# $P(x) = x^4 + ax^3 +bx^2 +cx+d$

Let $P(x) = x^4 + ax^3 +bx^2 +cx+d$ where $a, b, c, d$ are integers.

$P(x)$ is divided by $x-2012, x-2013, x-2014, x-2015, x-2016$ and has the remainders $24, -6, 4, -6, 24$ respectively.

What is the remainder when $P(x)$ is divided by $x-2017$ ?

$P(2012) = 24, \;P(2013) = -6, \;P(2014) = 4, \;P(2015) = -6, \;P(2016) = 24$

By Lagrange interpolation,

$P(2017) = \sum{24\cdot\frac{(2017-2013)(2017-2014)(2017-2015)(2017-2016)}{(2012-2013)(2012-2014)(2012-2015)(2012-2016)}} = 24+30+40+60+120=274$

By symmetry at $x=2014$ we can write $\ P(x) = 4 + a(x\!-\!2014)^2 + b(x\!-\!2014)^4$.

Evaluation implies that $\ a=-15,\ b=5\$ so $\,P(2017)=4-15(9)+5(81)=274$

• The problem assumes that the leading coefficient of $P(x)$ is equal to 1, so I think you should divide your $P(x)$ with 5? Then we get $P(2017)=\frac{274}{5}$? Not sure though... – Milan Mar 6 '17 at 15:44
• I made a mistake in calculation. Now my answer is the same as yours. Your method is interesting, please explain how you get $\ P(x) = 4 + a(x\!-\!2014)^2 + b(x\!-\!2014)^4$ ? – carat Mar 6 '17 at 15:47
• @Milan But the 5 values determine a unique quartic, so the problem must be misstated if it assumes the polynomial is monic. – Bill Dubuque Mar 6 '17 at 16:18
• @carat The symmetry shows it is an even polynomial function of $\,x-2014.\ \$ – Bill Dubuque Mar 6 '17 at 16:20
• Yeah, I got that wrong...Thanks for clearing it up! – Milan Mar 6 '17 at 16:25