Find $a+2b+3c$ Given $(x-1)^3$ is a Factor of $x^{10}+ax^2+bx+c$

[Albert1]

if $(x-1)^3 $ is a factor of $x^{10}+ax^2+bx+c$

please find :$a+2b+3c$

 

[mente oscura]

if $(x-1)^3 $ is a factor of $x^{10}+ax^2+bx+c$

please find :$a+2b+3c$

Hello.

I know that the idea is not very brilliant, but:

Spoiler

$$\dfrac{x^{10}+ax^2+bx+c}{x-1}=x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+(a+1)x+(a+b+1)+\dfrac{a+b+c+1}{x-1}$$

Therefore:

$$a+b+c+1=0$$. (1)

$$\dfrac{x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+(a+1)x+(a+b+1)}{x-1}=x^8+2x^7+3x^6+4x^5+5x^4+6x^3+7x^2+8x+(a+9)+\dfrac{2a+b+10}{x-1}$$

Therefore:

$$2a+b+10=0$$. (2)

$$\dfrac{x^8+2x^7+3x^6+4x^5+5x^4+6x^3+7x^2+8x+(a+9)}{x-1}=x^7+3x^6+6x^5+10x^4+15x^3+21x^2+28x+36+\dfrac{a+45}{x-1}$$

Therefore:

$$a+45=0$$. (3)

For (1), (2) and (3):

$$a=-45$$

$$b=80$$

$$c=-36$$

$$a+2b+3c=7$$

Regards.

 

[Opalg]

A slightly different method:
[sp]
Let $$f(x) = x^{10} + ax^2 + bx + c$$. If $(x-1)^3$ is a factor of $f(x)$ then $x=1$ is a triple zero of $f(x)$, which means that $f(x)$ and its first two derivatives must all vanish when $x=1$. Therefore $$f(1) = 1 + a + b + c = 0, \qquad(1)$$ $$f'(1) = 10 + 2a + b = 0, \qquad(2)$$ $$f''(1) = 90 + 2a = 0 \qquad(3).$$ These are the same three equations that mente oscura found, leading to the same solution $a+2b+3c = 7.$[/sp]

 

[kaliprasad]

Spoiler

$(x-1)^3$ is a factor of $P(x) = x^{10}+ax^2 + bx + c$
so $x^3$ is a factor of $P(x+1) = (x+1)^{10}+ a(x+1)^2+b(x+1) + c$
so in the above expression coefficient of $x^2$, x and constant term shall be zero
coefficient of $x^2$ = $\binom{10}{2}+ a = 0= 45 + a = 0 $
coefficient of $x$ = 10 + 2a + b = 0
constant term = 1 + a + b + c = 0

these are same equations as above 2 that is mente oscuras's and opalg's and so the ans are same as well