The quotient of s and three
1 answer:
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Answer:
D.)
Step-by-step explanation:
The zero's are referencing when y=0, note that when y=0 they are talking about the x-intercepts. You can graph the function and see when the graph crosses the x-axis or solve for the x-values. I will solve it via factoring and so:
Multiply the outer coefficients, in this case 1 and 6, and 1×6=6. Now let's think about all the factors of 6 we have: 6×1 and 2×3. Now is there a way that if we use any of these factors and add/subtract them they will return the middle term 5? Actually we can say 6-1=5 and 2+3=5. Let's try both.
First let's use 6 and -1 and so:
Notice how we have (x+6) and (x-6), these factors do not match so this is incorrect.
Now let's try 2 and 3 and so:
Notice how the factors (x+3) matched up so this is a factor and so is (x+2), now to solve for the zero's let's make f(x)=0 and solve each factor separately:
Case 1:
Case 2:
So your zero's are when x=-2 and x=-3.
D.) x=-3 and x=-2 because the graph crosses the x-axis at -3 and -2.
~~~Brainliest Appreciated~~~