Monic polynomial help

**jvignacio** · March 12th 2009, 06:09 AM

Find the monic polynomial of degree 5 which has 1, 1 + i and 2 − i as three of its roots..

any help on this question id appreciate.

**red_dog** · March 12th 2009, 06:23 AM

I think the polynomial has real coefficients. Then

$x_1=1, \ x_2=1+i, \ x_3=1-i, \ x_4=2-i, \ x_5=2+i$

$P=(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)=$

$=(x-x_1)[(x^2-(x_2+x_3)x+x_2x_3]\cdot[x^2-(x_4+x_5)x+x_4x_5]=$

$=(x-1)(x^2-2x+2)(x^2-4x+5)$

**jvignacio** · March 12th 2009, 06:37 AM

Originally Posted by red_dog

I think the polynomial has real coefficients. Then

$x_1=1, \ x_2=1+i, \ x_3=1-i, \ x_4=2-i, \ x_5=2+i$

$P=(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)=$

$=(x-x_1)[(x^2-(x_2+x_3)x+x_2x_3]\cdot[x^2-(x_4+x_5)x+x_4x_5]=$

$=(x-1)(x^2-2x+2)(x^2-4x+5)$

wow looks very complicated

didnt understand really.

**HallsofIvy** · March 12th 2009, 08:10 AM

Originally Posted by jvignacio

wow looks very complicated

didnt understand really.

Did you replace $x_1$ , $x_2$ , $x_3$ , and $x_4$ with 1, 1+ i, 1- i, 2- i and 2+ i, respectively?

First, the reason red-dog said, "I think the polynomial has real coefficients" is that if you allow complex coefficients, there are an infinite number of such polynomials. Assuming complex coefficients, all roots must be in complex conjugate pairs. That is why 1- i and 2+ i must also be roots. Of course, it follows (You've solved polynomial equations by factoring haven't you. Same idea here). Saying "monic" polynomial means the leading coefficient (the coefficient of $x^5$ here) is 1.

That is why the polynomial must be the product, (x-1)(x-1-i)(x-1+i)(x-2+i)(x-2-i). That is the polynomial. Whether you are required to multiply that out depends on the particular statement of the question.

If you do, it's simplest to multiply the complex conjugates first: $(x-1-i)(x-1+i)$ can be done as "sum and difference product": $((x-1)-i)((x-1)+i)= (x-1)^2- i^2= x^2- 2x+ 1+ 1= x^2- 2x+ 2$

(x-2+i)(x-2-i)= ((x-2)+i)((x-2)-i= $(x-2)^2- i^2= x^2- 4x+ 4+ 1= x^2- 4x+ 5$

The polynomial you want is $(x-1)(x^2- 2x+ 2)(x^2- 4x+ 5)$
Can you multiply that out?

**jvignacio** · March 14th 2009, 03:17 AM

Originally Posted by HallsofIvy

Did you replace $x_1$ , $x_2$ , $x_3$ , and $x_4$ with 1, 1+ i, 1- i, 2- i and 2+ i, respectively?

First, the reason red-dog said, "I think the polynomial has real coefficients" is that if you allow complex coefficients, there are an infinite number of such polynomials. Assuming complex coefficients, all roots must be in complex conjugate pairs. That is why 1- i and 2+ i must also be roots. Of course, it follows (You've solved polynomial equations by factoring haven't you. Same idea here). Saying "monic" polynomial means the leading coefficient (the coefficient of $x^5$ here) is 1.

That is why the polynomial must be the product, (x-1)(x-1-i)(x-1+i)(x-2+i)(x-2-i). That is the polynomial. Whether you are required to multiply that out depends on the particular statement of the question.

If you do, it's simplest to multiply the complex conjugates first: $(x-1-i)(x-1+i)$ can be done as "sum and difference product": $((x-1)-i)((x-1)+i)= (x-1)^2- i^2= x^2- 2x+ 1+ 1= x^2- 2x+ 2$

(x-2+i)(x-2-i)= ((x-2)+i)((x-2)-i= $(x-2)^2- i^2= x^2- 4x+ 4+ 1= x^2- 4x+ 5$

The polynomial you want is $(x-1)(x^2- 2x+ 2)(x^2- 4x+ 5)$
Can you multiply that out?

yeah umm is it $x^5 - 7x^4 + 21x^3 -25x^2 + 20x - 10$ ?

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