f is divisible by g . g = ( x - 1 )^3 f = x^4 + ( m - 1 )*x^3 + ( n + 2 )*x^2 + x + p - 3

2 Answers | Add Yours

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

f(x) =x^4+(m-1)x^3+(n+2)x^2+x+p^3.

 g(x) = (x-1)^3.

f(x) is is divisible by (x-1)^3.

Solution :

Since f(x)  the 4th degree is divisible by  (x-1)^3, w

 we can write, f(x) as product of (x-1)^3 and ax+b

f(x)  = g(x) { ax+b) Or

 x^4 + ( m - 1 )*x^3 + ( n + 2 )*x^2 + x + p - 3 = (x-1)^3 (ax+b).

This is an  identtity,  so we can equate coefficients of  equal powers on both sides.

x^4 = ax^4. Or a = 1. 

 constant term p-3 = -b, or -b =3-p

Therefore ,

 x^4 + ( m - 1 )*x^3 + ( n + 2 )*x^2 + x + p - 3 = (x-1)^3*(x+3-p)

Now diffrentiate both sides:

4x^3+3(m-1)x^3 +2(n+2)x+1 =  3(x-1)^2 (x+3-p) +(x-1)^3 ..(1)

If x = 0, 1 =  3 (-1)^2(3-p)+(-1)^3 = 3(3-p) -3 = 9-3p-3 = 6-p.

1= 6-p.

p = 6-1 = 5.

We rewrite (1) with p = 5.

 4x^3+3(m-1)x^2+2(n+2)x+1 = 3(x-1)^2(x-2)+(x-1)^3...........(2).

Put x = 1, then 4+3(m-1)+2(n+2)+1 = 0

4+3m-3 +2n+4+1 = 0

3m+2n = -6.................(3)

Differentiate  both sides of (2):

 {4x^3+3(m-1)x^2+2(n+2)x+1}' = {3(x-1)^2(x-2)+(x-1)^3}'

12x^2 + 6(m-1)x +2(n+2) =  {(x-1)^2 (x-1+3x-6)}' = {(x-1)^2*(4x-7)}' = 2(x-1)(4x-7) +(x-1)^2*(4)

12x^2+6(m-1)x +2(n+2) = (x-1){ 2(4x-7)+4(x-1)} = (x-1)(12x-18)

12x^2+6(m-1)x +2(n+2) = 12x^2 -30x +18

Therefore 6(m-1) = -30 and  2(n+2) = 18

m-1 = -30/6 = -5

m= -5+1 = -4.

2(n+2) = 18

n+2= 18/2 = 9

n = 9-2 = 7.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We notice that f is not defined, because we don't know the coefficients of it's expression.

We know that f(x) is divided by g(x)=(x-1)^3.

We'll write the rule of division with reminder:

f = g*c + r

If f is divided by g, that means that the reminder is 0.

f = g*c

We'll calculate the root of g = ( x - 1 )^3

( x - 1 )^3 = 0

x - 1 = 0

x = 1

 that means tha

We notice that if we'll substitute the roots of g(x) in the expression of f(x), we'll clear f(x).

We alsonotice that the value "1" is the multiple root, of 3rd order, of g(x), meaning that:

f(1)=0 <=> f(1)=1+m-1+n+2+1+p-3=0 <=> m+n+p=0

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

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

f"'(1) different from 0

3m+2n-6m-2n=-6+10

We'll combine like terms from the left side:

-3m=4

We'll divide by -3:

m=-4/3

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

-4+2n=-6

 2n=-6+4

 2n=-2

We'll divide by 2:

 n=-1

 m+n+p=0

(-4/3)+(-1)=-p

p=7/3

f(x) = x^4 + ( -4/3 - 1 )*x^3 + ( -1 + 2 )*x^2 + x + 7/3 - 3

f(x) = x^4 + -7x^3/3 + x^2 + x - 2/3

We’ve answered 318,967 questions. We can answer yours, too.

Ask a question