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if the quadratic 3x^2+bx+10 can be written in the form a(x+m)^2+n, where m and n are integers, what is the largest integer that must be a divisor of b?

Dec 26, 2018

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$$3x^2 + b x + 10 = \\ 3\left(x^2 + \dfrac b 3 x + \dfrac{10}{3}\right) = \\ 3\left(\left(x+\dfrac b 6\right)^2 - \dfrac{b^2}{36}+\dfrac{10}{3}\right) = \\ 3\left(x+\dfrac b 6\right)^2+\left(10-\dfrac{b^2}{12}\right)$$

$$m = \dfrac b 6\\ \text{so suppose that 6 is the largest integer that must be a divisor of }b\\ \dfrac{b^2}{36} = \dfrac{(6k)^2}{36} = \dfrac{36k^2}{36}=k^2 \in \mathbb{Z}\\ \text{so }n \in \mathbb{Z}\\ \text{So the number you are after is }6$$

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Dec 26, 2018