Question

5x-(4-2x+x^2)(x+2)+x(x-1)(x+1)=0​

Answer 1

5х-(4-2х+х^2)(х+2)+х(х-1)(х+1)=0

5х-(4х+8-2х^2-4х+х^3+2х^2)+х(х^2-1)=0

Сокращаем и приводим подобные:

5х-8-х^3+х^3-х=0

4х-8=0 |:2

Х=2

Answer 2

To solve the given equation, we'll start by expanding and simplifying the expression on the left-hand side (LHS) of the equation:

LHS: 5x - (4 - 2x + x^2)(x + 2) + x(x - 1)(x + 1) = 0

Step 1: Expand the second term (using FOIL method) (4 - 2x + x^2)(x + 2) = 4x + 8 - 2x^2 - 4x + x^3 + 2x^2 = x^3 + 6 - 2x

Step 2: Expand the third term (using FOIL method) x(x - 1)(x + 1) = x(x^2 + x - x - 1) = x(x^2 - 1) = x^3 - x

Now, let's substitute the expanded terms back into the original equation:

5x - (x^3 + 6 - 2x) + (x^3 - x) = 0

Combine like terms:

5x - x^3 - 6 + 2x + x^3 - x = 0

Simplify:

6x - 6 = 0

Now, we want to isolate the variable x. Let's add 6 to both sides of the equation:

6x = 6

Finally, divide both sides by 6 to solve for x:

x = 6/6

x = 1

So, the solution to the equation is x = 1.