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
X=5y+10 this is the answer
Answer:
The "check your answer" probably means check your answer by graphing.
Step-by-step explanation:
Answer:
x=0 y=3
Step-by-step explanation:
Answer:
(a) The unit circle is centered at (0,0) with a radius of 1.
(b) The equation of a circle of radius <em>r</em>, with a center located at (0,0):
<em>x</em>²<em>+ y</em>² <em>= r</em>².
(c) (i) P(1,0)
(ii) P(0,1)
(iii) P(-1,0)
(iv) P(0,-1)
Step-by-step explanation:
