Elimination 6x-6y=0 -6x-3y=27
1 answer:
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Answer:
3
Step-by-step explanation:
The fundamental theorem of algebra tells you the number of zeros is equal to the degree of the polynomial.
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<h3>application</h3>The given degree-3 polynomial has 3 zeros.
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<em>Additional comment</em>
Descarte's rule of signs tells you the absence of sign changes among the coefficients means that there are no positive real roots. Changing the sign of the one odd-degree term gives one sign change, so there is one negative real root. The other two zeros are complex.
one negative real zero, and two complex zeros
A graph confirms that the only real zero is negative and irrational, about -7.04.
The graph would have to look like the attachment I attached to this comment. If you can't see it then the graph will go through the y-intercept of -2 and the x-intercept of 3.
Explanation: Already we can tell that the line will go through -2 as the y-intercept. So we have to take the slope and apply it to the -2. The slope is in rise over run form. We rise 2 times. And then go to the right 3 times. We can do the same but backwards. Go down 2 times. and then go left 3 times. Do it twice in both ways and you have your line.
Explanation: Already we can tell that the line will go through -2 as the y-intercept. So we have to take the slope and apply it to the -2. The slope is in rise over run form. We rise 2 times. And then go to the right 3 times. We can do the same but backwards. Go down 2 times. and then go left 3 times. Do it twice in both ways and you have your line.