SOLUTION: Solve the equation by finding all real roots.
x^3-7x+15x-9=0
i have this
1 -7 15 -9 /(divided by) 1.
x^2-6x+9
(x+1) (x^2-6x+9)
it says factor again and where is
Algebra ->
Rational-functions
-> SOLUTION: Solve the equation by finding all real roots.
x^3-7x+15x-9=0
i have this
1 -7 15 -9 /(divided by) 1.
x^2-6x+9
(x+1) (x^2-6x+9)
it says factor again and where is
Log On
Question 134881This question is from textbook
: Solve the equation by finding all real roots.
x^3-7x+15x-9=0
i have this
1 -7 15 -9 /(divided by) 1.
x^2-6x+9
(x+1) (x^2-6x+9)
it says factor again and where is the soultion This question is from textbook
You can put this solution on YOUR website! Solve the equation by finding all real roots.
x^3-7x+15x-9=0
i have this
1 -7 15 -9 /(divided by) 1.
x^2-6x+9
(x+1) (x^2-6x+9)
it says factor again and where is the soultion
---------------------------
1)....1....-7....15....-9
......1....-6....9....|..0
So 1 is a Real zero
----------------------
Factorx^2-6x+9 to get (x-3)^2
So 3 is a Real zero with multiplicity two.
=========================
Cheers,
Stan H.
(