# SOLUTION: 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

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 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 youAnswer 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