# Find the constant term of polynomial

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 Im not sure it leads somewhere.

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

• @jayant98 It's OK as it is. – Parcly Taxel Feb 26 at 17:33
• It has the same constant term as $W\!+\!2x,\,$ which has roots $1,2,3,4,5$ so has constant term $(-7)(-5!)$ $\ \$ – Bill Dubuque Feb 26 at 20:26

## 1 Answer

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 $$(-7)(-1)(-2)(-3)(-4)(-5)=840$$.

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