Asked by Sam
Form a polynomial f(x) with real coefficients, with leading coefficient 1, having the given degree and zeros.
Degree: 4; zeros: -4+3i and 1; multiplicity of 2
Degree: 4; zeros: -4+3i and 1; multiplicity of 2
Answers
Answered by
Damon
if -4 + 3 i
then -4 - 3 i is also a zero
so
(x + 4 - 3 i) (x + 4 + 3 i) ( x-1) (x-1) = y
then -4 - 3 i is also a zero
so
(x + 4 - 3 i) (x + 4 + 3 i) ( x-1) (x-1) = y
Answered by
Reiny
alternative way:
f(x) = (x-1)^2 (a quadratic)
complex roots always come in conjugate pairs, so -4 - 3i is another
sum of the complex roots = -8
product of the complex roots = (-4+3i)(-4 - 3i)
= 16 - 9i^2 = 25
so the quadratic part is x^2 + 8x + 25
f(x) = (x^2 + 8x + 25)(x-1)^2
of course you could also expand
f(x) = <b>(x - (-4+3i))(x - (-4-3i))</b>(x-1)^2 to get the same thing
this is how Damon did it
f(x) = (x-1)^2 (a quadratic)
complex roots always come in conjugate pairs, so -4 - 3i is another
sum of the complex roots = -8
product of the complex roots = (-4+3i)(-4 - 3i)
= 16 - 9i^2 = 25
so the quadratic part is x^2 + 8x + 25
f(x) = (x^2 + 8x + 25)(x-1)^2
of course you could also expand
f(x) = <b>(x - (-4+3i))(x - (-4-3i))</b>(x-1)^2 to get the same thing
this is how Damon did it
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