Question

List the possible rational roots, then find the actual rational roots.

Question: x^3-5x^2+7x-2=0

steps completed so far: (x-2)(x^2-3x+1)=0

confused as to how to get the roots

1/2 (3-sqrt(5)) and 1/2 (3+sqrt(5))

Answer 1

well, you're only asked of the "possible rational roots"
the likely ones
recall the "rational root test" \(\bf x^3-5x^2+7x-2=0\implies {\color{red}{ 1}}x^3-5x^2+7x-{\color{blue}{ 2}}=0\\ \quad \\
\textit{possible roots}\implies \pm \cfrac{{\color{blue}{\textit{factors of constant term} }}}{{\color{red}{ \textit{factors of leading term coefficient}}}}\)

Answer 2

± 2/1 ?

Answer 3

well, all factors so \(\bf \pm\cfrac{{\color{blue}{ 2,1}}}{{\color{red}{ 1}}}\)

Answer 4

\(\bf \pm \cfrac{2}{1},\quad \pm \cfrac{1}{1}\)

Answer 5

ohhhh okay! all of the factors! so then what is the next step after that?

Answer 6

next step, you go to the fridge and get a root bear and some doritos =)

Answer 7

well, that's all you're asked, so :)

Answer 8

i have to still find the actual rational roots still

Answer 9

ohhh yea... just notice that

Answer 10

ohh so then we have to go through 2, -2, 1, and -1 and find out which ones work?

Answer 11

right, the likely suspect would be x-2, thus +2

Answer 12

perfect! thank you!!

Answer 13

\(\bf (x-2)(x^2-3x+1)=0\) you'd need the quadratic formula to factor out the quadratic \(\bf (x^2-3x+1=0\\ \quad \\

x= \cfrac{ - (-3) \pm \sqrt { (-3)^2 -4(1)(1)}}{2(1)}\)

Answer 14

so are those answers considered rational ? you get (3±√5)/2