$$ is the same as evaluating this new expression at $$0.$$ What do you obtain?
• Hint 5
What manipulations can you perform on exponents when the bases are equal?
• Hint 6
Specifically, try writing the product as a single number raised to a power, where the latter is an expression.
• Hint 7
Looking 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.

If you have queries or suggestions about the content on this page or the CSAT Practice Platform then you can write to us at . Please do not write to this address regarding general admissions or course queries.