What are the zeros of f(x) = x^2 - x- 12? Please explain if you can.

4 and -3

Step-by-step explanation:

So our function is f(x) = x² - x -12

We need to find factors such that 0 = x² - x -12

So, it looks something like this 0 = (x - a)(x - b)

We know that the first term is x², which can only be obtained if we multiply x by another x.

Now, we need to find the other terms a and b. We know that a × b = -12

The correct answer would be that a = 4 and b = -3

0 = (x - 4)(x + 3)

We know this because, if we multiply this out, we get x² + 3x -4x -12, which can be simplified to x² -x -12.

Our zeroes are the values of x that this equation 0 = (x - 4)(x + 3) is true. What values of x will make this zero?

4 and -3

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)

is there a question here?

step-by-step explanation: