Question

find the quadratic polynomial whose zeroes are reciprocal of the polynomial p(x)=x^2-x-2

Answer 1

[0010001111]... Hello User... [1001010101]
Here's your answer...

${x}^{2} - x - 2 \\ = {x}^{2} - 2x + x - 2 \\ = (x - 2)(x + 1)$
To find zeroes of this polynomial, we equate it to 0.
$(x - 2)(x + 1) = 0 \\ x = 2 \: or \: - 1$
The reciprocals of these will be 1/2 and - 1.
The new polynomial will be...
${x}^{2} - ( \frac{1}{2} - 1)x + \frac{1}{2} \times ( - 1) \\ = {x}^{2} + \frac{1}{2} x - \frac{1}{2}$
This polynomial can also be written as...
$2 {x}^{2} + x - 1$
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