Question 1209693: What is the remainder when the polynomial
g(x) = x^3 - 14x^2 - 67x - 90 + 5x^3 - 20x^2 + 7x - 135
is divided by x - 1?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! To find the remainder when g(x) is divided by x - 1, we can use the Remainder Theorem. The Remainder Theorem states that if we divide a polynomial g(x) by (x - c), the remainder is g(c). In this case, we are dividing by x - 1, so c = 1.
First, let's simplify g(x) by combining like terms:
g(x) = (x^3 + 5x^3) + (-14x^2 - 20x^2) + (-67x + 7x) + (-90 - 135)
g(x) = 6x^3 - 34x^2 - 60x - 225
Now, we need to find g(1):
g(1) = 6(1)^3 - 34(1)^2 - 60(1) - 225
g(1) = 6 - 34 - 60 - 225
g(1) = -313
Therefore, the remainder when g(x) is divided by x - 1 is -313.
Final Answer: The final answer is $\boxed{-313}$
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