When #P(x)=x^32x^2+ax+b# is divided by #(x2)#, remainder is#1# and when divided by #(x+1)#, remainder is #28#. Find #a# and #b#?
we will use the remainder theorem that states if
add the eqns
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using remainder theorem, when
Subtracting (b) from (a), we get
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To find the values of (a) and (b), we can use the Remainder Theorem, which states that if a polynomial (P(x)) is divided by (x  c), the remainder is (P(c)).
Given that (P(x) = x^3  2x^2 + ax + b) is divided by (x  2) with a remainder of 1, we have (P(2) = 1). Similarly, when (P(x)) is divided by (x + 1) with a remainder of 28, we have (P(1) = 28).
Now, we can substitute the values of (x) into (P(x)) and solve for (a) and (b):

When (x = 2): [P(2) = 2^3  2(2)^2 + 2a + b = 1] Simplifying: [8  8 + 2a + b = 1] [2a + b = 1] (Equation 1)

When (x = 1): [P(1) = (1)^3  2(1)^2  a + b = 28] Simplifying: [1 + 2  a + b = 28] [a + b = 27] (Equation 2)
Now, we have a system of two equations (Equation 1 and Equation 2) with two variables ((a) and (b)). We can solve this system to find the values of (a) and (b).
Adding Equation 1 and Equation 2: [2a + b + (a + b) = 1 + 27] [a + 2b = 28]
Now, we have one equation in terms of (a) and (b). We need another equation to solve for both (a) and (b).
Let's multiply Equation 2 by 2: [2(a + b) = 2(27)] [2a + 2b = 54]
Now, we can add this to our previous equation: [(a + 2b) + (2a + 2b) = 28 + 54] [a + 2b  2a + 2b = 82] [0a + 4b = 82] [4b = 82] [b = \frac{82}{4} = 20.5]
Now, we can substitute (b = 20.5) into Equation 1 to solve for (a): [2a + 20.5 = 1] [2a = 1  20.5 = 19.5] [a = \frac{19.5}{2} = 9.75]
So, the values of (a) and (b) are (a = 9.75) and (b = 20.5).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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