find the remainder when (x^40 -3) is divided by (x+1)
asdads
Well, to find the remainder when (x^40 - 3) is divided by (x+1), we can use long division or synthetic division. But let me tell you a funny story first!
Why did the math book go to therapy?
Because it had too many problems!
Now, back to the question. Let's use synthetic division to divide (x^40 - 3) by (x+1):
-1 | 1 0 0 0 ... 0 -3
| -1 1 -1 ... 1
-------------------------------
1 -1 1 -1 ... 1 -2
The remainder when (x^40 - 3) is divided by (x+1) is -2.
To find the remainder when a polynomial is divided by another polynomial, we can use the Remainder Theorem. According to the Remainder Theorem, if a polynomial f(x) is divided by (x - a), the remainder is equal to f(a).
In this case, we want to find the remainder when (x^40 - 3) is divided by (x + 1). To do this, we evaluate the polynomial (x^40 - 3) at x = -1.
Substituting x = -1 into (x^40 - 3) gives us:
(-1)^40 - 3 = 1 - 3 = -2
Therefore, the remainder when (x^40 - 3) is divided by (x + 1) is -2.
To find the remainder when dividing a polynomial by another polynomial, you can use polynomial long division. Here's how you can solve the problem step by step:
Step 1: Set up the long division. Write the dividend (x^40 - 3) inside the long division symbol (⌈⌉) and the divisor (x + 1) to the left.
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x + 1| x^40 + 0x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
Step 2: Divide the first term of the dividend (x^40) by the first term of the divisor (x). The result is x^39.
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x + 1| x^40 + 0x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
- x^40 - x^39
Step 3: Multiply the divisor (x + 1) by the quotient (x^39). Place the product (x^39 * x) below the dividend.
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x + 1| x^40 + 0x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
- x^40 - x^39
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- x^39
Step 4: Subtract the product (x^39 * x) from the terms of the dividend. Combine like terms.
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x + 1| x^40 + 0x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
- x^40 - x^39
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0x^40 - x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
Step 5: Repeat steps 2-4 for the next term in the dividend, which is -x^39 (- x^39).
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x + 1| x^40 + 0x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
- x^40 - x^39
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0x^40 - x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
- 0x^40 - 0x^39
Step 6: Carry on repeating steps 2-4 until you have considered all terms in the dividend.
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x + 1| x^40 + 0x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
- x^40 - x^39
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0x^40 - x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
- 0x^40 - 0x^39
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- x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
Step 7: Continue the process until you have considered all the terms.
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x + 1| x^40 + 0x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
- x^40 - x^39
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0x^40 - x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
- 0x^40 - 0x^39
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- x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
+ x^39 + x^38
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0x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
- 0x^39 - 0x^38
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0x^39 + 0x^38 + ... + 0x^1 + (-3)x^0
Step 8: Reach the remainder, which is 0x^39 + 0x^38 + ... + 0x^1 + (-3)x^0. Since the remainder is zero, we can conclude that (x^40 - 3) is perfectly divisible by (x + 1).
Therefore, the remainder when (x^40 - 3) is divided by (x + 1) is zero.
Use the Remainder theorem
let f(x) = x^40 - 3
if f(x) is divided by x+1 , the f(-1) is the remainder,
so f(-1) = ......