Imaginary Numbers

P

plac6636

New member
Joined
Aug 29, 2005
Messages
19
Ok here's the question:
For the equation x^3-x^2-7x+15=0 show that 2-i is a solution by substituting it for x.

So I have figured out that the equation after you substitute the 2-i you get:
(2-i)^3 - (2-i)^2 -7(2-i) + 15 = 0

I know how to compute all but the first part of the equation now. For the last part I came up with -(4 - 4i + i^2 )- 14 + 7i +15 = 0 (I know I can simplify more but I wanted to know how to do the (2-i)^3 first. Can you give me some hints? I tried to foil 2 sets and then foil that answer with the third but that doesn't make sense.
 
I'm sorry, but I'm unclear on what you mean by "the first part" and "the last part" of the equation. Do you mean "the left-hand side" and "the right-hand side", respectively? (I don't think so, but....)

Please reply with clarification of what you are asking. Thank you.

Eliz.
 
Sorry...
I solved the -(2-i)^2 - 7(2-i) + 15 = 0 "parts" to get the -(4-4i+i^2)-14+7i+15=0...I just don't know what to do with the (2-i)^3 "part"...
 
Oh; so when you say you "solved this equation", you mean you "expanded these terms of the expression", a very different thing.

To multiply out (2 - i)<sup>3</sup>, multiply 2 - i by 2 - i, and then multiply the result by 2 - i.

Eliz.
 
ok...
So I FOILED the first two and got (4-4i+i^2)...so I have a 2-i left. Do you FOIL again? That doesn't make sense. There is where I got lost before.
 
This is why I hate it when educators rely on "FOIL": it leaves students having no idea how to do any multiplication other than one binomial by one binomial.

To learn how to multiply general polynomials, try here:

. . . . .FreeMathHelp lesson on Multiplying Polynomials

Eliz.
 
I say BAN “FOIL” altogether!! What a useless concept.
Teachers who use it are just contributing to the ignorance of algorithmic thinking.
(a+b)<SUP>3</SUP>=(a+b)(a+b)(a+b)=(a<SUP>2</SUP>+2ab+b<SUP>2</SUP>)(a+b)=a<SUP>3</SUP>+a<SUP>2</SUP>b+2a<SUP>2</SUP>b+2ab<SUP>2</SUP>+b<SUP>3</SUP>=a<SUP>3</SUP>+3a<SUP>2</SUP>b+3ab<SUP>2</SUP>+b<SUP>3</SUP>.

To find (2−i)<SUP>3</SUP> let a=2 and b=−i.
 
You must log in or register to reply here.
Top