Respuesta :
Answer:
The equation of a polynomial of degree 3, with zeros 1, 2 and -1 is [tex]x^{3}-2 x^{2}-x+2=0[/tex]
Solution:
Given, the polynomial has degree 3 and roots as 1, 2, and -1. And f(0) = 2.
We have to find the equation of the above polynomial.
We know that, general equation of 3rd degree polynomial is
[tex]F(x)=x^{3}-(a+b+c) x^{2}+(a b+b c+a c) x-a b c=0[/tex]
where a, b, c are roots of the polynomial.
Here in our problem, a = 1, b = 2, c = -1.
Substitute the above values in f(x)
[tex]F(x)=x^{3}-(1+2+(-1)) x^{2}+(1 \times 2+2(-1)+1(-1)) x-1 \times 2 \times(-1)=0[/tex]
[tex]\begin{array}{l}{\rightarrow x^{3}-(3-1) x^{2}+(2-2-1) x-(-2)=0} \\ {\rightarrow x^{3}-(2) x^{2}+(-1) x-(-2)=0} \\ {\rightarrow x^{3}-2 x^{2}-x+2=0}\end{array}[/tex]
So, the equation is [tex]x^{3}-2 x^{2}-x+2=0[/tex]
Let us put x = 0 in f(x) to check whether our answer is correct or not.
[tex]\mathrm{F}(0) \rightarrow 0^{3}-2(0)^{2}-0+2=2[/tex]
Hence, the equation of the polynomial is [tex]x^{3}-2 x^{2}-x+2=0[/tex]
