Question

Find all the zeros of the polynomial function P(x) x^4-4x^3 -5x^2+38x- 30

Answer 1

The zeros of the polynomial function are 1, -3, 3+i and 3-i.

Explanation:

The given function is

`P(x)=x^4-4x^3-5x^2+38x-30`

We can find a zero of the given function by hit and trial method.

Substitute x=1 in the given function.

`P(1)=(1)^4-4(1)^3-5(1)^2+38(1)-30`

`P(1)=0`

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.

`P(-3)=(-3)^4-4(-3)^3-5(-3)^2+38(-3)-30`

`P(-3)=0`

The value of function is 0 at x=-3 it means (x+3) is a factor of given function.

`(x-1)(x+3)=x^2 + 2 x - 3`

Divide the given function by
`x^2 + 2 x - 3`, to get remaining factors.

`(x^4-4x^3-5x^2+38x-30)/(x^2 + 2 x - 3)=x^2 - 6 x + 10`

So, the factor form of given function is

`P(x)=(x - 1) (x + 3) (x^2 - 6 x + 10)`

Quadratic formula: If a quadratic equation is
`ax^2+bx+c=0`, then

`x=(-b\pm √(b^2-4ac))/(2a)`

Using quadratic formula find the zeroes of
`x^2 - 6 x + 10`.

`x=(-(-6)\pm √((-6)^2-4(1)(10)))/(2(1))`

`x=(6\pm √(-4))/(2)`

`x=(6\pm 2i)/(2)`
`[\because √(-1)=i]`

`x=3\pm i`

Therefore the zeros of the polynomial function are 1, -3, 3+i and 3-i.