Finding the sum of the coefficients of polynomial of degree 21

Question

Problem : Find the sum of the coefficients of the polynomial$ p(x) =(3x-2)^{17}(x+1)^4$

Solution :

$ p(x) =(3x-2)^{17}(x+1)^4 $

$= (a_0+a_1x+....a_{17}x^{17})(b_0+b_1x+....b_4x^4)$ for some $a_i; b_j$

= $(c_0+c_1x+...c_{21}x^{21})$ for some $c_k$

Sum of the coefficients of p(x) =$c_0+c_1+c_2+....c_{21}$ In solution it is given that the sum of the coefficients can be obtained by putting x =1 , my question is why x =1?

Please explain this concept.. else is clear in this problems... Thanks..

The degree of this polynomial is $21$, not $17$ – Ross Millikan May 10 '13 at 03:44 — Ross Millikan, May 10 '13 at 03:44

score 5 · Answer 1 · answered May 10 '13 at 03:42

5

If $P(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \cdots + a_{0}$, and we want to find $a_{n} + a_{n-1} + \cdots + a_{1} + a_{0}$, then we see that:

$P(1) = a_{n}(1)^{n} + a_{n-1}(1)^{n-1} + \cdots + a_{1}(1) + a_{0} = a_{n}(1) + a_{n-1}(1) + \cdots + a_{1}(1) + a_{0} = a_{n} + a_{n-1} + \cdots + a_{1} + a_{0}$

answered May 10 '13 at 03:42

Alex Wertheim

20,278

2

The intuitive way to say it is that setting $x=1$ gets rid of all the $x$'s – Ross Millikan May 10 '13 at 03:45

Finding the sum of the coefficients of polynomial of degree 21

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