# Math, factoring

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.

1. 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 )

x ^ 4 - 3 x ^ 3 - 3 x ^ 2 + 11 x - 6 = ( x - 3 ) ( x + 2 ) ( x - 1 ) ( x - 1 )

All this mean:

x = - 2 , x = 3 and a double root x = 1

posted by Bosnian
2. 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

posted by Steve

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