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?

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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.

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