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

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

BAFFLED.

Top 2 Answers

A few days ago

Anonymous

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).

4 years ago

houchins

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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Factoring – precalc?

Factoring – precalc?

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