How do you find a polynomial function f with real coefficients of the indicated degree that possesses the given zeros: degree 2; 4+3i?

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Answer 1 · 2017-07-02T16:56:01

#f(x) = x^2 - 8x + 25#

We know that the function is a degree-2 polynomial, so the function has two solutions, one of which is given as #4 + 3i#.

Know that if one imaginary species is a solution to a polynomial equation, its conjugate is also a solution.

Therefore, the other solution to this equation is the conjugate of #4 + 3i#, which is #color(blue)(4-3i#.

Now that we know the solutions to the polynomial equation, let's derive a function, by setting them equal to zero like so:

#(x - (4 +3i))(x - (4 -3i)) = 0#

which is the same as

#(x - 4 - 3i)(x - 4 + 3i) = 0#

Factoring out the polynomial gives us

#x^2 - 4x + 3x i - 4x + 16 - 12i - 3x i + 12i - 9i^2 = 0#

Combining like terms..

#x^2 - 8x + 16 - 9i^2#

#i^2 = -1#:

#x^2 - 8x + 25#

The equation of the function with zeros #4 +- 3i# is thus

#color(blue)(f(x) = x^2 - 8x + 25#

Answer 2 · 2017-07-02T22:13:27

# x^2-8x+25= 0 #

Complex roots of real valued polynomials always appear in conjugate pairs, so if one root of #f(x)=0# is:

# alpha = 4+3i #

Then another root is:

# beta = 4-3i #

Se we are looking for a polynomial of degree #2#, ie a quadratic with roots #alpha# and #beta#.

So then:

Sum #=alpha+beta#
#" " = (4+3i) + (4-3i)#
#" " = 8#

Product #=alphabeta#
#" " = (4+3i)(4-3i)#
#" " = 16-9i^2#
#" " = 25#

And so the required equation is:

# x^2-("sum of roots")x + ("product of roots") = 0 #
# :. x^2-8x+25= 0 #