Monic polynomial question (1 Viewer)

[pink_rabbit_<3]

Hey guys im stuck on this one polynomial question..please help!

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

thanx heaps

 

[tommykins]

P(x) = x^5 + bx^4 + cx^3 + dx^2 + ex + f ( there is no a as a is one since its monic)

I'm also assuming the coefficients are real, please state so other wise.

Roots are 1, 1+i, and 2-i - since the coefficients are real, the roots are also 1-i, 2+i (conjugate complex roots when a polynomial has real coefficients).

Sum of roots = 1 + (1+i) + (2-i) + (1-i) + (2+i) = -b = 6

therefore b = -6

Product of roots = 1(1+i)(1-i)(2+i)(2-i) = (1-i²)(4-i²)(1) = (2)(5)(1) = 10 = -f

Therefore f = -10.

Do sum of roots two at a time, then 3 at a time, then 4 at a time for c,d,e respectively.

Or you could just expand the polynomial in its linear factors

P(x) = (x-1)(x+(2+i)(x+(2-i)(x+(1+i)(x+(1-i)

Obviously all this would only work if the coeffecients of P(x) is all real.

 

[pink_rabbit_<3]

omg wow thanks so much! do u have time for a quick trig question too? lol thanks heaps...
find the exact values of sin (n/8) and cos (n/8) (n represents pie) lol

 

[namburger]

omg wow thanks so much! do u have time for a quick trig question too? lol thanks heaps...
find the exact values of sin (n/8) and cos (n/8) (n represents pie) lol

Is there a lead-up Q?

 

[pink_rabbit_<3]

lead up Q? whats that...argh i still cant get the monic polynomial question...

 

[YannY]

why are you doing this? havent you finished your hsc?

 

[SpinCobra]

Lead up question.

As in..

A) Prove blah blah blah.
B) Hence solve.. blah blah blah.

 

[pink_rabbit_<3]

why are you doing this? havent you finished your hsc?

haha yeh ive finished but im trying to help a friend...lol besides its similar to the maths im doing in my first year in uni so i would like to know how to do it as well...

 

[pink_rabbit_<3]

Lead up question.

As in..

A) Prove blah blah blah.
B) Hence solve.. blah blah blah.

oh thanks i see...no theres no lead up namburger....

 

[YannY]

omg wow thanks so much! do u have time for a quick trig question too? lol thanks heaps...
find the exact values of sin (n/8) and cos (n/8) (n represents pie) lol

cos2x=1-2(sinx)^2

therefore sinx=sqrt[(1-cos2x)/2]
sin(pi/8)=sqrt[(1-cospi/4)/2]
=sqrt[(1-1/sqrt2)/2]

do this same for cos(pi/8) where cos2x=2[(cosx)^2]-1 and cospi/4=1/sqrt2

No need for a lead up question this is simply a 3u question

 

[tommykins]

lead up Q? whats that...argh i still cant get the monic polynomial question...

Do you have the answer? please check if B and F are correct.

P(x) = x^5 + bx^4 + cx^3 + dx^2 + ex + f ( there is no a as a is one since its monic)

Roots are 1, 1+i, and 2-i,1-i, 2+i (conjugate complex roots when a polynomial has real coefficients).

Sum of roots 2 at a time = 1(1+i) + 1(2-i) + 1(1-i) + 1(2+i) + (1+i)(2-i) + (1+i)(1-i) + (1+i)(2+i) + (2-i)(1-i) + (2-i)(2+i) => expand it out, do the addition/subtraction of the real and imaginary parts, and the result will = c

Also note, the values alternate, so sum of roots = -b, sum of roots 2 at a time = c, sum of roots 3 at a time = -d, sum of roots 4 at a time = e.

It'd take too long for me to do as I have an assignment, but just post if you need any other help.

 

[pink_rabbit_<3]

oh i see...thanks a lot yanny...but should i simplify the sin (pie/8) after or should i just leave it to how u left it....

 

[YannY]

oh i see...thanks a lot yanny...but should i simplify the sin (pie/8) after or should i just leave it to how u left it....

Well i cant see how you would simplify such an expression but if you can go ahead. Oh and by simplifying if you mean leave it in decimal form then dont, thats not simplifying it.

 

[pink_rabbit_<3]

Do you have the answer? please check if B and F are correct.

P(x) = x^5 + bx^4 + cx^3 + dx^2 + ex + f ( there is no a as a is one since its monic)

Roots are 1, 1+i, and 2-i,1-i, 2+i (conjugate complex roots when a polynomial has real coefficients).

Sum of roots 2 at a time = 1(1+i) + 1(2-i) + 1(1-i) + 1(2+i) + (1+i)(2-i) + (1+i)(1-i) + (1+i)(2+i) + (2-i)(1-i) + (2-i)(2+i) => expand it out, do the addition/subtraction of the real and imaginary parts, and the result will = c

Also note, the values alternate, so sum of roots = -b, sum of roots 2 at a time = c, sum of roots 3 at a time = -d, sum of roots 4 at a time = e.

It'd take too long for me to do as I have an assignment, but just post if you need any other help.

ok thanks alot but shouldnt the sum of roots be equal to -7? not -6....
nah i dont have the answer....sorry for disturbing you

 

[tommykins]

Yes it is -7, my mistake.