Respuesta :
Answer:
The roots of the polynomial equation [tex]x^{3}-4x^{2}+x+26=0[/tex] are[tex]x=-2, 3+2i\ and\ 3-2i[/tex].
Step-by-step explanation:
The polynomial provided is:
[tex]x^{3}-4x^{2}+x+26=0[/tex]
As the polynomials highest degree is 3 there will be 3 roots of the polynomial equation.
The first root can be determined by the hit-and-trial method.
For x = 2,
[tex]x^{3}-4x^{2}+x+26=0\\(2)^{3}-4(2)^{2}+(2)+26=0\\20\neq 0[/tex]
For x = - 2
[tex]x^{3}-4x^{2}+x+26=0\\(-2)^{3}-4(-2)^{2}+(-2)+26=0\\0=0[/tex]
Thus, one of the roots of the polynomial [tex]x^{3}-4x^{2}+x+26=0[/tex] is [tex]x=-2[/tex] or one factor is [tex](x+2)[/tex].
Now divide he polynomial with this factor as follows:
[tex]\frac{x^{3}-4x^{2}+x+26}{x+2} =x^{2}-6x+13[/tex]
The resultant polynomial is a quadratic equation.
Solve the equation [tex]x^{2}-6x+13[/tex] as follows:
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac} }{2a} \\=\frac{-(-6)\pm\sqrt{(-6)^{2}-4\times1\times13} }{2\times1} \\=\frac{6\pm\sqrt{-16} }{2} \\=\frac{6\pm4i }{2} \\=3\pm2i[/tex]
Thus, the roots of the polynomial equation [tex]x^{3}-4x^{2}+x+26=0[/tex] are[tex]x=-2, 3+2i\ and\ 3-2i[/tex].
Answer:
3 +or- 2i, -2
Step-by-step explanation:
Looked up the answer for a test but couldn't find the right one so I just guessed and this is what the correct answer was when I went back and checked. Hope this helps!
