few needing help with

SMITH4

New member
Joined
Jan 26, 2006
Messages
8
Here is the first one. I think the answer is A

Divide x^4 + x^2 + 1 by x^2 - x + 1 . The quotient is

a. x^2 - x - 1
b. x^2 + x + 1
c. x^2 - x + 1
d. x^2 - 2x - 1
e. none of these

Second one, I think the answer is c.

(x + 1)(2x - 3)(3x - 4) = Ax^3 + Bx^2 + Cx + D but only if C =

a. -11
b. -5
c. 6
d. 12
e. none of these
 
Hello, SMITH4!

Both your answers are wrong . . .

\(\displaystyle \text{Divide }x^4\,+\,x^2\,+\,1\text{ by }x^2\,-\,x\,+\,1.\;\text{The quotient is:}\)

\(\displaystyle (a)\;x^2\,-\,x\,-\,1\;\;(b)\;x^2\,+\,x\,+\,1\;\;(c)\;x^2\,-\,x\,+\,1\;\;(d)\;x^2\,-\,2x\,-\,1\;\;(e)\text{ none of these}\)
Did you do the division? . . . You should get: \(\displaystyle \,(b)\;x^2\,+\,x\,+\,1\)

You could have checked your answr by multiplying.

[In fact, you could have found the right answer by multiplying!]


\(\displaystyle (x\,+\,1)(2x\,-\,3)(3x\,-\,4)\:=\:Ax^3\,+\,Bx^2\,+\,Cx\,+\,D,\,\text{ but only if }C\,=\,\)__

\(\displaystyle a)\;\)-\(\displaystyle 11\;\;\;b)\;\)-\(\displaystyle 5\;\;\;c)\;6\;\;\;d)\;12\;\;\;e)\text{ none of these}\)
Did you multiply that out?
It comes out to: \(\displaystyle \,6x^3\,-\,11x\,-\,5x\,+\,12\)

The answer is: \(\displaystyle \,(b)\;\)-\(\displaystyle 5\)
 
Top