algebra question

Question

Consider the following.

f(x) = x⁴ – 4x³ + 8x² + 20x – 65

(a) Write the polynomial as the product of factors that are irreducible over the rationals. (Hint: One factor is x² – 5.)

f(x) =  

(b) Write the polynomial as the product of linear and quadratic factors that are irreducible over the reals.

f(x) =  

 

 

(c) Write the polynomial in completely factored form.

f(x) =  

 

 

Answer 1

Here is the way it is supposed to be done!!!

P = {set of factors of 65}

= {+-1, +-5, +- 13, +-65}

Q = {+-1}

P/Q = P = {+-1, +-5, +- 13, +-65}

is the set of POSSIBLE rational solutions.

Now you have to plug in each one of these

to see if any are a solution.

Naturally, none of them work. So now you have to

find the solutions numerically, by examining the

graph of the polynomial.

When you do, you will see that sqrt(5) is a solution.

Substituting sqrt(5) gives:

25 - 20*sqrt(5) + 40 + 28*sqrt(5)- 65 = 0

Then -sqrt(5) is also a solution since the coefficients

are rational. So x + sqrt(5) and x-sqrt(5) are factors

as given

Therefore (x-sqrt(5))(x +sqrt(5)) = x^2 - 5 is a factor

So now you divide the polynomials:

x^2 - 4x + 13

__________________________________

x^2 + 0x - 5 | x^4 - 4x^3 + 8x^2 + 20x - 65

x^4 - 0x^3 - 5x^2

----------------------------------------

-4x^3 + 13x^2 + 20x

-4x^3 + 0x^2 + 20x

-----------------------

13x^2 - 65

13x^2 - 65

x^2 - 4x + 13 = 0

x = [4 +or- sqrt ( (-4)^2 - 4(1)(13) )] / 2

= [ 4 +or- sqrt( 16 - 52) ] / 2

= [ 4 +or- sqrt(-36) ] / 2

= [ 4 +or- 6i ] / 2

= 2 +or- 3i

check:

x=2+3i

x^2 = (2+3i)(2+3i) = 4 + 6i + 6i + (-9) = -5 + 12i

x^3 = x^2 * x

= (-5 + 12i) ( 2+3i)

= -10 -15i + 24i + 36(-1)

= -46 + 9i

x^4 = x^3 * x

= (-46 + 9i)(2+3i)

= -92 + -138i + 18i + -27

= -119 + -120i

x^4 - 4x^3 + 8x^2 + 20x - 65

-119 + -120i -4(-46 +9i) + 8 ( -5 + 12i) + 20( 2 + 3i) - 65

-119 + -120i + 184 - 36i + -40 + 96i + 40 + 60i - 65

imaginary numbers cancel out as +and- 156i

real numbers cancel out as +and- 184

So 2+3i is a solution. Therefore 2-3i is also a solution.

The complete solution set is {+or- sqrt(5) , 2 +or- 3i }

In factored form, the polynomial is

(x^2-5)(x^2-4x+13) =

(x-sqrt(5))(x+sqrt(5)) ((x - 2)+3i)((x-2)-3i)