If <em>x</em> + 1 is a factor of <em>p(x)</em> = <em>x</em>³ + <em>k</em> <em>x</em>² + <em>x</em> + 6, then by the remainder theorem, we have
<em>p</em> (-1) = (-1)³ + <em>k</em> (-1)² + (-1) + 6 = 0 → <em>k</em> = -4
So we have
<em>p(x)</em> = <em>x</em>³ - 4<em>x</em>² + <em>x</em> + 6
Dividing <em>p(x)</em> by <em>x</em> + 1 (using whatever method you prefer) gives
<em>p(x)</em> / (<em>x</em> + 1) = <em>x</em>² - 5<em>x</em> + 6
Synthetic division, for instance, might go like this:
-1 | 1 -4 1 6
... | -1 5 -6
----------------------------
... | 1 -5 6 0
Next, we have
<em>x</em>² - 5<em>x</em> + 6 = (<em>x</em> - 3) (<em>x</em> - 2)
so that, in addition to <em>x</em> = -1, the other two zeros of <em>p(x)</em> are <em>x</em> = 3 and <em>x</em> = 2
Answer:
1 month
Step-by-step explanation:
249 is the amount she spends for buying the kitten, (299-50=249) which she only has to pay once.
20 is the amount she has to spend every month, the rate of change.
The problem can be modeled with the general equation:
y = 20x + 249
In this equation, x the the time in months and y is the amount of money spent.
Substitute 250 for y.
y = 20x + 249
250 = 20x + 249 <= isolate x to get the number of months
1 = 20x
x = 1/20
Carol will only spent money every month, not for 1/20 of a month.
It will only take Carol 1 month to spent $250.
Check answer:
Substitute x for 1
y = 20x + 249
y = 20(1) + 249
y = 269
269 is more than 250 already.
Answer:
My calculator said error and i tried looking it up it said error too. I'm sorry I tried, you'll have to do it by hand
Step-by-step explanation:
3979.00-781.00= 3198.00 is how much money was taken in at each game
Mica should have added 2 to 3 instead of 1
