Answer :
A polynomial of degree 3 with real coefficients that satisfies the condition is, P(x) = -5 ( x - 3 ) ( [tex]x^{2}[/tex] + 1 ).
Based on the given conditions,
We want to find a third degree polynomial with zeros x = 3 and x = i and x = - i and P(2)= 25.
First, note that by the Complex Root Theorem, since x = 3 is a root, x = i is a root, x = -i must also be a root.
Hence, we will have the three factors,
P(x) = a ( x - ( 3 ) ) ( x - ( i ) ) ( x - ( -i ) )
Where a is the leading coefficient.
Expand and simplify the second and third factors:
( x - ( i ) ) ( x - ( -i ) ) = ( x - i ) ( x + i )
( x - ( i ) ) ( x - ( -i ) ) = x ( x - i ) + i ( x - i )
( x - ( i ) ) ( x - ( -i ) ) = ( [tex]x^{2}[/tex] - ix ) + ( ix - [tex]i^{2}[/tex] )
( x - ( i ) ) ( x - ( -i ) ) = [tex]x^{2}[/tex] + 1
Hence:
P(x) = a ( x - ( 3 ) ) ( [tex]x^{2}[/tex] + 1 ) (Equation-1)
Since P(2) = 25:
We can substitute value in equation-1,
P(x) = a ( x - ( 3 ) ) ( [tex]x^{2}[/tex] + 1 )
25 = a ( 2 - (3) ) ( [tex]2^{2}[/tex] + 1 )
25 = a ( -1 ) (4+1)
25 = a ( -1) (5)
25 = a (-5)
a = -25/5
a = -5
Hence,
The third degree polynomial function is ,
P(x) = -5 ( x - ( 3 ) ) ( [tex]x^{2}[/tex] + 1 )
P(x) = -5 ( x - 3 ) ( [tex]x^{2}[/tex] + 1 )
Therefore,
A polynomial of degree 3 with real coefficients that satisfies the condition is, P(x) = -5 ( x - 3 ) ( [tex]x^{2}[/tex] + 1 ).
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