A few days ago
Factoring – precalc?
The first question on a test I failed today was the most complex and disgusting question I’ve ever seen, I got no points and don’t think I was even close.
X^-2/3 *(x-8)^4+x^1/3(x-8)^3 -2x^-2/3(x-8)^3
or
X to the negative 2/3rd times x minus 8 to the 4th + x to the one third power times x minus 8 to the third minus 2x to the negative 2/3 times x-8 to the third
I
am
BAFFLED.
Top 2 Answers
A few days ago
Favorite Answer
Let s = x^(-2/3) * (x – 8)^4 + x^(1/3) (x – 8)^3 – 2x^(-2/3) (x-8)^3.
Extract the common factor of x^(-2/3) (x – 8)^3:
s = x^(-2/3) (x – 8)^3 [ (x – 8) + x – 2 ]
Combine terms in the square brackets:
s = x^(-2/3) (x – 8)^3 [ 2x – 10 ]
Factor out the 2 from the square brackets:
s = 2x^(-2/3)(x – 8)^3 (x – 5).
0
4 years ago
permit’s say you have a function f(x) = ax^2 + bx + d. i’m calling the final term “d” considering we already have a “c”. Use polynomial long branch to divide this by ability of x-c See what situations you need to get a the remainder of 0. Then attempt to generalize this for polynomials of degree “n”.
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