Question

(x − 2)(x − 3)(x − 5) = (x − 2)(x − 4)(x − 5).

Answer 1

$(x-2)(x-3)(x-5)=(x-2)(x-4)(x-5)\\\\(x-2)(x-5)\cdot \Big ((x-3)-(x-4)\Big )=0\\\\(x-2)(x-5)\cdot 1=0\\\\x_1=2\; ,\; \; x_2=5$

Answer 2

To solve this equation, we need to find the values of "x" that satisfy the equation. To do this, we can start by expanding both sides of the equation and then simplifying:

Expanding the left side: (x - 2)(x - 3)(x - 5) = (x^2 - 5x + 6)(x - 5) = x^3 - 5x^2 + 6x - 5x^2 + 25x - 30 = x^3 - 10x^2 + 31x - 30.

Expanding the right side: (x - 2)(x - 4)(x - 5) = (x^2 - 4x + 4)(x - 5) = x^3 - 5x^2 + 4x - 5x^2 + 25x - 20 = x^3 - 10x^2 + 29x - 20.

Now we have the equation: x^3 - 10x^2 + 31x - 30 = x^3 - 10x^2 + 29x - 20.

To solve for "x," we can cancel out the common terms on both sides (x^3 and -10x^2) since they are the same on both sides:

31x - 30 = 29x - 20.

Next, let's isolate the "x" term on one side:

31x - 29x = -20 + 30, 2x = 10.

Finally, divide both sides by 2 to solve for "x":

x = 10 / 2, x = 5.

So, the solution to the equation is x = 5.