Question

obtain all zeroes of x^4-5x^3-6x^2-32x+32 if two zeroes are 1 and 4​

Answer 1

Answer:

So, the other two roots of the polynomial equation are:

$2\sqrt{2}i,-2\sqrt{2}i$

Step-by-step explanation:

We are given a polynomial function as:

$f(x)=x^4-5x^3-6x^2-32x+32$

We are given two roots as: 1,4.

Now we know that for any polynomial equation of the type:

$ax^4+bx^3+cx^2+dx+e$

The sum of the roots is given by:

$\dfrac{-b}{a}$

and the product of roots is given by:

$\dfrac{e}{a}$

We are given two zeros as 1 and 4.

so, let g,h be the other two zeros of the polynomial equation.

Hence,The sum of zeros is given by:

$g+h+1+4=\dfrac{-(-5)}{1}\\\\g+h+5=5\\\\g+h=0$

$g=-h$

Also the product of zeros is given by:

$g\times h\times 1\times 4=\dfrac{32}{1}\\\\4gh=32\\\\gh=8$

Hence,

$(-h)(h)=8\\\\h^2=-8\\\\h=\sqrt{-8}\\ \\h=2\sqrt{2}i$

Hence,

$g=-h\\\\g=-2\sqrt{2}i$

So, the other two roots of the polynomial equation are:

$2\sqrt{2}i,-2\sqrt{2}i$