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Sagot :
Consider that the cubic function having zeroes a, b, c, is given by,
[tex]f(x)=x^3-(a+b+c)x^2+(ab+bc+ac)-abc[/tex]Given that the zeroes of the cubic function are -3, -1, and 2,
[tex]\begin{gathered} a=-3 \\ b=-1 \\ c=2 \end{gathered}[/tex]Substitute the values to obtain the cubic function,
[tex]\begin{gathered} f(x)=x^3-\mleft\lbrace\mleft(-3\mright)+\mleft(-1\mright)+2\mright\rbrace x^2+\mleft\lbrace\mleft(-3\mright)\mleft(-1\mright)+\mleft(-1\mright)2+\mleft(-3\mright)2\mright\rbrace-(-3)(-1)2 \\ f(x)=x^3-\mleft\lbrace-3-1+2\mright\rbrace^{}x^2+\mleft\lbrace3-2-6\mright\rbrace x-6 \\ f(x)=x^3-(-2)x^2+(-5)x-6 \\ f(x)=x^3+2x^2-5x-6 \end{gathered}[/tex]Thus, the required cubic function is obtained as,
[tex]f(x)=x^3+2x^2-5x-6[/tex]
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