Question

solve the equation in the real number system: x^4-8x^3+16x^2
+8x-17=0

Answer 1

by inspection 1 is a solution

Answer 2

because
\[1-8+16+8-17=0\]

Answer 3

so you know this factors as
\[(x-1)\times \text{something}\] and you find the something by division

Answer 4

yes...

Answer 5

always check to see if 1 works because it is easy. -1 is easy too. check it

Answer 6

then the answer is x^2-8x+17=0 right??

Answer 7

so you know this factors as
\[(x-1)(x^3-4x^2+9x+17)\] by division.

Answer 8

you can use synthetic division to divide by x - 1

Answer 9

sorry i made mistake.

Answer 10

it is
\[(x-1)(x^3-7x^2+9x+17)\]

Answer 11

that is the correct answer. now check the second factor and you will see that -1 is a zero

Answer 12

:).. ok...

Answer 13

i am sorry about that and you are right! it is
\[(x-1)(x+1)(x^2-8x+17)\] you had it all along!

Answer 14

my mistake i should have read your answer more carefully. you got it! third factor has no real zeros so you are done

Answer 15

you were right 6 lines up. done!

Answer 16

x^2-8x+17=0 right?

Answer 17

yes that is the last factor.

Answer 18

this has no real zeros

Answer 19

and since your instructions were "solve in real number system" you are done. real zeros are 1 and -1

Answer 20

good work!