Complex Polynomial Q (1 Viewer)

[Lindurr]

Consider f(t)=t^6+t^5-t^4-5t^3-6t^2-6t-4. Given -1+i is a root of 'f' and that 'f' also has two real integer roots,
factorise f into complex factors

Thankyou heaps

 

[deswa1]

<img src="http://latex.codecogs.com/gif.latex?\textup{Okay, we know that -1+i is a root, therefore -1-i is also a root as the coefficients are real.}\\ f(t)=t^6+t^5-t^4-5t^3-6t^2-6t-4\\ (t+1-i)(t+1+i)\textup{ are factors of f(t)}\\ \therefore t^2+2t+2 \textup{ is a factor of f(t)}\\ f(t)=(t^2+2t+2)(t^4-t^3-t^2-t-2)\\ \textup{We know that there are two integer roots. These can only come from 1,-1,2,-2.}\\ \textup{Testing values with the remainder theorem, f(-1)=f(2)=0}\\ \therefore (t+1)(t-2) \textup{ are factors}\\ f(t)=(t^2+2t+1)(t+1)(t-2)(t^2+1)\\ f(t)=(t+1)(t-2)(t+i)(t-i)(t+1+i)(t+1-i)" title="\textup{Okay, we know that -1+i is a root, therefore -1-i is also a root.}\\ f(t)=t^6+t^5-t^4-5t^3-6t^2-6t-4\\ (t+1-i)(t+1+i)\textup{ are factors of f(t)}\\ \therefore t^2+2t+2 \textup{ is a factor of f(t)}\\ f(t)=(t^2+2t+2)(t^4-t^3-t^2-t-2)\\ \textup{We know that there are two integer roots. These can only come from 1,-1,2,-2.}\\ \textup{Testing values with the remainder theorem, f(-1)=f(2)=0}\\ \therefore (t+1)(t-2) \textup{ are factors}\\ f(t)=(t^2+2t+1)(t+1)(t-2)(t^2+1)\\ f(t)=(t+1)(t-2)(t+i)(t-i)(t+1+i)(t+1-i)" />

 

[jet]

<img src="http://latex.codecogs.com/gif.latex?\textup{Okay, we know that -1+i is a root, therefore -1-i is also a root.}\\ f(t)=t^6+t^5-t^4-5t^3-6t^2-6t-4\\ (t+1-i)(t+1+i)\textup{ are factors of f(t)}\\ \therefore t^2+2t+2 \textup{ is a factor of f(t)}\\ f(t)=(t^2+2t+2)(t^4-t^3-t^2-t-2)\\ \textup{We know that there are two integer roots. These can only come from 1,-1,2,-2.}\\ \textup{Testing values with the remainder theorem, f(-1)=f(2)=0}\\ \therefore (t+1)(t-2) \textup{ are factors}\\ f(t)=(t^2+2t+1)(t+1)(t-2)(t^2+1)\\ f(t)=(t+1)(t-2)(t+i)(t-i)(t+1+i)(t+1-i)" title="\textup{Okay, we know that -1+i is a root, therefore -1-i is also a root.}\\ f(t)=t^6+t^5-t^4-5t^3-6t^2-6t-4\\ (t+1-i)(t+1+i)\textup{ are factors of f(t)}\\ \therefore t^2+2t+2 \textup{ is a factor of f(t)}\\ f(t)=(t^2+2t+2)(t^4-t^3-t^2-t-2)\\ \textup{We know that there are two integer roots. These can only come from 1,-1,2,-2.}\\ \textup{Testing values with the remainder theorem, f(-1)=f(2)=0}\\ \therefore (t+1)(t-2) \textup{ are factors}\\ f(t)=(t^2+2t+1)(t+1)(t-2)(t^2+1)\\ f(t)=(t+1)(t-2)(t+i)(t-i)(t+1+i)(t+1-i)" />

You need to say that -1-i is also a root because the polynomial has real coefficients. If it didn't have real coefficients, it wouldn't be true.

 

[Siddy123]

<img src="http://latex.codecogs.com/gif.latex?\textup{Okay, we know that -1+i is a root, therefore -1-i is also a root as the coefficients are real.}\\ f(t)=t^6+t^5-t^4-5t^3-6t^2-6t-4\\ (t+1-i)(t+1+i)\textup{ are factors of f(t)}\\ \therefore t^2+2t+2 \textup{ is a factor of f(t)}\\ f(t)=(t^2+2t+2)(t^4-t^3-t^2-t-2)\\ \textup{We know that there are two integer roots. These can only come from 1,-1,2,-2.}\\ \textup{Testing values with the remainder theorem, f(-1)=f(2)=0}\\ \therefore (t+1)(t-2) \textup{ are factors}\\ f(t)=(t^2+2t+1)(t+1)(t-2)(t^2+1)\\ f(t)=(t+1)(t-2)(t+i)(t-i)(t+1+i)(t+1-i)" title="\textup{Okay, we know that -1+i is a root, therefore -1-i is also a root.}\\ f(t)=t^6+t^5-t^4-5t^3-6t^2-6t-4\\ (t+1-i)(t+1+i)\textup{ are factors of f(t)}\\ \therefore t^2+2t+2 \textup{ is a factor of f(t)}\\ f(t)=(t^2+2t+2)(t^4-t^3-t^2-t-2)\\ \textup{We know that there are two integer roots. These can only come from 1,-1,2,-2.}\\ \textup{Testing values with the remainder theorem, f(-1)=f(2)=0}\\ \therefore (t+1)(t-2) \textup{ are factors}\\ f(t)=(t^2+2t+1)(t+1)(t-2)(t^2+1)\\ f(t)=(t+1)(t-2)(t+i)(t-i)(t+1+i)(t+1-i)" />

you use terry lee's book?