Need help! Given (2-ax)^8, and a is positive constant. Also coefficient of x^2=112. Find a and coefficient x^3

Answer 1

coefficient of x^2 is
C(8;2)*2^6*a^2=112

Hence a^2=1/16

a=1/4

coefficient of x^3 is
C(8;3)*2^5 *(-a)^3=?

=56*32*(-1/4)^3
=28

Answer=28

Answer 2

Generally, if you are to expand (A+X)^n, it will have the summation of (n+1) terms. (Note : I'm using the uppercase A and X, so as not to confuse you with the ones in your question).
Expand (A+X)^n= Summation of binomial expression:

(A+X)^n === C(n, r)A^r X^(n-r) ----------------eqn1

See, n=8

We need an expression that has a term x^2, then we must put (n-r) to be 2, i.e (n-r)=2, therefore, r must be 6;

i.e r=6

So we must find the coefficient of (A+X)^8:
(A+X)^8 = C(8,6).A^6.X^2

Now, A=2 and X = -ax, where C(8,6)=C(8,2)=(8.7)/(2.1)=28

Put our parameters inside eqn1 (above), get:

(2-ax)^8===(28)(2^6)(-ax)^2
=1792(a^2)(x^2)

Now we can see that coefficient of x^2 is 1792(a^2), this is the value we need to work with, and equate it to 112 that you mentioned in the question.

1792(a^2) = 112
16(a^2)=1

But since you said a is always positive, we choose the -ve one.

a=¼

End of 1st part.
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For the second part of the question:
Using the same expansion as above, setting (n-r)==3, where

r=5, and the 4th term, (think of 3+1), of the expansion is
-1792(a^3)(x^3), and the coefficient of x^3 is simply -1792(a^3)

Since, a=¼

then that coefficient of x^3 = -1792(¼^3) = -28
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If you dont understand this process, then expand the original expressiion in its longest form, then pick the coefficients as required:, i.e.

(2-ax)^8 = 2^8 + 8(-ax).2^7 + 28(-ax)^2.(2^6) + 56(-ax)^3.2^5 + 70(-ax)^4.2^4 + 56(-ax)^5.2^3 + 28(-ax)^6.2^2 + 8(-ax)^7.2 + (-ax)^8

=256 - 1024(ax) + 1792(ax)^2 - 1792(ax)^3 + 1120(ax)^4 - 448(ax)^5 + 112(ax)^6 -16(ax)^7 + (ax)^8

Easily you can picture the 3rd term as having coefficient with x^2.

i.e 1792(a^2)=112

a=¼

And also you can picture the 4th term as having coefficient with x^3.

i.e: -1792(a^3) = -28

End of question.