Coefficients in a Septic

There are $3$ ways to get a $x^4$ term in the expansion of $(x^2-x-2)^7$ : $\text{Zero } x^2 \text{ terms, four } (-x) \text{ terms, three } (-2) \text{ terms}$ $\text{One } x^2 \text{ term, two } (-x) \text{ terms, four } (-2) \text{ terms}$ $\text{Two } x^2 \text{ terms, zero } (-x) \text{ terms, five } (-2) \text{ terms}$ We can think of finding the coefficient of each of these by thinking of each $(x^2-x-2)$ in the expansion as boxes, and picking one of the terms from it (either $x^2, -x,$ $\text{ or }-2$ ). Thus, our coefficients, respectively, are $\dfrac{7!}{4!3!} \cdot (-x)^4 \cdot (-2)^3 = -280$ , $\dfrac{7!}{1!2!4!} \cdot x^2 \cdot (-x)^2 \cdot (-2)^4 = 1680$ , $\dfrac{7!}{2!5!} \cdot (x^2)^2 \cdot (-2)^5 = -672$ . Adding these all up yields $\boxed{728}$

if we will factorize $x^2-x-2 = (x-2)(x+1)$ , so $(x^2-x-2)^7 =(x-2)^7(x+1)^7$ . Then by expanding using the binomial theorem, we can calculate the coefficient of $x^4$ by multiplying the coefficients of $x$ in both expansions (e.g. $x^2$ in 1st * $x^2$ in 2nd + $x$ in 1st * $x^3$ in 2nd, etc) so by adding the products of the coefficients we will get 728 as the answer.

[Edits for clarity. Latex edits. - Calvin]

The general term for any binomial expression of 3 variables is

n! / {(p!)(q!)(r!)} (b^q)(c^r)..........................................(first equation)

where n is the exponent the binomial has been raised to p,q,r are the exponent variables of each of the 3 binomial terms a,b,c respectively.

Here n=7 (given),and 2p+q=4 (as we need to find the coefficients of x^4).Also a=x^2,b= -x and c= -2. Now make 3 columns of three variables (p,q,r). We need to find for what values of p,q,r; the above equations (n=7,and 2p+q=4 ) get satisfied.

We find three sets for (p,q,r) which are(0,4,3)(1,2,4)(2,0,5). for all these values of (p,q,r) we get the exponent of x as 4.

Thus we need to put these set of values in the equation

7! / {(p!)(q!)(r!)} ((-x)^q)((-2)^r)......................(this is the same as the first equation only with the variables a,b,c replaced by the appropriate values from the given binomial in the question)

and sum up the three values to get the final coefficient of x^4.

Putting (p,q,r)

as(0,4,3)we get -280

as(1,2,4)we get 1680

as(2,0,5)we get -672

Add these up to get +728.

The terms that will have degree four after expansion will come from multiplying out:

$x^2$ twice and -2 five times

$x^2$ once, -x twice, and -2 four times, and

-x four times and -2 three times

These are the only combinations of the seven factors whose degrees sum to four. There will be ${7 \choose 2}$ of the first kind, each with a coefficient of -32. There will be $7 \times {6 \choose 2}$ of the second kind, each with a coefficient of 16. And there will be ${7 \choose 4}$ of the third kind, each with a coefficient of -8. That means that the total coefficient of the $x^4$ term in the expansion will be the sum of all of these: 728.

First, I factored x^2-x-2 into x+1 and x-2. Since I already know (x+1)^7, using the Pascal’s triangle, I just need to work out x-2. I do this by listing the 7th row of Pascal’s triangle. (1 7 21 35 35 21 7 1) Then I label each of the numbers. (1 is zero, 7 is first, 21 is third, etc.) After that, I multiply each number by (-2)^label. Example: 1 (-2)^0 and 7 (-2)^1 and etc. Then, I take the coefficient of x^4 in x+1 and multiply it with the coefficient of x^0 in x-2. Similarly, I take the coefficient of x^3 in x+1 and multiply it with the coefficient of x^1 in x-2 and so on. When I add the results, I get 728.

