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Algebra Difficulty 4.2 AIME Find the answer Philippines

Problem:
The polynomial p(x)=x23x+1p(x) = x^{2} - 3x + 1 has zeros rr and ss and a quadratic polynomial q(x)q(x) has leading coefficient 11 and zeros r3r^{3} and s3s^{3}. Find q(1)q(1).

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:
We have r+s=3r + s = 3 and rs=1rs = 1. By Vieta's formulas, we compute
r3+s3=(r+s)33rs(r+s)=273(1)(3)=18 r^{3} + s^{3} = (r + s)^{3} - 3rs(r + s) = 27 - 3(1)(3) = 18
and r3s3=1r^{3} s^{3} = 1. Thus, we obtain q(x)=(xr3)(xs3)=x2(r3+s3)x+r3s3=x218x+1q(x) = (x - r^{3})(x - s^{3}) = x^{2} - (r^{3} + s^{3})x + r^{3} s^{3} = x^{2} - 18x + 1. Hence, we get q(1)=16q(1) = -16.

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