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There's fifth degree polynomial, it's first coefficient equals $-7$. $$-7x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0$$ Also: $$W(1)=-2$$ $$W(2)=-4$$ $$W(3)=-6$$ $$W(4)=-8$$ $$W(5)=-10$$ Find the value of constant term.

It could be solved by system of equations. But I think that there's an easier way to do it. I've tried to sum some of the given values, and erase other coefficients, but I`m not sure it leads somewhere.

Could someone help me solve this and help me to understand it?

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  • $\begingroup$ @jayant98 It's OK as it is. $\endgroup$ – Parcly Taxel Feb 26 at 17:33
  • $\begingroup$ It has the same constant term as $W\!+\!2x,\,$ which has roots $1,2,3,4,5$ so has constant term $(-7)(-5!)$ $\ \ $ $\endgroup$ – Bill Dubuque Feb 26 at 20:26
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The obvious linear function fitting the five given points is $-2x$. We split that out from the polynomial: $$W(x)=[-7x^5+a_4x^4+a_3x^3+a_2x^2+(a_1+2)x+a_0]-2x$$ It is clear that the square-bracketed expression must be 0 at $x=1,2,3,4,5$, so can be written as $$-7(x-1)(x-2)(x-3)(x-4)(x-5)$$ whose constant term, and thus $a_0$, is $(-7)(-1)(-2)(-3)(-4)(-5)=840$.

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  • $\begingroup$ Oh I thought about the linear -2x But I had no clue how to use it here Now it`s all clear Thank you very much! $\endgroup$ – a_man_with_no_name Feb 26 at 17:37

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