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The expression $6y^2-y-51$ can be rewritten as $(3Ay+B)(y-C)$, where $A$, $B$, and $C$ are positive integers. Find $(AC)^2-B$.

Guest Jul 10, 2020

geno3141 · Answer 1 · 2020-07-10T06:05:26

6y² - y - 51 = (3Ay + B)(y - C)

Multiplying out: (3Ay + B)(y - C) = 3Ay² + By - 3ACy - BC = 3Ay² + (B - 3AC)y - BC

Since 6y² - y - 51 = 3Ay² + (B - 3AC)y - BC

The y²-terms must be equal: 6y² = 3Ay² ---> 6 = 3A ---> A = 2

The y-terms must be equal: -y = (B - 3AC)y ---> -1 = B - 3AC

---> since A = 2 ---> -1 = B - 3(2)C ---> -1 = B - 6C

The constants must be equal: -51 = -BC --- BC = 51

Since we are dealing only with positive integers: B = 1 and C = 51

or B = 51 and C = 1

or B = 3 and C = 17

or B = 17 and C = 3

To get B - 6C = -1 we can choose B to be 17 and C to be 3 ---> 17 - 6(3) = 17 - 18 = -1

So: A = 2 and B = 17 and C = 3

I'll let you finish the problem ...