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The polynomial \(f(x)\) satisfies \(f(x+1)-f(x)=6x+4.\) Find the leading coefficient of \(f(x)\)

 

So far, I'm convinced that f(x) is a*(6x+4)+b, but I'm stuck. I tried 6, but the leading coefficient is not 6.

 

All help is appreciated! Thanks!

 Sep 9, 2020
 #1
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WLOG let f(x) be linear. So f(x)=ax+b. So \(f(x+1)=a(x-1)+b=ax-a+b\rightarrow ax-a+b-ax-b=-a\). Note that this is constant, so f(x) must be quadratic. Let f(x)=ax^2+bx+c. So \(f(x+1)=a(x+1)^2+b(x+1)+c=ax^2+2ax+bx+a+b+c\rightarrow \)\(f(x+1)-f(x)=2ax+c\implies 2a=6\rightarrow \boxed{a=3} \)

 Sep 9, 2020

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