Determine the sum of all coefficients of the polynomial
f ( x ) = ( 3 x 3 − 3 x − 1 ) 3 8 ⋅ ( 2 x + 1 ) 4 ⋅ ( x 4 − 3 x 2 + 5 x − 2 ) 7 7 .
Details and assumptions
After expanding out the expression for 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 , where a 0 , a 1 , a 2 , … , a n are numbers. The sum of coefficients of f ( x ) is a n + a n − 1 + … + a 1 + a 0 .
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Then how would you explain the expansion of ( x 2 − x + 1 ) 2 . If we substitute 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 , and the sum of its coefficients is 1 − 2 + 3 − 2 + 1 = 5 . What say?
I'm sorry but 1 - 2 + 3 - 2 + 1 = 1?
Sum of Coefficients = f(1) = 81
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 ) 3 8 × ( 2 ( 1 ) + 1 ) 4 × ( 1 4 − 3 ( 1 2 ) + 5 ( 1 ) − 2 ) 7 7
f ( 1 ) = ( [ − 1 3 8 ] × [ 3 4 ] × [ 1 7 7 ]
f ( 1 ) = 8 1
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 , easiest way is to determine 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 , 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 into this "factored" form. Outcome will be the same. So, when x = 1 we have: ( 3 × 1 3 − 3 × 1 − 1 ) 3 8 × ( 2 × 1 + 1 ) 4 × ( 1 4 − 3 × 1 2 + 5 × 1 − 2 ) 7 7 = = ( − 1 ) 3 8 × ( 3 ) 4 × ( 1 ) 7 7 = 1 × 8 1 × 8 1 = 8 1
Let x = 1 then ull get the answer
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.
Encontramos a soma dos coeficientes de um polinômio qualquer calculando f ( 1 ) , segue,
f ( 1 ) = ( 3 ⋅ 1 3 − 3 ⋅ 1 − 1 ) 3 8 ⋅ ( 2 + 1 ) 4 ⋅ ( 1 − 3 + 5 − 2 ) 7 7 f ( 1 ) = 1 3 8 ⋅ 3 4 ⋅ 1 7 7 f ( 1 ) = 1 ⋅ 8 1 ⋅ 1 f ( 1 ) = 8 1
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
We can get sum of all coefficients by putting x=1 thus simply calculating, we get f(1)= 81
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
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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