Question

Solve the equation.
x^3-2x^2-2x+4=0

Answer 1

\[x^3-2x^2-2x+4=0\]
\[(x^3-2x)-(2x^2-4)=0\]
Factorise each group in the brackets for me please.

Answer 2

@orbie

Answer 3

so would it be \[x(x ^{2}-2)-2(x ^{2}-2)=0\]?

Answer 4

Good job mate. Now the next part you should know what to do. Even a monkey would know what to do. You see you just grouped it and now you have x^2-2 in each term. So you would group them
\[(x^2-2)(?-?)=0\]

Answer 5

So would you know what goes in the next bracket with the question marks?

Answer 6

@orbie

Answer 7

(x-2)

Answer 8

Yep. Excellent.

Answer 9

Now what would be the x^2 value and x-value be?

Answer 10

x^2=2 and x=2? Right?

Answer 11

Correct. Now we want to find the x-values in x^2=2. We already know one value of x which is 2 right? Now we just have to square root the x^2 to get a positive and negative value for x.

Answer 12

so it would be x= sqrt(4) and then it would be +/- 2?

Answer 13

Nope. WE square root both sides. You square root the x^2. That becomes x. Now you square root the 2. what does that become?

Answer 14

Just the square root of 2?

Answer 15

square root 2 and what? When you square root something, there must have a negative value as well as a positive value.

Answer 16

so it would be -sqrt 2 and (positive)sqrt 2?

Answer 17

You shot the target right down. Excellent job mate.

Answer 18

Well Done!

Answer 19

Thank you! =]

Answer 20

You're welcome.