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Given that \(a\)  is a multiple of \(1428\), find the greatest common divisor of \(a^2+9a+24\) and \(a+4\).

 Feb 7, 2018
 #1
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Since 1428 = 2^2 * 3 * 7 * 17, and the first factor, or 4, is common to the polynomial

[a^2 + 9 a + 24 and  a + 4], then no matter what value "a" takes, the GCD of a*1428 and the polynomial will always be the same, namely: 2 x 2 = 4. 

 Feb 7, 2018
 #2
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We can use the Euclidean Algorithm. \(\begin{align*} &\text{gcd}\,(a^2+9a+24,a+4) \\ &\qquad=\text{gcd}\,(a^2+9a+24-(a+5)(a+4),a+4)\\ &\qquad=\text{gcd}\,(a^2+9a+24-(a^2+9a+20),a+4)\\ &\qquad=\text{gcd}\,(4,a+4). \end{align*}\) Since \(4\) is a factor of \(a\) and thus \(a+4\) , the greatest common divisor is \(\boxed{4}\).

 Feb 7, 2018

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