Question

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

Answer 1

Answer:

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

Step-by-step 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.

$\frac{x^4-4x^3-5x^2+38x-30}{x^2 + 2 x - 3}=x^2 - 6 x + 10$

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

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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

$x=\frac{-(-6)\pm \sqrt{(-6)^2-4(1)(10)}}{2(1)}$

$x=\frac{6\pm \sqrt{-4}}{2}$

$x=\frac{6\pm 2i}{2}$                   $[\because \sqrt{-1}=i]$

$x=3\pm i$

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