Use the given zero r=5+2i to find all the zeros of
f(x)= (x^3)-(7x^2)-(x)+(87)
thanks so much!
f(x)= (x^3)-(7x^2)-(x)+(87)
thanks so much!
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Since f(x) is a cubic polynomial with real integer coefficients and a complex root r1 = 5 + 2.i, it will have a second root r2 which is the complex conjugate of r1. Since cubics always have at least one real root, the third r3 must be real.
Hence we may write
f(x) = x³ - 7.x² - x + 87 = (x - 5 + 2.i).(x - 5 - 2.i).(x - r3)
. . .= [(x² - 5.x - 2.i.x) + (-5.x + 25 + 10.i) + (2.i.x - 10.i + 4).(x - r3)
. . .= (x² - 10.x + 29).(x - r3)
Comparing the product of the last two terms 29 and r3 in each bracket in the final expression with the constant term in f(x), we see that r3 = -3. To check the result, we expand
f(x) .= (x² - 10.x + 29).(x + 3) = (x³ - 10.x² + 29.x + 3.x² - 30.x + 87) = x³ - 7.x² - x + 87
as required. So the three roots are 5 ± 2.i and -3.
Hence we may write
f(x) = x³ - 7.x² - x + 87 = (x - 5 + 2.i).(x - 5 - 2.i).(x - r3)
. . .= [(x² - 5.x - 2.i.x) + (-5.x + 25 + 10.i) + (2.i.x - 10.i + 4).(x - r3)
. . .= (x² - 10.x + 29).(x - r3)
Comparing the product of the last two terms 29 and r3 in each bracket in the final expression with the constant term in f(x), we see that r3 = -3. To check the result, we expand
f(x) .= (x² - 10.x + 29).(x + 3) = (x³ - 10.x² + 29.x + 3.x² - 30.x + 87) = x³ - 7.x² - x + 87
as required. So the three roots are 5 ± 2.i and -3.
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divide by x^2 -10x +29 an result is x+3 then remaining zero is -3