Firstly, let's break down the problem into manageable steps and analyze the provided polynomial [tex]\( P(x) = x^3 - x^2 - 4x + 4 \)[/tex].
### (a) Solving [tex]\( P(x) = 0 \)[/tex]
Given the polynomial [tex]\( P(x) = (x - 2)(x - 1)(x + 2) \)[/tex]:
To find the roots, we set [tex]\( P(x) = 0 \)[/tex]:
[tex]\[
(x - 2)(x - 1)(x + 2) = 0
\][/tex]
The roots (the values of [tex]\( x \)[/tex] that make [tex]\( P(x) = 0 \)[/tex]) can be found by setting each factor equal to zero:
[tex]\[
x - 2 = 0 \implies x = 2
\][/tex]
[tex]\[
x - 1 = 0 \implies x = 1
\][/tex]
[tex]\[
x + 2 = 0 \implies x = -2
\][/tex]
Therefore, the solution set for [tex]\( P(x) = 0 \)[/tex] is:
[tex]\[
\{ -2, 1, 2 \}
\][/tex]
### (b) Solving [tex]\( P(x) < 0 \)[/tex]
To solve the inequality [tex]\( P(x) < 0 \)[/tex], we need to find the intervals where the polynomial is negative.
Since [tex]\( P(x) = (x - 2)(x - 1)(x + 2) \)[/tex], we can analyze the sign of the product over the intervals determined by the roots [tex]\( -2, 1, \)[/tex] and [tex]\( 2 \)[/tex]:
1. For [tex]\( x \in (-\infty, -2) \)[/tex]:
- All three factors are negative (negative [tex]\(\times\)[/tex] negative [tex]\(\times\)[/tex] negative = negative).
2. For [tex]\( x \in (-2, 1) \)[/tex]:
- The factor [tex]\( (x + 2) \)[/tex] is positive, while [tex]\( (x - 1) \)[/tex] and [tex]\( (x - 2) \)[/tex] are negative (positive [tex]\(\times\)[/tex] negative [tex]\(\times\)[/tex] negative = positive).
3. For [tex]\( x \in (1, 2) \)[/tex]:
- The factor [tex]\( (x - 1) \)[/tex] is positive, [tex]\( (x - 2) \)[/tex] is negative, and [tex]\( (x + 2) \)[/tex] is positive (positive [tex]\(\times\)[/tex] positive [tex]\(\times\)[/tex] negative = negative).
4. For [tex]\( x \in (2, \infty) \)[/tex]:
- All three factors are positive (positive [tex]\(\times\)[/tex] positive [tex]\(\times\)[/tex] positive = positive).
Therefore, [tex]\( P(x) < 0 \)[/tex] for the intervals:
[tex]\[
(-\infty, -2) \cup (1, 2)
\][/tex]
### (c) Solving [tex]\( P(x) > 0 \)[/tex]
Similarly, to solve the inequality [tex]\( P(x) > 0 \)[/tex], we look at where the polynomial is positive:
From our analysis above, [tex]\( P(x) > 0 \)[/tex] for the intervals:
[tex]\[
(-2, 1) \cup (2, \infty)
\][/tex]
In conclusion:
- (a) The solution for [tex]\( P(x) = 0 \)[/tex] is [tex]\(\{ -2, 1, 2 \}\)[/tex].
- (b) The solution set for [tex]\( P(x) < 0 \)[/tex] is [tex]\( (-\infty, -2) \cup (1, 2) \)[/tex].
- (c) The solution set for [tex]\( P(x) > 0 \)[/tex] is [tex]\( (-2, 1) \cup (2, \infty) \)[/tex].
