SOLUTION: Multiplication of algebraic expression 1. (2x^2-3x+1)(2x^2+3x-1) Thankyouuu

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Question 979439: Multiplication of algebraic expression
1. (2x^2-3x+1)(2x^2+3x-1)
Thankyouuu

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

%282x%5E2-3x%2B1%29%282x%5E2%2B3x-1%29

You can do it like you have to when the things you're
multiplying aren't similar. 

Use the formula

(A+B+C)(D+E+F) = AD+AE+AF+BD+BE+BF+CD+CE+CF

with 

A=2x^2
B=-3x
C=1
D=2x^2
E=3x
F=-1

AD = (2x^2)(2x^2) = 4x^4
AE = (2x^2)(3x) = 6x^3
AF = (2x^2)(-1) = -2x^2
BD = (-3x)(2x^2) = -6x^3
BE = (-3x)(3x) = -9x^2
BF = (-3x)(-1) = 3x
CD = (1)(2x^2) = 2x^2
CE = (1)(3x) = 3x
CF = (1)(-1) = -1

4x^4+6x^3-2x^2-6x^3-9x^2+3x+2x^2+3x-1

Combine like terms and end up with

4x^4-9x^2+6x-1

------------------------------------------------

But since the two factors are similar you can make them
become conjugates and it's a good bit easier to do it this way
when you can:

Factor -1 out of the last two terms in the first parentheses.
Factor +1 out of the last two terms in the second parentheses.

%282x%5E2-%283x-1%29%29%282x%5E2%2B%283x-1%29%29

Now they are conjugates.

So when you FOIL that out the OUTER product and the INNER
product cancel and you get:

4x%5E4-%283x-1%29%5E2

To square the last binomial multiply it by itself
and use FOIL:

4x%5E4-%283x-1%29%283x-1%29

4x%5E4-%289x%5E2-6x%2B1%29

4x%5E4-9x%5E2%2B6x-1

Easier, but in general you'll have to do them the long way.

Edwin