Finding $m+n+k$ Given $f(x)=x^4-29x^3+mx^2+nx+k$

[anemone]

Given that $f(x)=x^4-29x^3+mx^2+nx+k$, and $f(5)=11$, $f(11)=17$ and $f(17)=23$, find the sum of $m+n+k$.

 

[evinda]

Given that $f(x)=x^4-29x^3+mx^2+nx+k$, and $f(5)=11$, $f(11)=17$ and $f(17)=23$, find the sum of $m+n+k$.

Spoiler

$f(5)=11 \Rightarrow 625-3625+25m+5n+k=11 \Rightarrow 25m+5n+k=3011$

$f(11)=17 \Rightarrow -23958+121m+11n+k=17 \Rightarrow 121m+11n+k=23975$

$f(17)=23 \Rightarrow -58956+289m+17n+k=23 \Rightarrow 289m+17n+k=58979$

From the relations $25m+5n+k=3011, 121m+11n+k=23975 \text{ and }289m+17n+k=58979$,we get that:

$k=-3734,m=195,n=374$

Therefore, the sum $m+n+k$ is equal to $-3165$ .

 

[kaliprasad]

Spoiler

we have f(x) = x+ 6 for x= 5 11 and 17

so f(x)= (x-5)(x-11)(X-17)Q(x) +x + 6
ax f(x) is a 4th oder polynomal so Q(x) = linear say x+ a
coefficient of $x^3 = - 29 = -5 - 11 - 17 + a$ so a = 4

so f(x) = (x-5)(x-11)(x-17)(x+4) + x + 6
put x = 1 to get
1- 29 + m + n + k = (-4) *(-10) *(-16) * 5 + 1 + 6 = - 3200 +7 = - 3193

or m + n + k = -3193 + 28 = - 3165

 

[anemone]

Well done, evinda and thanks for participating! :)

To be completely honest with you, I initially looked at this problem in a much more complicated way and hence I thought this problem is a hard one for certain.

- - - Updated - - -

Spoiler

we have f(x) = x+ 6 for x= 5 11 and 17

so f(x)= (x-5)(x-11)(X-17)Q(x) +x + 6
ax f(x) is a 4th oder polynomal so Q(x) = linear say x+ a
coefficient of $x^3 = - 29 = -5 - 11 - 17 + a$ so a = 4

so f(x) = (x-5)(x-11)(x-17)(x+4) + x + 6
put x = 1 to get
1- 29 + m + n + k = (-4) *(-10) *(-16) * 5 + 1 + 6 = - 3200 +7 = - 3193

or m + n + k = -3193 + 28 = - 3165

Yes, that method works as well and thanks for participating, kali!