Find K if x + 1 is a factor of P(x) = Kx^2-x+2
1 answer:
Answer:
Step-by-step explanation:
According to the Factor Theorem, a binomial (<em>x</em> - <em>a</em>) is a factor of a polynomial P(x) if and only if P(a) = 0.
We have the polynomial:
And we want to find the value of <em>K</em> such that (<em>x</em> + 1) is a factor.
We can rewrite our factor as (<em>x</em> - (-1)). Thus, <em>a</em> = -1.
Then in orer for (<em>x</em> + 1) to be a factor, P(-1) must equal 0. Hence:
Simplify:
Solve for <em>K. </em>Hence:
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Answer:
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Step-by-step explanation:
Hi, there!
Since we can describe eccentricity as
a) Eccentricity close to 0
An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)
b) Eccentricity =5
An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:
c) Eccentricity close to 1
In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.
