Suppose f(x) is a polynomial of degree 4 or greater such that f(1)=2, f(2)=3, and f(3)=5.
Find the remainder when f(x) is divided by (x-1)(x-2)(x-3).
To find the remainder when dividing a polynomial f(x) by a quadratic factor (x - a)(x - b), we can use the remainder theorem. According to the theorem, if f(x) is divided by (x - a)(x - b), then the remainder will have the same value as f(a) and f(b).
In this case, we are dividing f(x) by (x - 1)(x - 2)(x - 3). We are given that f(1) = 2, f(2) = 3, and f(3) = 5. So we can substitute these values into our polynomial to find the remainders.
First, let's set up the polynomial f(x) as f(x) = ax^4 + bx^3 + cx^2 + dx + e. Since the degree of f(x) is 4 or greater, we include the terms up to x^4.
Using the given values, we can set up the following equations:
f(1) = a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e = 2
f(2) = a(2)^4 + b(2)^3 + c(2)^2 + d(2) + e = 3
f(3) = a(3)^4 + b(3)^3 + c(3)^2 + d(3) + e = 5
Simplifying these equations, we get:
a + b + c + d + e = 2
16a + 8b + 4c + 2d + e = 3
81a + 27b + 9c + 3d + e = 5
We now have a system of linear equations in five variables (a, b, c, d, and e). Solving this system would give us the coefficients for f(x), and therefore, we can find the polynomial itself.
However, since the question asks for the remainder when f(x) is divided by (x - 1)(x - 2)(x - 3), we need to find the remainder directly using the remainders at x = 1, 2, and 3.
Let's denote the remainder by R(x). According to the remainder theorem, R(x) will have the same value as f(x) at x = 1, 2, and 3.
Therefore, the remainder R(x) when f(x) is divided by (x - 1)(x - 2)(x - 3) will be:
R(x) = f(x) - f(1) = f(x) - 2
Using f(1) = 2, we can write the remainder as R(x) = f(x) - 2.
So, the remainder when f(x) is divided by (x - 1)(x - 2)(x - 3) is R(x) = f(x) - 2.
To find the remainder when f(x) is divided by (x-1)(x-2)(x-3), we can use polynomial division. The remainder will be a polynomial of degree at most 2.
First, let's express the divisor as a product of linear factors:
(x-1)(x-2)(x-3) = (x^2 - 4x + 3)(x-3) = x^3 - 7x^2 + 15x - 9
Now, let's perform the polynomial division. We divide f(x) by (x-1)(x-2)(x-3) to find the quotient and remainder.
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(x-1)(x-2)(x-3) | f(x)
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To find the quotient, we start by dividing the highest degree term of f(x) by the highest degree term of our divisor.
Dividing x^4 by x^3 gives us x, so the first term in the quotient is x.
x
Now, we multiply the divisor (x^3 - 7x^2 + 15x - 9) by x and subtract the result from f(x):
x(x^3 - 7x^2 + 15x - 9)
- (x^4 - 7x^3 + 15x^2 - 9x)
(0x^3 - 7x^2 + 15x - 9)
Next, we repeat the process by dividing the highest degree term of the new polynomial (0x^3 - 7x^2 + 15x - 9) by the highest degree term of the divisor (x^3).
Dividing -7x^2 by x^3 gives us -7x, so the second term in the quotient is -7x.
x - 7
Now, we multiply the divisor (x^3 - 7x^2 + 15x - 9) by -7x and subtract the result from the current remainder:
-7x(x^3 - 7x^2 + 15x - 9)
- (0x^3 + 0x^2 - 0x + 63x)
(-7x^2 - 48x - 9)
We repeat the process once again by dividing the highest degree term of the new polynomial (-7x^2 - 48x - 9) by the highest degree term of the divisor (x^3).
Dividing -7x^2 by x^3 gives us -7x, so the third term in the quotient is -7x.
x - 7 - 7x
Now, we multiply the divisor (x^3 - 7x^2 + 15x - 9) by -7x and subtract the result from the current remainder:
-7x(x^3 - 7x^2 + 15x - 9)
- (-7x^3 + 49x^2 - 105x + 63x)
(42x^2 - 168x - 9)
At this point, we have a new polynomial (42x^2 - 168x - 9) with degree 2. This is our remainder.
Thus, the remainder when f(x) is divided by (x-1)(x-2)(x-3) is 42x^2 - 168x - 9.