Question

Y^3-2y^2+y-2=0 решитееееееееее

Answer 1

$y^3-2y^2+y-2=0\\y^2(y-2)+(y-2)=0\\(y-2)(y^2+1)=0\\y^2+1=0 => y^2\neq -1\\y=2$

Answer 2

${y}^{2} (y - 2) + (y - 2) = 0$

$({y}^{2} + 1)(y - 2) = 0$

$y = 2$

Answer 3

To solve the equation $y^3 - 2y^2 + y - 2 = 0$, we can use various methods, such as factoring or using the rational root theorem. Let's try factoring the equation first.

Step 1: Factor by Grouping $y^3 - 2y^2 + y - 2 = 0$

Let's group the terms:

$(y^3 - 2y^2) + (y - 2) = 0$

Step 2: Factor out common terms from each group:

$y^2(y - 2) + (y - 2) = 0$

Step 3: Factor out the common factor, which is $y - 2$:

$(y - 2)(y^2 + 1) = 0$

Now, we have a quadratic equation $y^2 + 1 = 0$ and a linear equation $y - 2 = 0$.

Step 4: Solve the quadratic equation: $y^2 + 1 = 0$

The quadratic equation $y^2 + 1 = 0$ has no real solutions since the discriminant ($b^2 - 4ac$) is negative ($0^2 - 4(1)(1) = -4$).

Step 5: Solve the linear equation: $y - 2 = 0$

Adding 2 to both sides, we get:

$y = 2$

So, the only real solution to the original equation is $y = 2$.