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If \((ax+2)(bx+7) = 15x^2 + cx + 14\) for all values of \(x\) and \(a+b = 8\), w...


Question

If \((ax+2)(bx+7) = 15x^2 + cx + 14\) for all values of \(x\) and \(a+b = 8\), what are the possible values for \(c\)?

Options

A)
3 and 5
B)
6 and 35
C)
10 and 21
D)
31 and 41

The correct answer is D.

Explanation:

Choice D is correct. One can find the possible values of a and b in (ax+2)(bx+7) by using the given equation a + b = 8 and finding another equation that relates the variables a and b. Since (ax+2)(bx+7) = 15\(x^2\) + cx + 14, one can expand the left side of the equation to obtain ab\(x^2\) + 7ax + 2bx + 14 = 15\(x^2\) + cx + 14. Since ab is the coefficient of \(x^2\) on the left side of the equation and 15 is the coefficient of \(x^2\) on the right side of the equation, it must be true that ab = 15. Since a + b =8, it follows that b = 8 - a. Thus, ab = 15 can be rewritten as a(8-a) = 15, which in turn can be rewritten as \(a^2 - 8a + 15 = 0\). Factoring gives (a-3)(a-5) = 0. Thus, either a = 3 and b = 5, or a = 5 and b = 3. If a = 3 and b = 5, then (ax+2)(bx+7) = (3x+2)(5x+7) = 15\(x^2\)+31x+14. Thus, one of the possible values of c is 31. If a = 5, and b = 3, then (ax+2)(bx+7) = (5x+2)(3x+7) = 15\(x^2\) + 41x + 14. Thus, another possible value for c is 41. Therefore, the two possible values for c are 31 and 41.


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