How do I expand (1 + x)^{2}(1 - 5x)^{14} as a series of powers of x?

[BOAS]

Hello,

I have a problem regarding the binomial theorem and a number of questions about what I can and can't do.

Homework Statement

Write the binomial expansion of [itex](1 + x)^{2}(1 - 5x)^{14}[/itex] as a series of powers of [itex]x[/itex] as far as the term in [itex]x^{2}[/itex]

Homework Equations

The Attempt at a Solution

I know how to expand each bracket separately but I'm really unsure of how to proceed with one multiplied by the other.

Do I expand the first one, and have that as a factor of every term in the expansion of the second?

i.e [itex](1 + x)^{2} = 1 + x^{2} + 2x[/itex]

[itex](1 - 5x)^{14} \approx 1 - 70x - 455x^{2}[/itex]

[itex](1 + x)^{2}(1 - 5x)^{14} \approx 1(1 + x^{2} + 2x) - 70x(1 + x^{2} + 2x) - 455x^{2}(1 + x^{2} + 2x)[/itex]

[itex](1 + x)^{2}(1 - 5x)^{14} \approx 1 + 2x + x^{2} - 70x - 140x^{2} - 70x^{3} - 455x^{2} - 910x^{3} - 455x^{4}[/itex]

I get the feeling that this is wrong, but I can't find any similar examples in my textbook or notes. If this happens to be the correct method, have I included to high powers? The individual expansions only reach [itex]x^{2}[/itex], but when they are combined, clearly it goes higher.

Thanks for any help you can give!

 

[vela]

The ##x^2## term in the expansion for ##(1-5x)^{14}## should be positive. Your work is otherwise okay, but you didn't need to calculate the ##x^3## and ##x^4## terms. You just want to identify which products will result in terms of order ##x^2## or lower and keep track of those.

 

[BOAS]

The ##x^2## term in the expansion for ##(1-5x)^{14}## should be positive. Your work is otherwise okay, but you didn't need to calculate the ##x^3## and ##x^4## terms. You just want to identify which products will result in terms of order ##x^2## or lower and keep track of those.

Ah, I forgot to square the coefficient. It should be;

[itex](1 - 5x)^{14} \approx 1 - 70x + 2275x^{2}[/itex]

wrt the rest of your post, does that mean I should approximate the two expansions only to the 'x' terms? Or do as before and ignore the higher powers?

Thanks for the help.

 

[vela]

You have to keep up to at least the ##x^2## terms because they will contribute to the final result.

 

[BOAS]

You have to keep up to at least the ##x^2## terms because they will contribute to the final result.

So,

[itex](1 + x)^{2}(1 - 5x)^{14} \approx 1(1 + 2x + x^{2}) - 70x(1 + 2x + x^{2}) + 2275x^{2}(1 + 2x + x^{2})[/itex]

[itex](1 + x)^{2}(1 - 5x)^{14} \approx 1 + 2x + x^{2} - 70x - 140x^2 + 2275x^{2}[/itex]

[itex](1 + x)^{2}(1 - 5x)^{14} \approx 1 - 68x + 2136x^2[/itex]

I think this is what you meant when you said to keep track of the products that would give me [itex]x^2[/itex] and lower.

 

[vela]

Yup, and you can streamline it a bit further:
$$(1 + x)^{2}(1 - 5x)^{14} \approx 1(1 + 2x + x^{2}) - 70x(1 + 2x) + 2275x^{2}(1)$$

 

[BOAS]

Yup, and you can streamline it a bit further:
$$(1 + x)^{2}(1 - 5x)^{14} \approx 1(1 + 2x + x^{2}) - 70x(1 + 2x) + 2275x^{2}(1)$$

Cool - Thank you

 

[PeroK]

[itex](1 + x)^{2}(1 - 5x)^{14} \approx 1 - 68x + 2136x^2[/itex]

That's correct. Although it might be interesting to consider for what values of x that approximation is accurate!

 

[BOAS]

That's correct. Although it might be interesting to consider for what values of x that approximation is accurate!

My expansion for [itex](1 + x)^{2}[/itex] was exact, but not every term of it was used when mutliplying with the second expansion.

My expansion for [itex](1 - 5x)^{14}[/itex] holds provided that [itex] -1 < -5x < 1[/itex] so [itex] \frac{1}{5} > x > - \frac{1}{5}[/itex]

I don't know how I combine this information.