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.
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
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
To find the value of a and b, we can use the fact that the sum and product of the roots of a polynomial equation can be determined from the coefficients of the terms in the equation.
Given that two of the roots of the equation x^4 + ax^3 + ax^2 + 11x + b = 0 are 3 and -2, let's find the value of a and b.
1. Finding the value of a:
Since we know that sum of all the roots of the equation is equal to -a, we can write the equation as:
3 + (-2) + α + β = -a (where α and β are the other two roots of the equation)
Simplifying the equation, we have:
1 + α + β = -a
Now, in the equation x^4 + ax^3 + ax^2 + 11x + b = 0, the coefficient of x^3 term is a. As per the equation, the sum of all the roots is equal to -a. Substituting the known values, we get:
3 + (-2) + α + β = -a
1 + α + β = -a
Since two of the roots are given as 3 and -2, we can substitute these values into the equation to get:
1 + α + β = -a
1 + α + β = -(-3)
1 + α + β = 3
Simplifying further, we have:
α + β = 3 - 1
α + β = 2
So, we have found that α + β = 2.
2. Finding the value of b:
Since the product of all the roots of the equation is equal to b, we can write the equation as:
3 * (-2) * α * β = b
Substituting the known values, we get:
(-6) * α * β = b
So, we have found that b = (-6)αβ.
Therefore, we can conclude that a = -3 and b = -6αβ.
To find the other roots, you can use polynomial long division or synthetic division to divide the given polynomial by (x - 3) and (x + 2) one at a time. This will help us determine the quadratic equation for the two remaining roots. Solving this quadratic equation will give us the other two roots.