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If two zeros of the polynomial f(x) = x4 - 6x3 - 26x2 + 138x - 35 are 2 ± √3.Find the other zeros. (Ans:7, -5)
Ans: Let the two zeros are 2 +√3 and 2 - √3
Sum of Zeros = 2 + √3 + 2 - √3
= 4
Product of Zeros = ( 2+√3 )(2 - √3 )
= 4 - 3
= 1
Quadratic polynomial is x2 - (sum) x + Product
x4 - 4 x3 + x2
-----------------
-2x3 - 27x2 + 138x
- 2x3 + 8x2 - 2x
-----------------------
-35x2 + 140x - 35
------------------------
0
∴ x2 - 2x - 35 = 0
(x - 7)(x + 5) = 0
x = 7, -5
other two Zeros are 7 and -5
Solve 4 cos(t )= 3 on[-8,10]. Solution : Here the first step is identical to the problems in the previous section. First we need to isolate the cosine on one side by itself & t
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pendaraban dan
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