[(x^2 - (x+2)]^7

7C0 [(x^2)^0] [(-(x+2))^7] + 7C1 [(x^2)^1] [(-(x+2))^6] + 7C2 [(x^2)^2]*[-((x+2))^5] From other terms coefficient of x^4 not possible

Coefficient of x^4 from 1st term(x^4 required form (-x-2)) 1 1 [7C4(-x)^4 (-2) 3] = -280

Coefficient of x^4 from 2nd term(x^2 required form (-x-2)) 7 1 [6C2(-x)^2 (-2) 4] = 1680

Coefficient of x^4 from 3rd term(x^0 required form (-x-2)) 21 1 [5C0(-x)^0 (-2) 5] = -672

adding all three -280 + 1680 -672 = 728

Intuitive solution: There are three ways of making $x^4$ : $x^2\cdot x^2$ , $x^2\cdot x\cdot x$ , and $x\cdot x\cdot x\cdot x$ .

First way: We multiply two $x^2$ 's. The remaining $5$ must be multiplying the $-2$ , or else it would get over $x^4$ . There are $\binom{7}{2}=21$ ways to choose these. Therefore, the running total for the coefficient of $x^4$ is $21\cdot (-2)^5=-672$ .

Second way: we first pick the $x^2$ , with $7$ different ways to pick it. In the remaining $6$ spots, we pick two $x$ 's, with $\binom{6}{2}=15$ ways. The remaining $4$ spots are $-2$ . Therefore, the coefficient of $x^4$ for this case is $7\cdot 15\cdot (-2)^4=1680$ .

Third way: We pick four $x$ 's, with $\binom{7}{4}=35$ different ways. The remaining $3$ spots are $-2$ . Therefore, the coefficient of $x^4$ is $35\cdot (-2)^3=-280$ .

We ad these up to get the final coefficient of $x^4$ which equals $-672+1680+-280=\boxed{728}$ .

We can write the given expression as $(x^{2}-(x+2))^{7}$ and then try to expand using binomial theorem for $(a+b)^{n}$ . In our case, $a=x^{2}$ and $b=-(x+2)^{2}$ . We are concerned only about terms which give us $x^{4}$ . There are following $3$ cases which can give us $x^{4}$ :

(1) $x^{4}$ from the first term (i.e. from $a=x^{2}$ ) and constant term from the second (i.e. from $b=-(x+2)^{2}$ ). In this case, coefficient from the first term is $^{7}C_{2}$ and from the second term is $2^{5}$ . Also, since the power is $5$ for the second term, we get negative sign.

(2) $x^{2}$ from first and $x^{2}$ from second. As in first case, we get $^{7}C_{1}*^{6}C_{4}*2^{4}=7*15*16$ with positive sign.

(3) Constant from first and $x^{4}$ from second. Here we get $^{7}C_{3}*2^{3}$ with negative sign.

Note that we can't get odd powers of $x$ from the first term since we have $x^{2}$ in the first term.

Thus the coefficient is : $-672+1680-280 = \boxed{728}$

$(x^2 - x - 2)^7 = (x-2)^7 \cdot (x+1)^7$

$= \displaystyle \sum_{i=0}^7 {7 \choose i} x^{7-i} (-2)^i \cdot \displaystyle \sum_{j=0}^7 {7 \choose j} x^j$

Choosing all $x^4$ terms : ${7 \choose 7} (-2)^7x^0\cdot {7 \choose 3}x^4 + {7 \choose 6} (-2)^6x^1\cdot {7 \choose 4}x^3 + ...$

Summing their coefficients, we get 728.

just use multinomial theorem

be careful for negative signs!!!

(x^2-x-2)^7=(x-2)^7(x+1)^7=(x^7+\binom{7}{1}(-2)x^6+\binom{7}{2}(-2)^2x^5+\binom{7}{3}(-2)^3x^4+\binom{7}{4}(-2)^4x^3+\binom{7}{5}(-2)^5x^2+\binom{7}{6}(-2)^6x^1+\binom{7}{6}(-2)^7)\times(x^7+\binom{7}{1}x^6+\binom{7}{2}x^5+\binom{7}{3}x^4+..............+1) we need only the coefficient of x^4 =(\binom{7}{3})(1)+(\binom{7}{4})(\binom{7}{6})+(\binom{7}{5})^2+(\binom{7}{6})(\binom{7}{4})+(\binom{7}{3}) solving we get 728

It can be factorized~ (x+1)^7(x-2)^7 Now two binomals..so find the coefficients of x^4 in both by binomial theorem and add.