Binomial Expansion Practice Problems: Multiplying Binomials

[Meezus]

Homework Statement

Homework Equations

The Attempt at a Solution

After using this formula I got 1-3x + 6x² for (1-x)^-3 and 1+x - 1/8 ∙ 4x² for (1+2x)^1/2 for the second part of the question I'd assume that you're supposed to multiply the equations? I think this done by timing the whole of second bracket by first number (1) then second number (-3x) and then third number (6x².from the second bracket.

 

[blue_leaf77]

1-3x + 6x2

Check again the sign of the second term.

 

[Meezus]

Check again the sign of the second term.

I don't see where I'm going wrong here? (-3)(-4)(x)²(1/2) = 6x²

 

[blue_leaf77]

The second term, not the third one.

 

[PeroK]

for the second part of the question I'd assume that you're supposed to multiply the equations? I think this done by timing the whole of second bracket by first number (1) then second number (-3x) and then third number (6x².from the second bracket.

No one says "timing". The word is "multiplying". But, yes, that's just the distributive law.

 

[Meezus]

The second term, not the third one.

Sorry, I'm not seeing the mistake? -3(x) = -3x correct?

 

[blue_leaf77]

The formula for the binomial expansion uses a plus sign in front of $x$ to define the equation: $(1+x)^n$. While in the question, the sign of $x$ is minus.

 

[Meezus]

The formula for the binomial expansion uses a plus sign in front of $x$ to define the equation: $(1+x)^n$. While in the question, the sign of $x$ is minus.

Oh I was just looking at it as +(-3) which is same as just -3 so I can't use this formula?

 

[Meezus]

No one says "timing". The word is "multiplying". But, yes, that's just the distributive law.

My mistake, would you mind checking to see if I did the first part correct?

 

[blue_leaf77]

You can use that formula, but the sign of each $x$ appearing in the RHS (the infinite sum) must be adjusted accordingly. I will rewrite the formula using different variable notation for the sake of clarity
$$
(1+u)^{-3} = 1-3u+6u^2-10u^3+\ldots
$$
what if you change $u\rightarrow -x$ as you have in the problem?

 

[Meezus]

You can use that formula, but the sign of each $x$ appearing in the RHS (the infinite sum) must be adjusted accordingly. I will rewrite the formula using different variable notation for the sake of clarity
$$
(1+u)^{-3} = 1-3u+6u^2-10u^3+\ldots
$$
what if you change $u\rightarrow -x$ as you have in the problem?

ooohh it all becomes negative? I see! So its 1-3x - 6x2?

 

[blue_leaf77]

ooohh it all becomes negative?

No, that's not how you go about this.
Ok rather than an arrow sign, I will use an equality sign. Make a change $u=-x$ in that formula. Does it help this time?

 

[PeroK]

ooohh it all becomes negative? I see! So its 1-3x - 6x2?

It's not the binomial theorem that is the problem, it's your basic algebra, especially handling negative signs. I suggest you do a bit of revision of this.

 

[Meezus]

No, that's not how you go about this.
Ok rather than an arrow sign, I will use an equality sign. Make a change $u=-x$ in that formula. Does it help this time?

1-3-x + 6-x² - 10-x³

 

[blue_leaf77]

I know what you did there is simply replacing $u$ with $-x$ without giving heed to the rules of writing the accepted mathematical expression. What does -3-x in that expression mean?

 

[Meezus]

I know what you did there is simply replacing $u$ with $-x$ without giving heed to the rules of writing the accepted mathematical expression. What does -3-x in that expression mean?

I'm not sure how it should be expressed. -3-x = -(x+3). Should it be 1 - (3+x)? (IGNORE)

Its (-3)(-x) so it should be 3x because its -3 * -x. sorry

 

[blue_leaf77]

Its (-3)(-x) so it should be 3x because its -3 * -x. sorry

Yes, it should have been $+3x$.