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September 10th, 2019, 10:47 AM  #1 
Newbie Joined: Jan 2017 From: London Posts: 17 Thanks: 0  What is this question asking for? Polynomial
Question: Find S as a polynomial in x where $\displaystyle S=(x1)^4+4(x1)^3+6(x1)^2+4(x1)+1 $ Understanding what is being asked here would be a useful first step Do I need to expand the brackets or what? I understand a polynomial to be an expression involving sums of powers of a variable e.g. x and the powers have to be non negative to qualify as a polynomial Last edited by mathsonlooker; September 10th, 2019 at 11:06 AM. 
September 10th, 2019, 10:58 AM  #2 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 
I think there's something missing in the wording. "Where," makes sense when followed by an equation, inequality, or condition, not an expression. But yes, my guess is they just want you to expand that and write the expression as $\displaystyle S=a_4 x^4 + a_3 x^3 +a_2 x^2 + a_1 x + a_0$ (find $\displaystyle a_0, a_1, ...$). 
September 10th, 2019, 11:07 AM  #3 
Newbie Joined: Jan 2017 From: London Posts: 17 Thanks: 0 
Sorry my bad. It's $\displaystyle S=(x1)^4+4(x1)^3+6(x1)^2+4(x1)+1 $ 
September 10th, 2019, 11:25 AM  #4 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 
Yes, just expand the brackets and combine like terms.

September 10th, 2019, 11:28 AM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403 
Basically they want to you expand the whole thing out and then regroup all the terms in powers of $x$ That being said. The structure of the coefficients (they are binomial coefficients) suggests some sort of mathematical sleight of hand to make this problem less brain numbing. $S = \sum \limits_{k=0}^4 \dbinom{4}{k}(x1)^k \cdot 1^{4k}$ Using the binomial theorem this is $S = ((x1) + 1)^4 = x^4$ Last edited by romsek; September 10th, 2019 at 11:36 AM. 
September 10th, 2019, 11:30 AM  #6 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90  
September 10th, 2019, 11:35 AM  #7 
Senior Member Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403  
September 10th, 2019, 11:44 AM  #8 
Newbie Joined: Jan 2017 From: London Posts: 17 Thanks: 0  
September 10th, 2019, 11:47 AM  #9  
Newbie Joined: Jan 2017 From: London Posts: 17 Thanks: 0  Quote:
 

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