Question

1. the zeros of the function $f(x) = x^2 - 5x - 6$ are

a. $-1$ and $6$
b. $1$ and $-6$
\t\t\t\t\t\t\t\t\t\t\tc. $2$ and $-3$
\t\t\t\t\t\t\t\t\t\t\td. $-2$ and $3$

1. the zeros of the function $f(x) = 2x^2 - 4x - 6$ are

a. $3$ and $-1$
\t\t\t\t\t\t\t\t\t\t\tc. $-3$ and $1$
b. $3$ and $1$
\t\t\t\t\t\t\t\t\t\t\td. $-3$ and $-1$

1. what are the zeros of the function $f(x) = x^2 - 13x - 30?$

a. $-10$ and $3$
\t\t\t\t\t\t\t\t\t\t\tc. $-15$ and $2$
b. $10$ and $-3$
\t\t\t\t\t\t\t\t\t\t\td. $15$ and $-2$

1. what are the zeros of the function $f(x) = (x + 2)^2 - 25?$

a. $-2$ and $5$
\t\t\t\t\t\t\t\t\t\t\tc. $-5$ and $2$
b. $7$ and $-3$
\t\t\t\t\t\t\t\t\t\t\td. $3$ and $-7$

1. if $f(x) = 2x^2 + x - 3$, which equation can be used to determine the zeros of the function?

a. $0 = (2x - 3)(x + 1)$
\t\t\t\t\t\t\t\t\t\t\tc. $0 = 2x(x + 1) - 3$
b. $0 = (2x + 3)(x - 1)$
\t\t\t\t\t\t\t\t\t\t\td. $0 = 2x(x - 1) - 3(x + 1)$

Explanation:

Question 1

Step1: Set \( f(x) = 0 \)

To find the zeros of \( f(x)=x^{2}-5x - 6 \), we set \( f(x)=0 \), so we have the equation \( x^{2}-5x - 6=0 \).

Step2: Factor the quadratic

We factor the quadratic expression. We need two numbers that multiply to \( - 6 \) and add up to \( -5 \). The numbers are \( -6 \) and \( 1 \) since \( (-6)\times1=-6 \) and \( -6 + 1=-5 \). So, \( x^{2}-5x - 6=(x - 6)(x + 1) \).

Step3: Solve for \( x \)

Set each factor equal to zero:

• \( x - 6=0\Rightarrow x = 6 \)
• \( x+1=0\Rightarrow x=-1 \)

Step1: Set \( f(x) = 0 \)

For \( f(x)=2x^{2}-4x - 6 \), set \( f(x) = 0 \), so \( 2x^{2}-4x - 6=0 \). We can divide through by \( 2 \) to simplify: \( x^{2}-2x - 3=0 \).

Step2: Factor the quadratic

We need two numbers that multiply to \( -3 \) and add up to \( -2 \). The numbers are \( -3 \) and \( 1 \). So, \( x^{2}-2x - 3=(x - 3)(x + 1) \).

Step3: Solve for \( x \)

Set each factor equal to zero:

• \( x - 3=0\Rightarrow x = 3 \)
• \( x + 1=0\Rightarrow x=-1 \)

Step1: Set \( f(x) = 0 \)

For \( f(x)=x^{2}-13x - 30 \), set \( f(x)=0 \), so \( x^{2}-13x - 30=0 \).

Step2: Factor the quadratic

We need two numbers that multiply to \( -30 \) and add up to \( -13 \). The numbers are \( -15 \) and \( 2 \) since \( (-15)\times2=-30 \) and \( -15 + 2=-13 \). So, \( x^{2}-13x - 30=(x - 15)(x + 2) \).

Step3: Solve for \( x \)

Set each factor equal to zero:

• \( x - 15=0\Rightarrow x = 15 \)
• \( x + 2=0\Rightarrow x=-2 \)

Answer:

a. \( - 1 \) and \( 6 \)

Question 2