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Algebra Difficulty 3.8 AMC 10/12 Find the answer Philippines

Problem:
Let ff be a polynomial function that satisfies f(x−5)=−3x2+45x−108f(x-5) = -3x^{2} + 45x - 108. Find the roots of f(x)f(x).

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

Solution

Solution:
Let y=x−5y = x - 5, so x=y+5x = y + 5.

Then f(y)=−3(x)2+45x−108=−3(y+5)2+45(y+5)−108f(y) = -3(x)^{2} + 45x - 108 = -3(y+5)^{2} + 45(y+5) - 108.

Expand (y+5)2(y+5)^{2}:
(y+5)2=y2+10y+25(y+5)^{2} = y^{2} + 10y + 25

So:
f(y)=−3(y2+10y+25)+45y+225−108f(y) = -3(y^{2} + 10y + 25) + 45y + 225 - 108
=−3y2−30y−75+45y+225−108= -3y^{2} - 30y - 75 + 45y + 225 - 108
=−3y2+15y+(225−108−75)= -3y^{2} + 15y + (225 - 108 - 75)
=−3y2+15y+42= -3y^{2} + 15y + 42

So f(y)=−3y2+15y+42f(y) = -3y^{2} + 15y + 42.

The roots of f(x)f(x) are the values of xx such that f(x)=0f(x) = 0:

−3x2+15x+42=0-3x^{2} + 15x + 42 = 0

Divide both sides by −3-3:
x2−5x−14=0x^{2} - 5x - 14 = 0

Factor:
(x−7)(x+2)=0(x - 7)(x + 2) = 0

So the roots are x=7x = 7 and x=−2x = -2.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.