Question

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`?

Answer 1

The roots are `x=1/2` (double root), `x=3` (double root).

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 `P(x)` by a linear polynomial `(x-a)` is equal to `P(a)`

So then the factor follows as a special case of the remainder theorem. The factor theorem states that a polynomial `P(x)` has a factor `(x - a)` if and only `P(a)=0`

We know that `c=1/2` is one root, So applying the factor theorem it must be that `(2x-1)` is a factor of `f(x)`

We can then use polynomial division to divide `f(x)` by `(2x-1)` to reduce the quartic expression we started with to:

`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 `y=2x^3-13x^2+24x-9`) we can see that `x=3` is a root of the cubic. In reality we have to try various simple values `+-1, +-2,...` in the vain hope that we find a further root (and hence a factor).

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 `(x-3)` must be a factor also (of `f(x)` and `C(x)`). Having found a factor of the cubic C(x) we can use polynomial division again to reduce the cubic to:

`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 `x=1/2` (double root), `x=3` (double root).

We can confirm this graphically:

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