Expansion: Given that the coefficient of x² in the expansion of (1+px)^4 * (1+qx)^3 is -6 and that...?

Question

Given that the coefficient of x² in the expansion of

( 1 + px )^4 * ( 1 + qx )^3

is -6, and that

p + q = 1,

find the values of p and q.

The answers are:

p = 2, q = -1

or

p = -2, q = 3.

I know how to expand the equations... choose and everything. But apparently there's a less tedious way to find out the coefficient of x² with multiple expansions.

For a simpler one, such as (1+4x)(2-x)^6, I know that you get two terms with x²:

(1)......1*15(2^4)(-x)² = 240x²
(2)......4x*(6)(2^5)(-x) = -768x²

(1)+(2) = -528x²

And hence you get the coefficient of x².

But for (1+px)^4 * (1+qx)^3, it's obviously different because both expansions have powers and henceforth I'll have to multiply certain terms here and there to get the final coefficient of x² ... it's too complicated, though, if I'll have to expand everything. Is there a simpler way to do it, and could you provide a working? Thanks!

I won't say I'm *sure* of these answers, because the answers I got were p = -3/2 and q = 5/2, as well. These're just the answers my Math lecturer gave me. Yes, I've used the Binomial Theorem, but this is just a three-marks question, so I don't think I'd need to expand it so much. The expansions for the first three terms of both equations are

(1+px)^4 = 1 + 4px + 6p²x² + ...
(1+qx)^3 = 1 + 3qx + 3q²x² + ...

The first part of the question was to find the terms of both equations including and/or up to the x² term in said equations. I omitted it because I had no problem with that, but it seems now that you'll have to use the information from this part to work out the answer for the second part, hence my question about the x² and how you'd find the values of p and q from there.

Thanks again!

Answer 1

I would use Pascal's triangle. You only need to expand part of each power.

(1 + px)^4 = 1 + 4px + 6p²x² + ...
(1 + qx)^3 = 1 + 3qx + 3q²x² + ...

Now there are only three terms that need to be added to get the coefficient of x².

6p²(1) + 4p(3q) + 1(3q²) = -6

Now substitute q = 1 - p.

6p² + 12p(1 - p) + 3(1 - p)² + 6 = 0
6p² + 12p - 12p² + 3 - 6p + 3p² + 6 = 0
-3p² + 6p + 9 = 0
p² - 2p - 3 = 0
(p + 1)(p - 3) = 0
p = -1, 3

q = 1 - p
q = 2, -2

So the answers are

(p, q) = (-1, 2) or (3, -2).

Answer 2

how do we get x^2?
multiply x by x
multiply constant by x^2
hence those are the important terms from your expansion and you can go directly to the term needed.
The term in x from ( 1 + px )^4 is 4C1(px)
and the term in x from ( 1 + qx )^3 is 3C1(qx)
we get 12pqx^2
The constant term in ( 1 + px )^4 is 1
and the x^2 term from ( 1 + qx )^3 is 3C2(qx)^2
we get 3q^2x^2
The constant term in ( 1 + qx )^3 is 1
and the x^2 term from ( 1 + px )^4 is 4C2(px)^2
to get 6p^2x^2
hence, the term in x^2 in the product is
[6p^2 + 3q^2 +12pq]x^2
we know 6p^2 + 3q^2 +12pq = -6
hence 2p^2 + 4pq + q^2 = -2
we also know p + q = 1 thus q=1-p
substitute into other equation and solve for p and q.
we get p = 3, q = -2 or p = -1, q = 2

Answer 3

( 1 + px )^4 * ( 1 + qx )^3
Using the binomial expansion theorem and extracting only the x^2 terms,
6p^2x^2 + 12pqx^2 + 3q^2x^2 = - 6x^2
2p^2 + 4pq + q^2 = - 2
p + q = 1
q = 1 - p
2p^2 + 4p(1 - p) + (1 - p)^2 = - 2
2p^2 + 4p - 4p^2 + 1 - 2p + p^2 = - 2
- p^2 + 2p + 3 = 0
p^2 - 2p - 3 = 0
(p + 1)(p - 3) = 0
p = - 1, 3
q = 2, -2

(p,q) = (- 1, 2), (3, - 2)

Answer 4

Have you tried using binomial theorem?