Question

(x²+2x-15)(x²-4x+3)(x-1)≤0

Answer 1

Ответ смотри в вложении

Answer 2

To solve the inequality $(x^2+2x-15)(x^2-4x+3)(x-1) \leq 0$, we need to find the values of $x$ that make the expression on the left-hand side less than or equal to zero.

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

1. $x^2+2x-15 = 0$: Factoring the quadratic equation gives $(x+5)(x-3) = 0$, so the critical points are $x = -5$ and $x = 3$.

2. $x^2-4x+3 = 0$: Factoring the quadratic equation gives $(x-1)(x-3) = 0$, so the critical points are $x = 1$ and $x = 3$.

3. $x-1 = 0$: Solving for $x$ gives $x = 1$.

Step 2: Plot these critical points on a number line:

pythonCopy code
... -5 ----- 1 ----- 3 ...

Step 3: Test each interval between the critical points to determine the sign of the expression:

• For $x < -5$: Pick $x = -6$ (arbitrary value less than -5) $(x^2+2x-15)(x^2-4x+3)(x-1) = ((-6)^2 + 2(-6) - 15)((-6)^2 - 4(-6) + 3)((-6) - 1) = (0)(45)(-7) = 0$ Since the product is zero, we include this interval.

• For $-5 < x < 1$: Pick $x = 0$ (arbitrary value between -5 and 1) $(x^2+2x-15)(x^2-4x+3)(x-1) = ((0)^2 + 2(0) - 15)((0)^2 - 4(0) + 3)((0) - 1) = (-15)(3)(-1) = 45$ Since the product is positive, we exclude this interval.

• For $1 < x < 3$: Pick $x = 2$ (arbitrary value between 1 and 3) $(x^2+2x-15)(x^2-4x+3)(x-1) = ((2)^2 + 2(2) - 15)((2)^2 - 4(2) + 3)((2) - 1) = (1)(-5)(1) = -5$ Since the product is negative, we include this interval.

• For $x > 3$: Pick $x = 4$ (arbitrary value greater than 3) $(x^2+2x-15)(x^2-4x+3)(x-1) = ((4)^2 + 2(4) - 15)((4)^2 - 4(4) + 3)((4) - 1) = (1)(7)(3) = 21$ Since the product is positive, we exclude this interval.

Step 4: Write the solution:

The solution to the inequality is $x \in (-\infty, -5] \cup (1, 3]$.