Respuesta :
Answer:
The zeros of the polynomial function are 1, -3, 3+i and 3-i.
Step-by-step explanation:
The given function is
[tex]P(x)=x^4-4x^3-5x^2+38x-30[/tex]
We can find a zero of the given function by hit and trial method.
Substitute x=1 in the given function.
[tex]P(1)=(1)^4-4(1)^3-5(1)^2+38(1)-30[/tex]
[tex]P(1)=0[/tex]
The value of function is 0 at x=1 it means (x-1) is a factor of given function.
Substitute x=-3 in the given function.
[tex]P(-3)=(-3)^4-4(-3)^3-5(-3)^2+38(-3)-30[/tex]
[tex]P(-3)=0[/tex]
The value of function is 0 at x=-3 it means (x+3) is a factor of given function.
[tex](x-1)(x+3)=x^2 + 2 x - 3[/tex]
Divide the given function by [tex]x^2 + 2 x - 3[/tex], to get remaining factors.
[tex]\frac{x^4-4x^3-5x^2+38x-30}{x^2 + 2 x - 3}=x^2 - 6 x + 10[/tex]
So, the factor form of given function is
[tex]P(x)=(x - 1) (x + 3) (x^2 - 6 x + 10)[/tex]
Quadratic formula: If a quadratic equation is [tex]ax^2+bx+c=0[/tex], then
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Using quadratic formula find the zeroes of [tex]x^2 - 6 x + 10[/tex].
[tex]x=\frac{-(-6)\pm \sqrt{(-6)^2-4(1)(10)}}{2(1)}[/tex]
[tex]x=\frac{6\pm \sqrt{-4}}{2}[/tex]
[tex]x=\frac{6\pm 2i}{2}[/tex] [tex][\because \sqrt{-1}=i][/tex]
[tex]x=3\pm i[/tex]
Therefore the zeros of the polynomial function are 1, -3, 3+i and 3-i.
