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Graph the following polynomial function by hand. Then solve the equation and inequalities.

[tex]\[
\begin{aligned}
P(x) & = x^3 - x^2 - 4x + 4 \\
& = (x-2)(x-1)(x+2)
\end{aligned}
\][/tex]

(a) [tex]\(P(x) = 0\)[/tex]
(b) [tex]\(P(x) \ \textless \ 0\)[/tex]
(c) [tex]\(P(x) \ \textgreater \ 0\)[/tex]

(a) The solution set for [tex]\(P(x) = 0\)[/tex] is [tex]\(\{1, 2, -2\}\)[/tex].
(Use a comma to separate answers as needed.)

(b) The solution set for [tex]\(P(x) \ \textless \ 0\)[/tex] is [tex]\(\square\)[/tex].
(Type your answer in interval notation.)

Sagot :

GinnyAnswer
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].
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