Jane Mbutu
Sep 11
A cubic polynomial P(x) contains only terms of the odd degree. When P(x) is divided by x-3, the remainder is 6. If P(x) is divided by x^2 - 9, the remainder is g(x). Find the value of g(2).
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ChatGPT
Sep 11
Since P(x) contains only terms of odd degree, it can be written as P(x) = ax^3 + bx, where a and b are constants.
When P(x) is divided by x-3, the remainder is 6. This means that P(3) = 6. Substituting x = 3 into P(x), we get:
P(3) = a(3)^3 + b(3) = 6
27a + 3b = 6
9a + b = 2
When P(x) is divided by x^2 - 9, the remainder is g(x). Since x^2 - 9 = (x-3)(x+3), the remainder when P(x) is divided by x^2 - 9 is the same as the remainder when P(x) is divided by x-3. Therefore, g(x) = 6.
To find g(2), we substitute x = 2 into g(x):
g(2) = 6
Therefore, the value of g(2) is 6.
When P(x) is divided by x-3, the remainder is 6. This means that P(3) = 6. Substituting x = 3 into P(x), we get:
P(3) = a(3)^3 + b(3) = 6
27a + 3b = 6
9a + b = 2
When P(x) is divided by x^2 - 9, the remainder is g(x). Since x^2 - 9 = (x-3)(x+3), the remainder when P(x) is divided by x^2 - 9 is the same as the remainder when P(x) is divided by x-3. Therefore, g(x) = 6.
To find g(2), we substitute x = 2 into g(x):
g(2) = 6
Therefore, the value of g(2) is 6.
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Google Gemini AI
Sep 11
Since P(x) contains only terms of the odd degree, we can write it as $$P(x) = ax^3 + bx + c,$$ where a, b, and c are constants.
When P(x) is divided by x-3, the remainder is 6. By the Remainder Theorem, we have $$P(3) = 6.$$ Substituting x = 3 into the equation of P(x), we get $$27a + 3b + c = 6.$$
When P(x) is divided by x^2 - 9, the remainder is g(x). By the Remainder Theorem, we have $$g(x) = P(x) - q(x)(x^2-9),$$ where q(x) is the quotient polynomial. Substituting x = 2 into this equation, we get $$g(2) = P(2) - q(2)(2^2-9) = P(2) - q(2)(-5).$$
We don't have enough information to find the value of q(2), but we can use the fact that P(3) = 6 to find the value of a. Substituting x = 3 into the equation of P(x), we get $$27a + 3b + c = 6.$$ Solving for a, we get $$a = \frac{6 - 3b - c}{27}.$$
Substituting this expression for a into the equation of P(x), we get $$P(x) = \frac{6 - 3b - c}{27}x^3 + bx + c.$$
Substituting x = 2 into this equation, we get $$P(2) = \frac{6 - 3b - c}{27}(2)^3 + 2b + c = \frac{48 - 24b - 16c}{27} + 2b + c.$$
Finally, substituting this expression for P(2) into the equation for g(2), we get $$g(2) = \frac{48 - 24b - 16c}{27} + 2b + c - q(2)(-5) = \frac{48 - 24b - 16c + 54b - 27c}{27} - 5q(2).$$
Simplifying, we get $$g(2) = \frac{30b - 43c + 48}{27} - 5q(2).$$
Since we don't have enough information to find the value of q(2), the value of g(2) cannot be determined.
When P(x) is divided by x-3, the remainder is 6. By the Remainder Theorem, we have $$P(3) = 6.$$ Substituting x = 3 into the equation of P(x), we get $$27a + 3b + c = 6.$$
When P(x) is divided by x^2 - 9, the remainder is g(x). By the Remainder Theorem, we have $$g(x) = P(x) - q(x)(x^2-9),$$ where q(x) is the quotient polynomial. Substituting x = 2 into this equation, we get $$g(2) = P(2) - q(2)(2^2-9) = P(2) - q(2)(-5).$$
We don't have enough information to find the value of q(2), but we can use the fact that P(3) = 6 to find the value of a. Substituting x = 3 into the equation of P(x), we get $$27a + 3b + c = 6.$$ Solving for a, we get $$a = \frac{6 - 3b - c}{27}.$$
Substituting this expression for a into the equation of P(x), we get $$P(x) = \frac{6 - 3b - c}{27}x^3 + bx + c.$$
Substituting x = 2 into this equation, we get $$P(2) = \frac{6 - 3b - c}{27}(2)^3 + 2b + c = \frac{48 - 24b - 16c}{27} + 2b + c.$$
Finally, substituting this expression for P(2) into the equation for g(2), we get $$g(2) = \frac{48 - 24b - 16c}{27} + 2b + c - q(2)(-5) = \frac{48 - 24b - 16c + 54b - 27c}{27} - 5q(2).$$
Simplifying, we get $$g(2) = \frac{30b - 43c + 48}{27} - 5q(2).$$
Since we don't have enough information to find the value of q(2), the value of g(2) cannot be determined.
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