From Cubic To Degree 6
If x 3 + 2 x 2 + 3 x = 4 , what is x 3 ( x 3 − 1 5 x 2 − 4 3 x − 9 5 ) ?
The answer is -64.
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3 solutions
I multiplied the given expression to get each term of the desired expression one at a time. I got lucky.
x 6 + 2 x 5 + 3 x 4 − 4 x 3 = 0 − 1 7 x 5 − 3 4 x 4 − 5 1 x 3 + 6 8 x 2 = 0 − 1 2 x 4 − 2 4 x 3 − 3 6 x 2 + 4 8 x = 0 − 1 6 x 3 − 3 2 x 2 − 4 8 x + 6 4 = 0
Add these equations to get the value of this expression to be -64.
Nice approach. The act of successively removing terms is akin to long division. If you look at the terms that you multiply by, it's exactly x 3 − 1 7 x 2 − 1 2 x − 1 6 as in my solution.
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Thanks, I didn't realize it until now. The proof of the lemma you posted will make an interesting exercise.
Put the value of x^3 from the first equation in the problem and then you will get a 4 degree polynomial. Multiply the first equation by x and then remove fourth power of x. Then you will get a three degree polynomial. Now when you would try to reduce it you would only get a constant left that is -64
I adopted this approach.
Performing long division / euclidean algorithm , we get that
x 6 − 1 5 x 5 − 4 3 x 4 − 9 5 x 3 = ( x 3 − 1 7 x 2 − 1 2 x − 1 6 ) ( x 3 + 2 x 2 + 3 x − 4 ) − 6 4 .
Hence, when x 3 + 2 x 2 + 3 x − 4 = 0 (2nd term of the factor), the expression is equal to -64.
Meta discussion about creating the problem:
How was I able to pull out x 3 as a term? Was it just luck?
Hint: More generally, show that if f ( x ) is a polynomial of degree n , then for every k there exists a polynomial g ( x ) of degree at most n and a constant K k such that
f ( x ) ∣ x k g ( x ) + K k
Then, I applied f ( x ) = x 3 + 2 x 2 + 3 x − 4 with k = 3 to create this problem.