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
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.
2 answers
Divide by (x^2-x-6) 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^4-3x^3-3x^2+11x-6
= (x+2)(x-3)(x^2-2x+1)
The other roots are 1,1
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^4-3x^3-3x^2+11x-6
= (x+2)(x-3)(x^2-2x+1)
The other roots are 1,1