Where R is the median between Q and L:
From my understanding of a triangle's centroid, it divides an angle bisector into parts of 2/3 and 1/3. In the given problem, these divisions are NS and SR. Therefore, twice SR would be equal to NS. From here, we can get the value of X, to solve for SR.
NS = 2SR
(x + 10) = 2(x + 3)
x + 10 = 2x + 6
x = 4
Therefore, SR = (x + 3) = 7
Write the coeeficientes of the polynomial in order:
| 1 - 5 6 - 30
|
|
|
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After some trials you probe with 5
| 1 - 5 6 - 30
|
|
5 | 5 0 30
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1 0 6 0 <---- residue
Given that the residue is 0, 5 is a root.
The quotient is x^2 + 6 = 0, which does not have a real root.
Therefore, 5 is the only root. You can prove it by solving the polynomial x^2 + 6 = 0.
Area = area of rectangle - area of 2 hemispheres
= 24 * 32 - pi * 12^2
= 315.6 sq units
