Polynomials

[BobbyJones]

This question says the 7th degree polynomial x^7 - 3x^6 - 7x^4 + 21x^3 - 8x +24 has a factor of (x-3)


a) Divide x^7 - 3x^6 - 7x^4 + 21x^3 - 8x +24 bu x-3, and put in form (x-3)(ax^6+bx^3+c)

---I've done this part

b) By putting z = x^3, find all the factors, real or complex of the 6th degree polynomial and thus express x^7 - 3x^6 - 7x^4 + 21x^3 - 8x +24 as the product of seven linear factors.

----I've managed to find the real factors but cant find the Imaginary. Can someone show me how to work out the imaginary factors please.

Thankyou.

 

[soroban]

Hello, BobbyJones!

\(\displaystyle \text{The polynomial }f(x) \:=\:x^7 - 3x^6 - 7x^4 + 21x^3 - 8x +24\) .\(\displaystyle \text{ has a factor of }(x-3).\)

\(\displaystyle \text{a) Divide }f(x)\text{ by }x-3\text{, and put in the form: }(x-3)(ax^6+bx^3+c)\)

--- I've done this part


b) By putting z = x^3, . This is unnecessary.
. . find all the factors, real or complex of the 6th degree polynomial
. . and thus express \(\displaystyle f(x)\) as the product of seven linear factors.

--- I've managed to find the real factors but cant find the Imaginary.
. . Can someone show me how to work out the imaginary factors please?
. . Thankyou.

\(\displaystyle f(x)\) factors: .\(\displaystyle (x-3)(x^6 - 7x^3 - 8) \:=\:0\)

Hence, one of the factors is: \(\displaystyle (x - 3)\)


We have: .\(\displaystyle x^6 - 7x^3 - 8 \:=\:0\)

Factor: .\(\displaystyle (x^3 + 1)(x^3-8) \:=\:0\)

And we have two cubic equations to solve . . .


\(\displaystyle x^3 + 1 \:=\:0 \quad\Rightarrow\quad (x+1)(x^2 - x + 1) \:=\:0\)

. . Hence: .\(\displaystyle (x + 1)\) is a factor.

.\(\displaystyle x^2-x+1\:=\:0 \quad\Rightarrow\quad x \:=\:\dfrac{1 \pm i\sqrt{3}}{2} \)
. . Hence: ..\(\displaystyle \left(x - \frac{1 + i\sqrt{3}}{2}\right)\,\text{ and }\,\left(x - \frac{1-i\sqrt{3}}{2}\right)\,\text{ are factors.}\)


\(\displaystyle x^3 - 8 \:=\:0 \quad\Rightarrow\quad (x - 2)(x^2 +2x + 4) \:=\:0\)

. . Hence: .\(\displaystyle (x-2)\) is a factor.

\(\displaystyle x^2 + 2x + 4 \:=\:0 \quad\Rightarrow\quad x \:=\:\dfrac{\text{-}2 \pm\sqrt{-12}}{2} \:=\:\dfrac{\text{-}2\pm2i\sqrt{3}}{2} \:=\:\text{-}1 \pm i\sqrt{3}\)

. . Hence: .\(\displaystyle \left(x - \left[\text{-}1 + i\sqrt{3}\right]\right)\,\text{ and }\,\left(x - \left[\text{-}1 - i\sqrt{3}\right]\right)\,\text{ are factors.} \)


Therefore: .\(\displaystyle f(x) \;=\;(x-3)(x+1)(x-2)\left(x - \frac{1\pm i\sqrt{3}}{2}\right)\big(x - [\text{-}1 \pm i\sqrt{3}]\big)\)


By the way, this is not a Differential Equation problem.