f(x)= x^4 -11x^3 +41x^2 -61x+30

Let us try and substitute with x=1

f(1) = 1-11+41-61 +30= 0

Then x1=1 is one of the solutions, and (x-1) is a factor for f(x)

Then f(x)= (x-1)Q(X)

==> Q(X)= f(x)/(x-1) = x^3 -10x^2 +31x-30

Now let us substitute with x=2

Q(2)= 8-40 +62 -30= 0

Then x2=2 is another zero for f(x) , and (x-2) is a factor for f(x).

Now f(x) = (x-1)(x-2) P(x)

p(x) = f(x) /(x-1)(x-2) = x^2 -8x +15

Factorize:

==> P(x) = x^2 -8x+15 = (x-3)(x-5)

Then f(x) = (x-1)(x-2)(x-3)(x-5)

Then the solutions are:

x1=1 , x2=2 , x3=3 , and x4= 5

To find the zeros of f(x)

f(x)= x^4 -11x^3 +41x^2 -61x +30.

Solution:

The constant term /coefficient of x^4 = 30/1 = 1*2*3*5

So, we can see whether for any of these vvalues or their -ves f(x) vanishes.

The put x=1, then f(1) = 1-11+41-61+30 = 0. So x-1 is a factor.

Put x=2, then f(2) = 16-11*8+41(4)-61(2)+30 = (16+164+30)-((88+122) = 210-210 =0. So x-2 is a factor.

So f(x) (x-1)(x-2)Q(x)

In a like manner, for x=3 , 81-11(27)+41(9)-61(3)+30= (81+369+30)-(297+183) = 480 - 480 = 0. So x-3 is a factor.

So we write:

f(x) = (x-1)(x-2)(x-3) (x-k) , where k is to be detrmined. Put x= 0 on both sides and we get:

f(0) = o^411*0^3+41*0^2-61*0+30 = (0-1)(0-2)(0-3)(0-k).. Or

30 = 6k. Or k = 30/6 = 5.

Therefore , 1,2,3 and 5 are the zeros of f(x).