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Sagot :
Given:
The polynomial is
[tex]p(x)=x^3-2x^2+2x[/tex]
To find:
The real and complex zeros of the equation.
Solution:
We have,
[tex]p(x)=x^3-2x^2+2x[/tex]
For zeros, p(x)=0.
[tex]x^3-2x^2+2x=0[/tex]
[tex]x(x^2-2x+2)=0[/tex]
[tex]x(x^2-2x+2)=0[/tex]
[tex]x=0\text{ and }x^2-2x+2=0[/tex]
The real value of x is 0. The equation [tex]x^2-2x+2=0[/tex] will give complex roots. Here, a=1, b=-2 and c=2.
Using quadratic formula, we get
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(1)(2)}}{2(1)}[/tex]
[tex]x=\dfrac{2\pm \sqrt{4-8}}{2}[/tex]
[tex]x=\dfrac{2\pm \sqrt{-4}}{2}[/tex]
On further simplification, we get
[tex]x=\dfrac{2\pm \sqrt{-1}\sqrt{4}}{2}[/tex]
[tex]x=\dfrac{2\pm 2i}{2}[/tex]
[tex]x=\dfrac{2(1\pm i)}{2}[/tex]
[tex]x=1\pm i[/tex]
Therefore, the real zero is 0 and the complex zeros are 1+i and 1-i.
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