Question 61721: Can you help me please
Find all three zeros of f(x)=x^3-4X^2+13x+50 given that 3+4i is a zero
thank you
Answer by jai_kos(139)   (Show Source):
You can put this solution on YOUR website! f(x)=x^3-4X^2+13x+50
SOlution:
F(x) = x^3 - 4x^2 + 13 x + 50 ---->(1)
Check the last number, which is 50
Now find the factors of 50,
That is 50 ---- 1 , 2 , 5 ,25
So take all the numbers ---- 1 , -1 ,2 ,-2 ,5,-5,25 , -25
Now take the first factor 1. put in equation(1), we get
F(1) = 1^3 - 4*1^2 + 13*1+ 50 = 48
F(1) is not equal to zero.
Like this find for different numbers,
Then we find that F(-2) is equal to zero.
So x = -2 is a zero. ------>(2)
Now rewrite the above equation we find that...
x+2 = 0
Now take this and divide equation (1),
x^2 - 6x + 25
---------------------------
(x+2) | x^3 - 4x^2 + 13 x + 50
Subtract, we get
| x^3 + 2x^2
| - 6x^2 +13x
| - 6x^2 - 12x Subtract , we get
| 25x + 50
25x + 50 Subtract, we get
0
We get x^2 - 6x + 25
Simplify this quadratic equation, we get
x = - (-6) +- sqrt(6^2 - 4* 1* 25
2
x = 3 + 4i and x = 3 - 4i
Therefore the zeroes are x = 2 and x = 3 + 4i , x = 3 - 4i
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