Question

(x+3)^2(x-2)(x+5)^3 <0

Answer 1

Ответ:

8

Объяснение:

(x^2-3x-2x+6)-(x^2-5x+2x-10)=0

x^2-3x-2x+6-x^2+5x-2x+10=0

-2x+16=0

-2x=-16 |:(-2)

x=8

ответ: 8

Answer 2

To solve the inequality $(x+3)^2(x-2)(x+5)^3 < 0$, we can follow these steps:

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

1. $x+3 = 0 \Rightarrow x = -3$
2. $x-2 = 0 \Rightarrow x = 2$
3. $x+5 = 0 \Rightarrow x = -5$

Step 2: Create intervals on the number line using the critical points: $-\infty, -5, -3, 2, \infty$.

Step 3: Test a value from each interval in the inequality $(x+3)^2(x-2)(x+5)^3 < 0$ to determine the sign of the expression within that interval.

1. Test $x = -6$ (in the interval $-\infty, -5$): $((-6)+3)^2((-6)-2)((-6)+5)^3 = (-3)^2(-8)(-1)^3 = -72 < 0$

2. Test $x = -4$ (in the interval $-5, -3$): $((-4)+3)^2((-4)-2)((-4)+5)^3 = (-1)^2(-6)(1)^3 = 6 > 0$

3. Test $x = 0$ (in the interval $-3, 2$): $(0+3)^2((0)-2)((0)+5)^3 = (3)^2(-2)(5)^3 = -150 < 0$

4. Test $x = 3$ (in the interval $2, \infty$): $(3+3)^2((3)-2)((3)+5)^3 = (6)^2(1)(8)^3 = 3072 > 0$

Step 4: Analyze the sign of the expression within each interval:

• In the interval $-\infty, -5$, the expression is negative.
• In the interval $-5, -3$, the expression is positive.
• In the interval $-3, 2$, the expression is negative.
• In the interval $2, \infty$, the expression is positive.

Step 5: Determine the intervals where the expression is negative:

The expression is negative in the intervals $(- \infty, -5)$ and $(-3, 2)$.

Step 6: Write the solution in interval notation:

The solution to the inequality $(x+3)^2(x-2)(x+5)^3 < 0$ is $x \in (- \infty, -5) \cup (-3, 2)$. This means that the inequality is satisfied when $x$ is any real number in the intervals $(- \infty, -5)$ and $(-3, 2)$.