Thread: Solve f (x) = 0

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

solve 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 ?

Originally Posted by gilagila

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

solve 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 ?

Have you written this equation correctly? Is it or is it something else? Because if you factorise this, you should get ...

Originally Posted by Prove It

Have you written this equation correctly? Is it or is it something else? Because if you factorise this, you should get ...

sorry , due to not sure how to use the quadratic and cubic sign, just change it to ^2 and ^3.

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 .

So you know immediately that x+2= 0, or x- 3= 0, or 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 . That does not have any real number roots but you can find the complex roots by using the quadratic formula.

Originally Posted by HallsofIvy

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 .

So you know immediately that x+2= 0, or x- 3= 0, or 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 . That does not have any real number roots but you can find the complex roots by using the quadratic formula.

Is it meaning that the x = 3, -2 and (5+√(-19))/2 or (5- √(-19))/2 <- using formula calculation ?
meaning if f(x) = 0, then x = { 3, 2, (5+√(-19))/2 and (5- √(-19))/2}
am i right ?

Originally Posted by gilagila

Is it meaning that the x = 3, -2 and (5+√(-19))/2 or (5- √(-19))/2 <- using formula calculation ?
meaning if f(x) = 0, then x = { 3, 2, (5+√(-19))/2 and (5- &√(-19))/2}
am i right ?

Both x = (5+√(-19))/2 and x = (5- √(-19))/2 are solutions, although, as HallsofIvy mentioned, both of those are complex numbers. That's because for all real number values of x. To see that, graph

Originally Posted by SammyS

Both x = (5+√(-19))/2 and x = (5- √(-19))/2 are solutions, although, as HallsofIvy mentioned, both of those are complex numbers. That's because for all real number values of x. To see that, graph

ok. Thanks for the reply