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If (p(x)=x^2-11x+30) and (q(x)=p(x-2)), what are the zeroes of (q(x))?

Answer and explanation

Correct answer: (7,8)

The direct answer is option A, 7 and 8. First find the zeroes of \\(p(x)=x^2-11x+30\\). It factors as \\(p(x)=(x-5)(x-6)\\), so its zeroes are 5 and 6. Now \\(q(x)=p(x-2)\\). To find a zero of q, set \\(p(x-2)=0\\). The input of p, namely x-2, must therefore equal one of p's zeroes: \\(x-2=5\\) or \\(x-2=6\\). Solving gives \\(x=7\\) or \\(x=8\\). Thus option A is correct. Option B, 3 and 4, shifts the zeroes in the wrong direction by subtracting 2. Option C, 5 and 6, gives the original p zeroes and ignores the replacement x-2. Option D, 2 and 11, uses unrelated numbers from the expression and is not obtained by solving. A useful cue: replacing x by x-a shifts every zero to the right by a.

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Transformed-ZeroesPolynomialExpert

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