Polynomial Special Product Formula

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jpanknin

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The formula is [imath](a - b)^2 = a^2 - 2ab + b^2[/imath]. For the following equation, why does the second term (-1) not get put into the formula as a (-1)?

[imath](3y - 1)^2 = (3y)^2 - 2(3y)(-1) + (-1)^2 = 9y^2 + 6y + 1[/imath]. The correct answer is [imath]9y^2 - 6y + 1[/imath], but this can only happen if the (-1) is substituted into the formula as a (1) instead of a (-1).

Or should I look at it as [imath](a - b)^2 = a^2 + 2a(-b) + (-b)^2[/imath] and [imath](3y - 1)^2 = (3y)^2 + 2(3y)(-1) + (-1)^2[/imath]?

The FOIL and distributive methods work fine, but my question is how does the second term in (3y - 1) get substituted into the identity?
 
jpanknin said:
[imath](3y - 1)^2 = (3y)^2 - 2(3y)(-1) + (-1)^2 = 9y^2 + 6y + 1[/imath]. The correct answer is [imath]9y^2 - 6y + 1[/imath], but this can only happen if the (-1) is substituted into the formula as a (1) instead of a (-1).
Click to expand...
b=1 works fine.
 
jpanknin said:
The formula is [imath](a - b)^2 = a^2 - 2ab + b^2[/imath]. For the following equation, why does the second term (-1) not get put into the formula as a (-1)?

[imath](3y - 1)^2 = (3y)^2 - 2(3y)(-1) + (-1)^2 = 9y^2 + 6y + 1[/imath]. The correct answer is [imath]9y^2 - 6y + 1[/imath], but this can only happen if the (-1) is substituted into the formula as a (1) instead of a (-1).

Or should I look at it as [imath](a - b)^2 = a^2 + 2a(-b) + (-b)^2[/imath] and [imath](3y - 1)^2 = (3y)^2 + 2(3y)(-1) + (-1)^2[/imath]?

The FOIL and distributive methods work fine, but my question is how does the second term in (3y - 1) get substituted into the identity?
Click to expand...
If a were 3y and b were -1, then you'd have [math](a - b)^2 = ((3y) - (-1))^2 =(3y)^2 - 2(3y)(-1) + (-1)^2 = 9y^2+6y+1.[/math] But that's [imath](3y+1)^2[/imath], not [imath](3y-1)^2.[/imath]

Many of us memorize only [imath](a + b)^2 = a^2 + 2ab + b^2[/imath], and for your expression would use [math](3y-1)^2 = ((3y) + (-1))^2 =(3y)^2 + 2(3y)(-1) + (-1)^2 = 9y^2-6y+1.[/math]
You can't do both at once! Either you add -1, or you subtract 1. Subtracting -1 is adding 1.

As I sometimes point out, in math we say exactly what we mean, and mean exactly what we say.
 
Dr.Peterson said:
If a were 3y and b were -1, then you'd have [math](a - b)^2 = ((3y) - (-1))^2 =(3y)^2 - 2(3y)(-1) + (-1)^2 = 9y^2+6y+1.[/math] But that's [imath](3y+1)^2[/imath], not [imath](3y-1)^2.[/imath]

Many of us memorize only [imath](a + b)^2 = a^2 + 2ab + b^2[/imath], and for your expression would use [math](3y-1)^2 = ((3y) + (-1))^2 =(3y)^2 + 2(3y)(-1) + (-1)^2 = 9y^2-6y+1.[/math]
You can't do both at once! Either you add -1, or you subtract 1. Subtracting -1 is adding 1.

As I sometimes point out, in math we say exactly what we mean, and mean exactly what we say.
Click to expand...
That second line is what I needed. Thank you.
 
jpanknin said:
But if [imath](a - b) = (a + (-b))[/imath], then how do you differentiate, especially in [imath](a - b)^2 = a^2 -2ab + b^2[/imath]?
Click to expand...
Read the rest of what I wrote!
[math](a - b)^2 = (a + (-b))^2 =a^2 +2a(-b) + (-b)^2= a^2 -2ab + b^2[/math]
Please explain what you mean by "differentiate".
 
jpanknin said:
But if [imath](a - b) = (a + (-b))[/imath], then how do you differentiate, especially in [imath](a - b)^2 = a^2 -2ab + b^2[/imath]?
Click to expand...
I assume you mean "how to distinguish" between the two versions. "Differentiate" has different meanings in math.

When using the subtraction version [imath](a - b)^2[/imath],
[imath](3y- 1)^2 \implies a = 3y \text{ and } b=1[/imath]

When using the addition version [imath](a + b)^2[/imath],
[imath](3y- 1)^2=(3y+(-1))^2 \implies a = 3y \text{ and } b=-1[/imath]
 
BeachBanana said:
I assume you mean "how to distinguish" between the two versions. "Differentiate" has different meanings in math.

When using the subtraction version [imath](a - b)^2[/imath],
[imath](3y- 1)^2 \implies a = 3y \text{ and } b=1[/imath]

When using the addition version [imath](a + b)^2[/imath],
[imath](3y- 1)^2=(3y+(-1))^2 \implies a = 3y \text{ and } b=-1[/imath]
Click to expand...
Ah, didn't even catch that "differentiate" meaning. Apologies.

This makes perfect sense with the two versions. Thank you.
 
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