# How do you find the rest of the zeros given one of the zero #c=1/2# and the function #f(x)=4x^4-28x^3+61x^2-42x+9#?

##### 1 Answer

The roots are

#### Explanation:

You can't necessarily, but it may help. Let me explain;

The remainder theorem states that the remainder of the division of a polynomial

So then the factor follows as a special case of the remainder theorem. The factor theorem states that a polynomial **factor**

We know that

We can then use polynomial division to divide

#f(x) = (2x-1) xx "cubic expression" #

The BIG question is then can we further factorise that cubic?

The division (not shown) gives us.

#f(x)=(2x-1)(2x^3-13x^2+24x-9)#

By observation (ie I cheated and plotted the graph of

If we let:

#C(x) = 2x^3-13x^2+24x-9# , then

#C(3) = 2*3^3-13*3^2+24*3-9 = 54-117+72-9=0#

So again using the factor theorem

#C(x) = (x-3) xx "quadratic expression"# ,

and indeed we get:

# \ \ \ \ C(x) = (x-3)(2x^2-7x+3) #

# :. f(x) = (2x-1)(x-3)(2x^2-7x+3) #

Now we have a quadratic left, we can hopefully factorise and we get:

#(2x^2-7x+3) = (x-3)(2x-1)# :

# :. f(x) = (2x-1)(x-3)(x-3)(2x-1) #

# :. f(x) = (2x-1)^2(x-3)^2 #

And so the roots are

We can confirm this graphically:

graph{4x^4-28x^3+61x^2-42x+9 [-5, 5, -5, 5]}