Solve: (2 + x)(5 + x)(4 - x) < 0 (write answer in interval notation)
Which I think is xe (-infinity, -5) or (-2, 4)
and
Solve: (4x + 1)(3 - x) > or equal to 0 (write answer in interval notation)
Which im unsure
#1.
(2 + x)(5 + x)(4 - x) < 0
Equate each term inside the parenthesis to zero:
2 + x < 0
x < -2
5 + x < 0
x < -5
4 - x < 0
4 < x
x > 4
When x > 4, only the term (4 - x) is negative while the others are positive, so it satisfies the inequality.
When x < -5, say x = -6,
(2 - 6)(5 - 6)(4 + 6) < 0
-4 * -1 * 10 < 0
40 < 0
which is false. So x cannot be less than or equal to -5.
When x < -2 but greater than -5, say x = -3
(2 - 3)(5 - 3)(4 + 3) < 0
-1 * 2 * 7 < 0
-14 < 0
which is true. So -5 < x < -2. Combine it with x > 4, we'll have
(-5, -2) U (4, +infinity)
#2.
(4x + 1)(3 - x) ≥ 0
Same procedure we'll do. Equate each term to zero:
4x + 1 ≥ 0
4x ≥ -1
x ≥ -1/4
3 - x ≥ 0
3 ≥ x
We get the intersection. So x must be greater than or equal to -1/4 but less than or equal to 3:
[-1/4 , 3]
hope this helps~ `u`
After you get a bit of practice with this kind of stuff, you can feel confident using what you know about the graphs of polynomials.
(2 + x)(5 + x)(4 - x) < 0
You know this is a cubic, with three roots. The graph crosses the x-axis at each root. So, pick a very large negative value for x.
The product (-)(-)(+) is positive
So, you know that the graph crosses below the x-axis at -5 back up at -2 and back down at 4. So, f(x) < 0 on
(-5,-2)U(4,∞)
To solve the first inequality, (2 + x)(5 + x)(4 - x) < 0, we need to consider the sign of each factor.
Step 1: Set each factor equal to zero and find their roots:
- Set 2 + x = 0 -> x = -2
- Set 5 + x = 0 -> x = -5
- Set 4 - x = 0 -> x = 4
Step 2: Plot these three points on a number line:
- Place an open circle at -2
- Place an open circle at -5
- Place an open circle at 4
Step 3: Test the intervals between these points to determine the sign of the expression (2 + x)(5 + x)(4 - x):
- Test a value less than -5, like x = -10
- Substitute x = -10 into the expression: (2 + x)(5 + x)(4 - x) = (-8)(-5)(14) = 560
- Since the product is positive, the expression is positive in this range.
- Test a value between -5 and -2, like x = -3
- Substitute x = -3 into the expression: (2 + x)(5 + x)(4 - x) = (-1)(2)(7) = -14
- Since the product is negative, the expression is negative in this range.
- Test a value between -2 and 4, like x = 0
- Substitute x = 0 into the expression: (2 + x)(5 + x)(4 - x) = (2)(5)(4) = 40
- Since the product is positive, the expression is positive in this range.
- Test a value greater than 4, like x = 10
- Substitute x = 10 into the expression: (2 + x)(5 + x)(4 - x) = (12)(15)(-6) = -1,080
- Since the product is negative, the expression is negative in this range.
Step 4: Determine the intervals where the expression is less than zero (negative):
- Based on the tests, the expression is negative in the interval (-5, -2)
Step 5: Write the answer in interval notation:
- Answer: x ∈ (-5, -2)
Now, moving on to the second inequality, (4x + 1)(3 - x) ≥ 0:
Step 1: Set each factor equal to zero and find their roots:
- Set 4x + 1 = 0 -> 4x = -1 -> x = -1/4
- Set 3 - x = 0 -> x = 3
Step 2: Plot these two points on a number line:
- Place a closed circle at -1/4
- Place a closed circle at 3
Step 3: Test the intervals between these points to determine the sign of the expression (4x + 1)(3 - x):
- Test a value less than -1/4, like x = -2
- Substitute x = -2 into the expression: (4x + 1)(3 - x) = (-7)(5) = -35
- Since the product is negative, the expression is negative in this range.
- Test a value between -1/4 and 3, like x = 1
- Substitute x = 1 into the expression: (4x + 1)(3 - x) = (5)(2) = 10
- Since the product is positive, the expression is positive in this range.
- Test a value greater than 3, like x = 5
- Substitute x = 5 into the expression: (4x + 1)(3 - x) = (21)(-2) = -42
- Since the product is negative, the expression is negative in this range.
Step 4: Determine the intervals where the expression is greater than or equal to zero (non-negative):
- Based on the tests, the expression is non-negative in the intervals (-∞, -1/4] and [3, ∞)
Step 5: Write the answer in interval notation:
- Answer: x ∈ (-∞, -1/4] ∪ [3, ∞)