solvef(x) = 2 (x-3) (x+2)^2 + 2 (x+2) (x-3)^3

f(x)= 0

if we factorise, we get

f(x) = 2 (x-3) (x+2) (x^2– 5x+ 11)

0= 2 (x-3) (x+2) (x^2 – 5x+ 11)

is it meaning that x = 3, and x = 2 ?? or ?

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- Oct 10th 2011, 05:45 AMgilagilaSolve f (x) = 0
solve*f*(*x*) = 2 (*x*-3) (*x*+2)^2 + 2 (*x*+2) (*x*-3)^3

*f(x)*= 0

if we factorise, we get

*f*(*x*) = 2 (*x*-3) (*x*+2) (*x^2*– 5*x*+ 11)

*0*= 2 (*x*-3) (*x*+2) (*x^*2 – 5*x*+ 11)

is it meaning that x = 3, and x = 2 ?? or ?

- Oct 10th 2011, 05:49 AMProve ItRe: Solve f (x) = 0
Have you written this equation correctly? Is it $\displaystyle \displaystyle f(x) = 2(x - 3)(x + 2)2 + 2(x + 2)(x - 3)3$ or is it something else? Because if you factorise this, you should get $\displaystyle \displaystyle f(x) = 2(x - 3)(x + 2)(2 + 3) = 2(x - 3)(x + 2)5 = 10(x - 3)(x + 2)$...

- Oct 10th 2011, 05:52 AMgilagilaRe: Solve f (x) = 0
- Oct 10th 2011, 05:53 AMHallsofIvyRe: Solve f (x) = 0
One of the things I am sure yuo have already learned is that if abc= 0 then at least one of a, b, or c must be 0.

Yes, you can factor $\displaystyle 2(x-3)(x+2)^2+ 2(x+2)(x-3)^3= 2(x+2)(x- 3)[(x+2)+ (x- 3)^2]= 2(x+2)(x-3)(x^2- 5x+ 11)= 0$.

So you know immediately that x+2= 0, or x- 3= 0, or $\displaystyle x^2- 5x+ 11= 0$ is 0. If x- 3= 0, then x= 3 so x= 3 is a root. If x+ 2= 0 then x= -2 (NOT 2) so x= -2 is a root. But there are two other roots- numbers that satisfy $\displaystyle x^2- 5x+ 11= 0$. That does not have any real number roots but you can find the complex roots by using the quadratic formula. - Oct 10th 2011, 05:59 AMgilagilaRe: Solve f (x) = 0
- Oct 10th 2011, 09:37 AMSammySRe: Solve f (x) = 0
- Oct 10th 2011, 04:26 PMgilagilaRe: Solve f (x) = 0