Question

1. $f(x) = x^2 - 36$
2. $m(x) = x^2 + 2x - 8$
3. $x^2 - 7x = 0$
4. keith determined the zeros of the function $f(x)$ to be -6 and 5. what could be keiths function?

a. $f(x) = (x + 5)(x + 6)$
b. $f(x) = (x + 5)(x - 6)$
c. $f(x) = (x - 5)(x + 6)$
d. $f(x) = (x - 5)(x - 6)$

Explanation:

Step1: Factor difference of squares

$f(x)=x^2-36=(x-6)(x+6)$

Step2: Solve for zeros

Set $(x-6)(x+6)=0$ → $x=6$ or $x=-6$

Step1: Factor quadratic trinomial

$m(x)=x^2+2x-8=(x+4)(x-2)$

Step2: Solve for zeros

Set $(x+4)(x-2)=0$ → $x=-4$ or $x=2$

Step1: Factor out common term

$x^2-7x=0 \implies x(x-7)=0$

Step2: Solve for zeros

Set $x(x-7)=0$ → $x=0$ or $x=7$

Step1: Relate zeros to factors

Zeros $x=-6$ and $x=5$ correspond to $(x+6)$ and $(x-5)$

Step2: Match to function

The function is $f(x)=(x-5)(x+6)$

Answer:

1. $x = 6$ and $x = -6$
2. $x = -4$ and $x = 2$
3. $x = 0$ and $x = 7$
4. c. $f(x) = (x - 5)(x + 6)$