If F ( X ) = X 2 + 2 X 2 − 5 X − 6 F(x) = X^2 + 2x^2 - 5x - 6 F ( X ) = X 2 + 2 X 2 − 5 X − 6 :(a) Find F ( − 1 F(-1 F ( − 1 ].(b) Express F ( X F(x F ( X ] As A Product Of Linear Factors.(c) Describe The Values Of X X X For Which F ( X F(x F ( X ] Is Positive.

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(a) Find $f(-1)$

To find the value of $f(-1)$, we need to substitute $x = -1$ into the function $f(x) = x^2 + 2x^2 - 5x - 6$.

$f(-1) = (-1)^2 + 2(-1)^2 - 5(-1) - 6$

$f(-1) = 1 + 2 + 5 - 6$

$f(-1) = 2$

Therefore, the value of $f(-1)$ is 2.

(b) Express $f(x)$ as a product of linear factors

To express $f(x)$ as a product of linear factors, we need to factorize the quadratic expression.

$f(x) = x^2 + 2x^2 - 5x - 6$

$f(x) = 3x^2 - 5x - 6$

We can factorize the quadratic expression by finding two numbers whose product is $3 \times -6 = -18$ and whose sum is $-5$. The two numbers are $-9$ and $2$.

$f(x) = 3x^2 - 9x + 2x - 6$

$f(x) = 3x(x - 3) + 2(x - 3)$

$f(x) = (3x + 2)(x - 3)$

Therefore, the function $f(x)$ can be expressed as a product of linear factors as $(3x + 2)(x - 3)$.

(c) Describe the values of $x$ for which $f(x)$ is positive

To describe the values of $x$ for which $f(x)$ is positive, we need to find the intervals where the function is positive.

We can find the intervals by finding the zeros of the function.

$f(x) = (3x + 2)(x - 3)$

To find the zeros, we need to set each factor equal to zero and solve for $x$.

$3x + 2 = 0$

$3x = -2$

$x = -\frac{2}{3}$

$x - 3 = 0$

$x = 3$

Therefore, the zeros of the function are $x = -\frac{2}{3}$ and $x = 3$.

To find the intervals where the function is positive, we need to test a value from each interval.

Interval 1: $(-\infty, -\frac{2}{3})$

Test value: $x = -1$

$f(-1) = (3(-1) + 2)(-1 - 3)$

$f(-1) = (-1)(-4)$

$f(-1) = 4$

Since $f(-1) > 0$, the function is positive in the interval $(-\infty, -\frac{2}{3})$.

Interval 2: $(-\frac{2}{3}, 3)$

Test value: $x = 0$

$f(0) = (3(0) + 2)(0 - 3)$

$f(0) = (2)(-3)$

$f(0) = -6$

Since $f(0) < 0$, the function is not positive in the interval $(-\frac{2}{3}, 3)$.

Interval 3: $(3, \infty)$

Test value: $x = 4$

$f(4) = (3(4) + 2)(4 - 3)$

$f(4) = (14)(1)$

$f(4) = 14$

Since $f(4) > 0$, the function is positive in the interval $(3, \infty)$.

Therefore, the function $f(x)$ is positive in the intervals $(-\infty, -\frac{2}{3})$ and $(3, \infty)$.

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Q: What is the value of $f(-1)$?

A: To find the value of $f(-1)$, we need to substitute $x = -1$ into the function $f(x) = x^2 + 2x^2 - 5x - 6$.

$f(-1) = (-1)^2 + 2(-1)^2 - 5(-1) - 6$

$f(-1) = 1 + 2 + 5 - 6$

$f(-1) = 2$

Therefore, the value of $f(-1)$ is 2.

Q: How do we express $f(x)$ as a product of linear factors?

A: To express $f(x)$ as a product of linear factors, we need to factorize the quadratic expression.

$f(x) = x^2 + 2x^2 - 5x - 6$

$f(x) = 3x^2 - 5x - 6$

We can factorize the quadratic expression by finding two numbers whose product is $3 \times -6 = -18$ and whose sum is $-5$. The two numbers are $-9$ and $2$.

$f(x) = 3x^2 - 9x + 2x - 6$

$f(x) = 3x(x - 3) + 2(x - 3)$

$f(x) = (3x + 2)(x - 3)$

Therefore, the function $f(x)$ can be expressed as a product of linear factors as $(3x + 2)(x - 3)$.

Q: What are the zeros of the function $f(x)$?

A: To find the zeros of the function, we need to set each factor equal to zero and solve for $x$.

$3x + 2 = 0$

$3x = -2$

$x = -\frac{2}{3}$

$x - 3 = 0$

$x = 3$

Therefore, the zeros of the function are $x = -\frac{2}{3}$ and $x = 3$.

Q: In which intervals is the function $f(x)$ positive?

A: To find the intervals where the function is positive, we need to test a value from each interval.

Interval 1: $(-\infty, -\frac{2}{3})$

Test value: $x = -1$

$f(-1) = (3(-1) + 2)(-1 - 3)$

$f(-1) = (-1)(-4)$

$f(-1) = 4$

Since $f(-1) > 0$, the function is positive in the interval $(-\infty, -\frac{2}{3})$.

Interval 2: $(-\frac{2}{3}, 3)$

Test value: $x = 0$

$f(0) = (3(0) + 2)(0 - 3)$

$f(0) = (2)(-3)$

$f(0) = -6$

Since $f(0) < 0$, the function is not positive in the interval $(-\frac{2}{3}, 3)$.

Interval 3: $(3, \infty)$

Test value: $x = 4$

$f(4) = (3(4) + 2)(4 - 3)$

$f(4) = (14)(1)$

$f(4) = 14$

Since $f(4) > 0$, the function is positive in the interval $(3, \infty)$.

Therefore, the function $f(x)$ is positive in the intervals $(-\infty, -\frac{2}{3})$ and $(3, \infty)$.

Q: What is the significance of the zeros of the function $f(x)$?

A: The zeros of the function $f(x)$ are the values of $x$ for which the function is equal to zero. In this case, the zeros are $x = -\frac{2}{3}$ and $x = 3$. The zeros are significant because they divide the number line into intervals where the function is either positive or negative.

Q: How do we use the zeros to determine the intervals where the function is positive?

A: To determine the intervals where the function is positive, we need to test a value from each interval. If the value is positive, then the function is positive in that interval. If the value is negative, then the function is not positive in that interval.

Q: What is the relationship between the zeros and the intervals where the function is positive?

A: The zeros of the function divide the number line into intervals where the function is either positive or negative. The intervals where the function is positive are the intervals that do not contain any zeros.

Q: How do we use the intervals where the function is positive to determine the values of $x$ for which the function is positive?

A: To determine the values of $x$ for which the function is positive, we need to find the intervals where the function is positive. The values of $x$ that are in these intervals are the values for which the function is positive.

Q: What is the significance of the intervals where the function is positive?

A: The intervals where the function is positive are significant because they determine the values of $x$ for which the function is positive.