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)   (Show Source): You can put this solution on YOUR website!



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.



Now they are conjugates.

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



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







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

Edwin

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