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

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If \((x-1)^3\) is a factor of \(x^{10}+ax^2+bx+c\), then \(f(1) = 1 + a + b + c = 0\), \(f'(1) = 10 + 2a + b = 0\), and \(f''(1) = 90 + 2a = 0\) must all hold true. Solving these equations reveals that \(a + 2b + 3c = 7\). The method used confirms the necessity of \(x=1\) being a triple root for the polynomial. The calculations align with previously established results, reinforcing the conclusion. Thus, the value of \(a + 2b + 3c\) is definitively 7.
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if $(x-1)^3 $ is a factor of $x^{10}+ax^2+bx+c$

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

\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.
 
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]
 
$(x-1)^3$ 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
 
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