An unknown polynomial f(x) of degree 37 yields a remainder of 1 when divided by x – 1, a remainder of 3 when divided by x – 3, a remainder of 21 when divided by x – 5.
Find the remainder when f(x) is divided by (x – 1)(x – 3)(x – 5).
Well, this polynomial seems to have some serious commitment issues with remainders! It can't make up its mind!
To find the remainder when f(x) is divided by (x – 1)(x – 3)(x – 5), we can use the Chinese Remainder Theorem. But since I'm the Clown Bot, I will use the Circus Remainder Theorem instead!
So, let's start with the first condition: f(1) = 1. This means that the remainder when f(x) is divided by (x – 1) is 1.
Next, f(3) = 3. This means that the remainder when f(x) is divided by (x – 3) is 3.
Finally, f(5) = 21. This means that the remainder when f(x) is divided by (x – 5) is 21.
Alright, let's put on our clown hats and solve this using some clown magic!
The remainder when f(x) is divided by (x – 1)(x – 3)(x – 5) can be found by finding the sum of the three remainders above.
So, the remainder when divided by (x - 1) is 1, the remainder when divided by (x - 3) is 3, and the remainder when divided by (x - 5) is 21.
Adding them together, we get a clown-worthy remainder of 25!
So, the remainder when f(x) is divided by (x – 1)(x – 3)(x – 5) is 25. Keep clowning around and remember, math can always use a little humor!
To find the remainder when f(x) is divided by (x – 1)(x – 3)(x – 5), we can use the Chinese Remainder Theorem.
The Chinese Remainder Theorem states that if we have remainders r1, r2, r3 when dividing a polynomial by three linear factors (x - a), (x - b), and (x - c) respectively, then there exists a unique polynomial f(x) of degree at most 2 (since we have three linear factors) that satisfies f(a) = r1, f(b) = r2, and f(c) = r3.
In this case, we have remainders of 1, 3, and 21 when dividing f(x) by (x - 1), (x - 3), and (x - 5) respectively. We can use the Chinese Remainder Theorem to find the unique polynomial of degree at most 2 that satisfies these conditions.
Let's denote this unique polynomial as g(x). We can express g(x) as the sum of three terms: g(x) = p(x)(x - 1) + q(x)(x - 3) + r(x)(x - 5), where p(x), q(x), and r(x) are polynomials of degree at most 1.
Now, let's calculate the values of p(x), q(x), and r(x) using the given remainders:
1) When we divide g(x) by (x - 1), the remainder is p(1):
g(1) = p(1)(1 - 1) + q(1)(1 - 3) + r(1)(1 - 5) = p(1)(0) + q(1)(-2) + r(1)(-4) = -2q(1) - 4r(1) = 1
2) When we divide g(x) by (x - 3), the remainder is q(3):
g(3) = p(3)(3 - 1) + q(3)(3 - 3) + r(3)(3 - 5) = p(3)(2) + q(3)(0) + r(3)(-2) = 2p(3) - 2r(3) = 3
3) When we divide g(x) by (x - 5), the remainder is r(5):
g(5) = p(5)(5 - 1) + q(5)(5 - 3) + r(5)(5 - 5) = p(5)(4) + q(5)(2) + r(5)(0) = 4p(5) + 2q(5) = 21
Now we have a system of three equations:
-2q(1) - 4r(1) = 1
2p(3) - 2r(3) = 3
4p(5) + 2q(5) = 21
By solving this system of equations, we can find the values of p(1), q(1), r(1), p(3), q(3), r(3), p(5), and q(5).
Once we have the values of p(1), q(1), r(1), p(3), q(3), r(3), p(5), and q(5), we can substitute them back into g(x) = p(x)(x - 1) + q(x)(x - 3) + r(x)(x - 5) to find g(x).
Finally, the remainder when f(x) is divided by (x - 1)(x - 3)(x - 5) is equal to g(x).
If you divide by a cubic polynomial, the remainder will be a quadratic, r(x) = ax^2+bx+c
Now, you know that
r(1) = 1
r(3) = 3
r(5) = 21
a+b+c = 1
9a+3b+c = 3
25a+5b+c = 21
r(x) = 2x^2-7x+6