$f(x) = x^n + a_{n-1}x^{n-1} \cdots + a_1x + a_0,$ 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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How else can you write a polynomial when you know all of its roots?
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… 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? -
… 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? -
Finding
$f(0)$is the same as evaluating this new expression at$0.$What do you obtain? -
What manipulations can you perform on exponents when the bases are equal?
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Specifically, try writing the product as a single number raised to a power, where the latter is an expression.
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Looking at the expression (sum) in the power, do you notice a familiar relationship between its terms?
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Seeing as the coefficient of
$x^n$is 1, we can write$f(x)=\prod_{i=0}^{n-1}(x - 2^i).$Thus,$$ \begin{aligned} f(0)&=\prod_{i=0}^{n-1}-2^i \\&= \prod_{i=0}^{n-1}(-1)\cdot2^i \\&= (-1)^n\;2^{\sum_0^{n-1}i} \\&= (-1)^n\;2^{n(n-1)/2}. \end{aligned} $$There are no duplicate roots since$f$is of degree$n$and we were told it has$n$distinct roots.