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: the MAT117 Midterm Form A Page 1 of 1
Instructor: Beth Jones
equation of
the line:
12. [6 pts] Find a polynomial with integer coefficients with leading coefficient 1, degree 3,
zeros x  2 and x  3i . Write the polynomial in the form
f ( x)  ax 3  bx 2  cx  d .
Solution:
Since this polynomial is of degree 3, there must be 3 zeros. We are given two of the
zeros of the polynomial ( x  2 and x  3i ). We must find the third zeros. We find this
based on the fact that one of the zeros is x  3i and the polynomial has integer
coefficients. When a polynomial has integer coefficients any complex zeros must occur
in conjugate pairs. This means that since x  3i then x  3i must also be a zero. We
thus have the 3 zeros x  3i , x  3i , and x  2 . We will now start to write the
polynomial. Each of the zeros comes from a factor of the polynomial. The factor that
produces the zero x  3i is x  3i (notice that this is x minus the zero). The factor that
produces the zero x  3i is x  (3i )  x  3i . The factor that produces the zero x  2
is x  (2)  x  2 . Now to find the polynomial, we will multiply the factors together.
This gives us the factored polynomial of
f ( x)  a( x  3i )( x  3i )( x  2)
where a is the leading coefficient. We are told that the leading coefficient of the
polynomial is 1. This means that we can replace a with 1 to get
f ( x)  1( x  3i )( x  3i )( x  2)
f ( x)  ( x  3i )( x  3i )( x  2)
We are not yet done since we must write the polynomial in the form
f ( x)  ax 3  bx 2  cx  d . This means that we need to multiply the factors together.
We will start by multiplying the two factors produced by the complex zeros.
f ( x)  ( x  3i )( x  3i)( x  2)
f ( x)  ( x 2  3ix  3ix  9i 2 )( x  2)
f ( x)  ( x 2  9i 2 )( x  2)
f ( x)  ( x 2  9(1))( x  2)
f ( x)  ( x 2  (9))( x  2)
f ( x)  ( x 2  9)( x  2)
Note that after we multiplied the two factors produced by the complex zeros, there are no
longer any complex numbers left in the expression. We will now multiply these two
expressions together.
f ( x)  ( x 2  9)( x  2)
f ( x)  x 3  2 x 2  9 x  18
Thus the polynomial that we are looking for is
f ( x)  x 3  2 x 2  9 x  18
Sp04A Copyright 2004 Department of Mathematics and Statistics, Arizona State University