# ProblemHow to determine the real parameters a and b so that the polynomial f=x^4 +(a-1)*x^3+(b+2)*x^2 + x +c-3, to be divided by the polynomial g=(x-1)^3?

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The problem provides the information that the polynomial f is divided by polynomial `(x - 1)^3` , hence, the reminder is equal to zero. Using reminder theorem yields:

`x^4 + (a - 1)*x^3 + (b + 2)*x^2 + x +c - 3 = (x - 1)^3(dx + e)`

You need to raise to cube `(x - 1)` such that:

`(x - 1)^3 = x^3 - 1 - 3x(x - 1) => (x - 1)^3 = x^3 - 1 - 3x^2 + 3x`

`x^4 + (a - 1)*x^3 + (b + 2)*x^2 + x + c - 3 = (x^3 - 1 - 3x^2 + 3x)(dx + e)`

Opening the brackets to the right side yields:

`x^4 + (a - 1)*x^3 + (b + 2)*x^2 + x + c - 3 = dx^4 - dx - 3dx^3 + 3dx^2 + ex^3 - e - 3ex^2 + 3ex`

Equating the coefficients of like powers yields:

`1 = d`

`a - 1 = -3d + e => a - 1 = -3 + e => a - e = -2`

`b + 2 = 3d - 3e => b + 2 = 3 - 3e => b + 3e = 1`

`1 = -d + 3e => 1 = -1 + 3e => 3e = 2 => e = 2/3`

`c - 3 = -e => c = 3 - e => c = 3 - 2/3 => c = 7/3`

`a = -2 + e => a = -2 + 2/3 => a = -4/3`

`b = 1 - 3e => b = 1 - 2 = > b = -1`

**Hence, evaluating a,b,c, under the given conditions, yields `a = -4/3, b = -1 ` and `c = 7/3.` **

If f(x) is divided by g(x)=(x-1)^3, that means that the roots of g(x) substituted in f(x), clear f(x). We've observed that the value "1" is the multiple root, of 3rd order, of g(x), meaning that:

f(1)=0 <=> f(1)=1+a-1+b+2+1+c-3=0 <=> a+b+c=0

f'(1)=0 <=> f'(1)=4*1+3*(a-1)+2*(b+2)+1=0<=>3a+2b=-6

f"(1)=0<=>f"(1)=12*1+6*(a-1)+2*(b+2)=0<=>6a+2b=-10

f"'(1) different from 0

3a+2b-6a-2b=-6+10

-3a=4, **a=-4/3**

3(-4/3)+2b=-6

-4+2b=-6, 2b=-6+4, 2b=-2, **b=-1**

a+b+c=0, **=(-4/3)+(-1)=-c**

**c=7/3**