Easy polynomial by me

Let $g(x)=P(x)-x$ .We see that $g(1)=g(2)=g(3)=0$ so $1,2,3$ are the roots of $g(x)$ .Which means that $g(x)$ is of the form $a(x-1)(x-2)(x-3)$ , $a$ being some constant(The polynomial is not monic,as it will be seen later).So: $a(x-1)(x-2)(x-3)=P(x)-x\\ P(x)=a(x-1)(x-2)(x-3)+x$ It is given that $P(4)=5$ ,so $a(4-1)(4-2)(4-3)+4=5\\ a(3)(2)(1)+4=5\\ 6a=1 \rightarrow a=\frac{1}{6}$ So the polynomial is $\frac{1}{6}(x-1)(x-2)(x-3)+x$ .Simply evaluating the polynomial at 6 gives the answer $\boxed{16}$

in these type of questions, you just construct your own polynomial that satisfies the given conditions of the question...

so, the polynomial can be worked out as :

$\frac{1}{6}$ $(x-1)(x-2)(x-3)$ + $x$

put the values of P(6) to get $\boxed{16}$

This solution is simple but requires a bit of brute force in solving a system of 4 linear equations. I'll post it anyway.

Let $P(x)$ be $ax^3 + bx^2 + cx + d$

Then:

$P(1)=a + b + c + d = 1$

$P(2)=8a+4b+2c+d=2$

$P(3)=27a+9b+3c+d=3$

$P(4)=64a+16b+4c+d=5$

The solution of the above system is

$(a,b,c,d)=(\frac {1} {6},-1,\frac {17} {6}, -1)$

$P(x)=\frac {x^3} {6} - x^2 + \frac {17x^2} {6} - 1$

$P(6) = 16$