Math, factoring
posted by Adeline
Two of the roots of x^4 +ax^3 +ax^2+11x+b=0 are 3 and 2.
1. Find the value of a and b
2. Find the other roots.
I believe that a = 3 and b = 6.
I am unsure of how to find the other roots.
Please show how you found the answer.

Bosnian
For x = 3
x ^ 4 + a x ^ 3 + a x ^ 2 + 11 x + b = 0
Put x = 3 in this equation:
3 ^ 4 + a * 3 ^ 3 + a * 3 ^ 2 + 11 * 3 + b = 0
81 + 27 a + 9 a + 33 + b = 0
36 a + b + 114 = 0
For x =  2
x ^ 4 + a x ^ 3 + a x ^ 2 + 11 x + b = 0
Put x =  2 in this equation:
(  2 ) ^ 4 + a * (  2 ) ^ 3 + a * (  2 ) ^ 2 + 11 * (  2 ) + b = 0
16  8 a + 4 a  22 + b = 0
 4 a + b  6 = 0
Now you must solve system:
36 a + b + 114 = 0
 4 a + b  6 = 0
The solutions are: a =  3 , b =  6
So equation x ^ 4 + a x ^ 3 + a x ^ 2 + 11 x + b = 0 becomes:
x ^ 4  3 x ^ 3  3 x ^ 2 + 11 x  6 = 0
We can take a polynomial of degree 4, such as:
( x  p ) ( x  q ) ( x  r ) ( x  s )
where: p , q , r , s are roots of polynomials
In this casse you know two roots : p = 3 and q =  2
Now ( x  p ) ( x  q ) ( x  r ) ( x  s ) becomes:
( x  3 ) [ x  (  2 ) ] ( x  r ) ( x  s ) =
( x  3 ) ( x + 2 ) ( x  r ) ( x  s ) =
( x ^ 2  x  6 ) ( x  r ) ( x  s ) = 0
If
x ^ 4  3 x ^ 3  3 x ^ 2 + 11 x  6 = ( x ^ 2  x  6 ) ( x  r ) ( x  s )
then you must do long division:
( x ^ 4  3 x ^ 3  3 x ^ 2 + 11 x  6 ) / ( x ^ 2  x  6 ) = x ^ 2  2 x + 1
This mean ( x  r ) ( x  s ) = x ^ 2  2 x + 1
If ( x  r ) ( x  s ) = 0 then x ^ 2  2 x + 1 also = 0
x ^ 2  2 x + 1 = 0
Solution: x = 1
Or:
x ^ 2  2 x + 1 = ( x  r ) ( x  s ) = ( x  1 ) ( x  1 )
Factors of your polynomial:
x ^ 4  3 x ^ 3  3 x ^ 2 + 11 x  6 = ( x  3 ) ( x + 2 ) ( x  1 ) ( x  1 )
All this mean:
Your polynomialal have 3 roots:
x =  2 , x = 3 and a double root x = 1 
Steve
Divide by (x^2x6) and you get
quotient: x^2+(a+1)x+(2a+7)
remainder: (8a+24)x + b+6(2a+7)
For the remainder to be zero, we need
a = 3
b = 6
So, x^43x^33x^2+11x6
= (x+2)(x3)(x^22x+1)
The other roots are 1,1
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