Go Function 3

Observe that, we have $f(x)=x-1$ for $x=2,3,4,5$ . Thus we can say that the $4$ th degree polynomial $f(x)-(x-1)$ has four roots $2,3,4,5$ . Hence, $f(x)-(x-1)=A(x-2)(x-3)(x-4)(x-5)$ where $A$ is constant. Since $f(x)$ is monic, we have $A=1$ . Putting $x=1$ on both sides of this identity, we have $f(1)-0=(-1)(-2)(-3)(-4)=24$ Thus, $1+a+b+c+d=24$ and hence, $a+b+c+d=23\hspace{10pt}\blacksquare$

I actually substituted everything. $\left\{\begin{array}{rcl}16+8a+4b+2c+d&=&1\\81+27a+9b+3c+d&=&2\\256+64a+16b+4c+d&=&3\\625+125a+25b+5c+d&=&4\end{array}\right.$ $a=-14,b=71,c=-153,d=119$ $a+b+c+d=\fbox{23}$

Let $2,3,4,5$ be the roots then $f(x)=(x-2)(x-3)(x-4)(x-5) +x-1$

On simplifying you will get the result as $23$ .

Solution 1:

By making odd and even terms in x cancel out, we can write

f(x+1)+f(x-1)-2f(x) = 12x² + 6ax + 2b + 2

So f(4)+f(2)-2f(3) = 0 = 108 + 18a + 2b + 2 and f(5)+f(3)-2f(4) = 0 = 192 + 24a + 2b + 2

=> 18a + 108 = 24a + 192 => 6a = -84 => a = -14

=> 2b - 18*14 + 110 = 0 => b = 71

Replacing we get

1 = 16 - 8 * 14 + 4 * 71 + 2c + d, 2 = 81 - 27 * 14 + 9 * 71 + 3c + d

=> c = -153 => d = 119

So a+b+c+d = -14+71-153+119 = 23

Solution 2:

By observation, f(x) = x-1, or f(x)-x+1 = 0, has roots 2, 3, 4, and 5. So f(x)-x+1 = (x-2)(x-3)(x-4)(x-5), or f(x) = (x-2)(x-3)(x-4)(x-5)+x-1. It follows that 1+a+b+c+d = f(1) = (-1)(-2)(-3)(-4)+1-1, a+b+c+d = 23.