polynomial help (1 Viewer)

[M@ster P]

Solve the equation (x-1)^3 - 4(x-1)^2 + 5(x-1) - 2 = 0

i expanded it all out and got the answer x^3 - 4x^2 + 13x -12 and then i subsituted the factors of 12 (-1,1,-2,2,-3,3,-4,4-6,6,-12,12) until i get an answer 0, but the problem was that there was not one factor that gave me 0. I also checked the expanding at it seemed correct.

Hi guys i did this question many times and i can't get the answer, the answer is x=2 and x=3

 

[tacogym27101990]

回复: polynomial help

Solve the equation (x-1)^3 - 4(x-1)^2 + 5(x-1) - 2 = 0

i expanded it all out and got the answer x^3 - 4x^2 + 13x -12

you got the expansion wrong
it should be x³-7x² +16x -12 =0

 

[lyounamu]

Solve the equation (x-1)^3 - 4(x-1)^2 + 5(x-1) - 2 = 0

i expanded it all out and got the answer x^3 - 4x^2 + 13x -12 and then i subsituted the factors of 12 (-1,1,-2,2,-3,3,-4,4-6,6,-12,12) until i get an answer 0, but the problem was that there was not one factor that gave me 0. I also checked the expanding at it seemed correct.

Hi guys i did this question many times and i can't get the answer, the answer is x=2 and x=3

You expanded it wrong. It should be x^3 - 7x^2 + 16x - 12 =0

EDIT: Got beaten by tacogym 27101990!

 

[M@ster P]

ty

 

[sicmacao]

Solve the equation (x-1)^3 - 4(x-1)^2 + 5(x-1) - 2 = 0

i expanded it all out and got the answer x^3 - 4x^2 + 13x -12 and then i subsituted the factors of 12 (-1,1,-2,2,-3,3,-4,4-6,6,-12,12) until i get an answer 0, but the problem was that there was not one factor that gave me 0. I also checked the expanding at it seemed correct.

Hi guys i did this question many times and i can't get the answer, the answer is x=2 and x=3

You don't need to expand at all, you can let y = x-1

Then the equation becomes
y^3-4y^2+5y-2=0

Try y=1, => 1-4+5-2=0

so y=1 is a solution of the equation.
By division of polynomial

y^3-4y^2+5y-2 = 0
(y-1)(y^2-3y+2) = 0
(y-1)^2 (y-2) = 0
y = 1 or 2

Since y = x-1
therefore, x = y+ 1
x = 2 or 3