Here is the work! The zeros of the graph are at -1 and 3
Answer: No, x+3 is not a factor of 2x^2-2x-12
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Explanation:
Let p(x) = 2x^2 - 2x - 12
If we divide p(x) over (x-k), then the remainder is p(k). I'm using the remainder theorem. A special case of the remainder theorem is that if p(k) = 0, then x-k is a factor of p(x).
Compare x+3 = x-(-3) to x-k to find that k = -3.
Plug x = -3 into the function
p(x) = 2x^2 - 2x - 12
p(-3) = 2(-3)^2 - 2(-3) - 12
p(-3) = 12
We don't get 0 as a result so x+3 is not a factor of p(x) = 2x^2 - 2x - 12
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Let's see what happens when we factor p(x)
2x^2 - 2x - 12
2(x^2 - x - 6)
2(x - 3)(x + 2)
The factors here are 2, x-3 and x+2
Z is greater than or equal to 7.
you would put a 7 near the front of the number line and put a colored in circle above it. then you would draw a line from the circle to the other end of the line
3/4 divided by 1/5 = 3 3/4
Hope that helps :)
Total cards = 52
Total red cards = 26
Total black cards = 26
P(one red and one black) = P(red and then black)+ P(black and then red)
P(one red and one black) =(26/52)(26/51) + (26/52)(26/51)
P(one red and one black) = 26/51 (Answer A)
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Answer: 26/51 (Answer A)
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