SOLUTION: Find a polynomial of the specified degree that satisfies the given conditions. Degree 5; zeros −3, −1, 3 and square of 3 ; integer coefficients and constan

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find a polynomial of the specified degree that satisfies the given conditions. Degree 5; zeros −3, −1, 3 and square of 3 ; integer coefficients and constan      Log On


   

Question 1126946: Find a polynomial of the specified degree that satisfies the given conditions.
Degree 5; zeros −3, −1, 3 and
square of 3
; integer coefficients and constant term 54
Found 2 solutions by Edwin McCravy, greenestamps:
Answer by Edwin McCravy(20062) About Me  (Show Source):
You can put this solution on YOUR website!
Find a polynomial of the specified degree that satisfies the given conditions.
Degree 5; zeros -3, -1, 3 and
square of 3
; integer coefficients and constant term 54
The zeros are:
 


Multiply left and right sides:

%28x%2B3%29%28x%2B1%29%28x-3%29%28x-sqrt%283%29%29%28x%2Bsqrt%283%29%29+=+0

Multiply both sides by a constant A:

A%28x%2B3%29%28x%2B1%29%28x-3%29%28x-sqrt%283%29%29%28x%2Bsqrt%283%29%29+=+0

FOIL out the last two parentheses on the left:

A%28x%2B3%29%28x%2B1%29%28x-3%29%28x%5E2-3%29+=+0

FOIL out the last two parentheses on the left again:

A%28x%2B3%29%28x%2B1%29%28x%5E3-3x-3x%5E2%2B9%29+=+0

Get the last parentheses in descending order:

A%28x%2B3%29%28x%2B1%29%28x%5E3-3x%5E2-3x%2B9%29+=+0

Multiply the last two factors in parentheses on the left:

A%28x%2B3%29%28x%5E4-2x%5E3-6x%5E2%2B6x%2B9%29=0

A%28x%5E5+%2B+x%5E4+-+12x%5E3+-+12x%5E2+%2B+27x+%2B+27%29=0

Distribute the A:

Ax%5E5+%2B+Ax%5E4+-+12Ax%5E3+-+12Ax%5E2+%2B+27Ax+%2B+27A%29=0

Since the constant term 27A must equal 54, we set
them equal:

                    27A = 54
                      A = 2

So Ax%5E5+%2B+Ax%5E4+-+12Ax%5E3+-+12Ax%5E2+%2B+27Ax+%2B+27A%29=0

becomes

2x%5E5+%2B+2x%5E4+-+12%282%29x%5E3+-+12%282%29x%5E2+%2B+27%282%29x+%2B+27%282%29%29=0

 or

2x%5E5+%2B+2x%5E4+-+24x%5E3+-+24x%5E2+%2B+54x+%2B+54+=+0

So the polynomial is the left side without the "= 0".

2x%5E5+%2B+2x%5E4+-+24x%5E3+-+24x%5E2+%2B+54x+%2B+54

Edwin

Answer by greenestamps(13206) About Me  (Show Source):
You can put this solution on YOUR website!


In order for the problem to make sense, I will assume that the last root you show is the square ROOT of 3, not the square of 3.

Then, since the polynomial is to have integer coefficients, the 5th root must be negative square root of 3.

The polynomial with those 5 roots is

a%28x%2B3%29%28x%2B1%29%28x-3%29%28x-sqrt%283%29%29%28x%2Bsqrt%283%29%29

The constant term of the polynomial is

a%283%29%281%29%28-3%29%28-3%29+=+27a

Since the constant term is 54, a must be 2. Then the polynomial (still in factored form) is

2%28x%2B3%29%28x%2B1%29%28x-3%29%28x-sqrt%283%29%29%28x%2Bsqrt%283%29%29

In expanded form the polynomial is