Sum these coefficients

Algebra Level 2

Determine the sum of all coefficients of the polynomial

f ( x ) = ( 3 x 3 3 x 1 ) 38 ( 2 x + 1 ) 4 ( x 4 3 x 2 + 5 x 2 ) 77 . f(x) = (3x^3-3x-1)^{38}\cdot(2x+1)^4\cdot(x^4-3x^2+5x-2)^{77}.

Details and assumptions

After expanding out the expression for f ( x ) f(x) given in the problem, one will arrive at a formula that looks like f ( x ) = a n x n + a n 1 x n 1 + + a 1 x + a 0 f(x)=a_n x^n+a_{n-1}x^{n-1}+\dotsc+a_1 x+a_0 , where a 0 , a 1 , a 2 , , a n a_0,a_1,a_2,\dotsc,a_n are numbers. The sum of coefficients of f ( x ) f(x) is a n + a n 1 + + a 1 + a 0 a_n + a_{n-1} + \ldots + a_1 + a_0 .


The answer is 81.

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10 solutions

Marco Massa
Nov 25, 2013

We can have the sum of the cofficients of f(x) without expanding out the expression for f(x) given in the problem. This because of the sum of the coefficients of a polynomial is the value of that polynomial for x=1. So we have f(1)=3^4 = 81

Then how would you explain the expansion of ( x 2 x + 1 ) 2 (x^2-x+1)^2 . If we substitute x = 1 x=1 in this expression we get sum of coefficients as 1. But the above expression expands out to be x 4 2 x 3 + 3 x 2 2 x + 1 x^4-2x^3+3x^2-2x+1 , and the sum of its coefficients is 1 2 + 3 2 + 1 = 5 1-2+3-2+1=5 . What say?

Krystal D'Souza - 7 years, 6 months ago

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You get 1, not 5.

Joel Lee - 7 years, 6 months ago

I'm sorry but 1 - 2 + 3 - 2 + 1 = 1?

Andre Chan - 7 years, 6 months ago
Niranjan Patil
Dec 18, 2013

Sum of Coefficients = f(1) = 81

Michael David Sy
Nov 25, 2013

It's a fact that the sum of the coefficients of a given polynomial P(x) is P(1). Substituting 1 into the equation would give us,

f ( 1 ) = ( 3 [ 1 3 ] 3 [ 1 ] 1 ) 38 × ( 2 ( 1 ) + 1 ) 4 × ( 1 4 3 ( 1 2 ) + 5 ( 1 ) 2 ) 77 f(1) = (3[1^3] - 3[1] - 1)^{38} \times (2(1) + 1)^4 \times (1^4 - 3(1^2) + 5(1) - 2)^{77}

f ( 1 ) = ( [ 1 38 ] × [ 3 4 ] × [ 1 77 ] f(1) = ( [-1^{38}] \times [3^4] \times [1^{77}]

f ( 1 ) = 81 f(1) = \boxed {81}

Matija Mišak
Nov 25, 2013

When we want to sum coefficients of polynomial f ( x ) = a n x n + a n 1 x n 1 + . . . + a 1 x + a 0 f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0 , easiest way is to determine f ( 1 ) f(1) , because f ( 1 ) = a n 1 n + a n 1 1 n 1 + . . . + 1 a 1 + a 0 = a n + a n 1 + . . . + a 1 + a 0 f(1)=a_n1^n+a_{n-1}1^{n-1}+...+1a_1+a_0=a_n+a_{n-1}+...+a_1+a_0 , exactly what we want. Now to our task. We have our polynomial written as product of 3 polynomials. Result will be polynomial with degree of 38+4+77=119. But you don't have to multiply all those numbers. It would take you too much time. Just pluck x = 1 x=1 into this "factored" form. Outcome will be the same. So, when x = 1 x=1 we have: ( 3 × 1 3 3 × 1 1 ) 38 × ( 2 × 1 + 1 ) 4 × ( 1 4 3 × 1 2 + 5 × 1 2 ) 7 7 = (3\times 1^3-3\times 1-1)^{38} \times (2\times 1+1)^4 \times (1^4-3\times 1^2+5\times 1-2)^77 = = ( 1 ) 3 8 × ( 3 ) 4 × ( 1 ) 7 7 = 1 × 81 × 81 = 81 =(-1)^38 \times (3)^4 \times (1)^77=1\times 81\times 81=81

Juss Lunz
Nov 25, 2013

Let x = 1 then ull get the answer

Francisco Udaundo
Nov 29, 2013

the sum of the coefficients of the first quantity is -1 which when raised to a positive even power gives positive 1. The sum of the coefficients of the second quantity is 3 which when raised to the 4th power gives 81. For the last quantity, the sum of the coefficients is 1 and when raised to any positive integer gives 1. The product of 1, 81 and 1 is 81.

Daniel Ferreira
Nov 28, 2013

Encontramos a soma dos coeficientes de um polinômio qualquer calculando f ( 1 ) f(1) , segue,

f ( 1 ) = ( 3 1 3 3 1 1 ) 38 ( 2 + 1 ) 4 ( 1 3 + 5 2 ) 77 f ( 1 ) = 1 38 3 4 1 77 f ( 1 ) = 1 81 1 f ( 1 ) = 81 \\ f(1) = (3 \cdot 1^3 - 3 \cdot 1 - 1)^{38} \cdot (2 + 1)^4 \cdot (1 - 3 + 5 - 2)^{77} \\\\ f(1) = 1^{38} \cdot 3^4 \cdot 1^{77} \\\\ f(1) = 1 \cdot 81 \cdot 1 \\\\ \boxed{f(1) = 81}

Danny Chapple
Nov 28, 2013

f(1)= the sum of the coefficients, e.g. let f(x) = x^2+4x+4 = (x+2)^2. therefore f(1) = (1)^2 + 4(1) + 4 = 9 = sum of coefficients. so f(1) of question above = (3(1)^3 - 3(1) - 1)^38 x (2(1) + 1)^4 x ((1)^4 - 3(1)^2 + 5(1) - 2)^77 = 81

shack would be proud

Kinar Patel - 7 years, 6 months ago
Satvik Golechha
Mar 11, 2014

We can get sum of all coefficients by putting x=1 thus simply calculating, we get f(1)= 81

Ryan Lahfa
Feb 26, 2014

It's quite a brute force solution but, by expanding this expression using Wolfram Alpha, we can get something quite big. (http://www.wolframalpha.com/input/?_=1393460775966&i=%283x^3+-+3x+-+1%29^38++%282x+%2b+1%29^4++%28x^4+-+3x^2+%2b+5x+-+2%29^77&fp=1&incTime=true)

Next, we can remove all x^... by using a regex. Python code: "k = re.split('x\^?[0-9]*', s)" where s is the expanded expression. And sum it up all: "sum(int(x) for x in k)" and we got 81.

81

Archit Khandelwal - 7 years, 2 months ago

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