Question

(х²-2х)(х²-х-6)(х²+3х)<=0

Answer 1

Вот здесь все расписанно

Answer 2

To solve the inequality $(x^2-2x)(x^2-x-6)(x^2+3x) \leq 0$, we need to find the intervals on the number line where the expression is less than or equal to zero.

Step 1: Find the critical points by setting each factor equal to zero and solving for $x$:

1. $x^2 - 2x = 0$: Factoring out $x$, we get $x(x - 2) = 0$, so $x = 0$ or $x = 2$.
2. $x^2 - x - 6 = 0$: Factoring the quadratic, we have $(x - 3)(x + 2) = 0$, so $x = 3$ or $x = -2$.
3. $x^2 + 3x = 0$: Factoring out $x$, we get $x(x + 3) = 0$, so $x = 0$ or $x = -3$.

Step 2: Create intervals on the number line using the critical points as boundaries:

• $(-\infty, -3)$
• $(-3, -2)$
• $(-2, 0)$
• $(0, 2)$
• $(2, 3)$
• $(3, \infty)$

Step 3: Test points in each interval to determine the sign of the expression within that interval:

• For $x < -3$, let's take $x = -4$ (a value less than -3): $((-4)^2 - 2(-4))((-4)^2 - (-4) - 6)((-4)^2 + 3(-4)) = (24)(30)(-4) < 0$ The expression is negative in this interval.

• For $-3 < x < -2$, let's take $x = -2.5$ (a value between -3 and -2): $((-2.5)^2 - 2(-2.5))((-2.5)^2 - (-2.5) - 6)((-2.5)^2 + 3(-2.5)) = (15.625)(20.125)(-3.125) < 0$ The expression is negative in this interval.

• For $-2 < x < 0$, let's take $x = -1$ (a value between -2 and 0): $((-1)^2 - 2(-1))((-1)^2 - (-1) - 6)((-1)^2 + 3(-1)) = (3)(6)(-2) < 0$ The expression is negative in this interval.

• For $0 < x < 2$, let's take $x = 1$ (a value between 0 and 2): $((1)^2 - 2(1))((1)^2 - (1) - 6)((1)^2 + 3(1)) = (-1)(-6)(4) > 0$ The expression is positive in this interval.

• For $2 < x < 3$, let's take $x = 2.5$ (a value between 2 and 3): $((2.5)^2 - 2(2.5))((2.5)^2 - (2.5) - 6)((2.5)^2 + 3(2.5)) = (3.125)(5.625)(10.625) > 0$ The expression is positive in this interval.

• For $x > 3$, let's take $x = 4$ (a value greater than 3): $((4)^2 - 2(4))((4)^2 - (4) - 6)((4)^2 + 3(4)) = (16)(6)(28) > 0$ The expression is positive in this interval.

Step 4: Determine the solution:

The inequality $(x^2-2x)(x^2-x-6)(x^2+3x) \leq 0$