Answer:
(k+3)(k+4)
The zeros are -3 and -4
Step-by-step explanation:
k2+7k+12
The middle number is 7 and the last number is 12.
This means we need two numbers that:
Add together to get 7
Multiply together to get 12
to fit in this equation (k+_)(k+_)
in this equation, it would be 3 and 4
3+4 = 7
3*4 = 12
Now we fill in the blanks to get
To get the zeros, use the equations
k+3=0 and k+4=0
You end up with
k=-3 and k=-4
Let numbers be x and y
ATQ
Adding both
Now putting value in eq(2)
the answer would be a number line with an open circle on -4, a closed circle on 3, and shading in between.
