Question

x^3-4x^2+8 use rational root theorem . I got the number two but I can't figure out the other two numbers

Answer 1

are you only suppose to find the rational zeros?

Answer 2

2| 1 -4 0 8
| 2 -4 -8
-----------------
1 -2 -4| 0

you can find other 0's by solving
\[x^2-2x-4=0\]

Answer 3

yeah! oh duh! thanks so much!

Answer 4

uaos;j whf except i don't know what im doing and now can't figure the rest out!

Answer 5

use the quadratic formula

Answer 6

ohhh . thanks!

Answer 7

pardon? we can't apply quadratic formula given that the equation is in third degree.. it's a polynomial. . . . .

Answer 8

\[x^2-2x-4=0\]
is 2nd degree

Answer 9

\[x=\frac{-(-2) \pm \sqrt{(-2)^2-4(1)(-4)}}{2(1)}\]

Answer 10

\[x=\frac{2 \pm \sqrt{4+16}}{2}=\frac{2 \pm \sqrt{20}}{2}=\frac{2 \pm 2 \sqrt{5}}{2}=1\pm \sqrt{5}\]

Answer 11

so we have the three solutions
the one you found which is x=2
and the other or irrational solutions (the ones i found)

Answer 12

in continuation to myinaniya. . .

therefore;
\[(x ^{2}-2x-4)(x-2)\] this are the factors derived from the original equation.

to continue; to make a perfect square root
\[(x ^{2}-2x-4)\] - to factor; \[x ^{2}-2x= 4\]\[x ^{2}-2x+1=4+1\]\[x ^{2}-2x+1=5\] \[(x-1)(x-1)=5\]

the factors are: (x-2); (x-1)(x-1)=5

to solve:
x-2=0 x-1=5 x-1=5
x=2 x=6 x=6

therefor roots are; (2,6,6)

Answer 13

no you solve x^2-2x-4=0 wrong

Answer 14

you can only do:
if you have a*b=0, then either a=0 or b=0 or both=0
-------------------
\[(x-1)(x-1)=5\]
\[(x-1)^2=5\]
however you can solve this by using the definition of absolute value
you need to take square root both sides to do this
\[x-1=\pm \sqrt{5}\]
now add 1 on both sides
\[x=1 \pm \sqrt{5}\]

Answer 15

thank you for the correction. . . .