# 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?

Dec 29, 2016

The roots are $x = \frac{1}{2}$ (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 $P \left(x\right)$ by a linear polynomial $\left(x - a\right)$ is equal to $P \left(a\right)$

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

We know that $c = \frac{1}{2}$ is one root, So applying the factor theorem it must be that $\left(2 x - 1\right)$ is a factor of $f \left(x\right)$

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

$f \left(x\right) = \left(2 x - 1\right) \times \text{cubic expression}$

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

The division (not shown) gives us.

$f \left(x\right) = \left(2 x - 1\right) \left(2 {x}^{3} - 13 {x}^{2} + 24 x - 9\right)$

By observation (ie I cheated and plotted the graph of $y = 2 {x}^{3} - 13 {x}^{2} + 24 x - 9$) we can see that $x = 3$ is a root of the cubic. In reality we have to try various simple values $\pm 1 , \pm 2 , \ldots$ in the vain hope that we find a further root (and hence a factor).

If we let:

$C \left(x\right) = 2 {x}^{3} - 13 {x}^{2} + 24 x - 9$, then
$C \left(3\right) = 2 \cdot {3}^{3} - 13 \cdot {3}^{2} + 24 \cdot 3 - 9 = 54 - 117 + 72 - 9 = 0$

So again using the factor theorem $\left(x - 3\right)$ must be a factor also (of $f \left(x\right)$ and $C \left(x\right)$). Having found a factor of the cubic C(x) we can use polynomial division again to reduce the cubic to:

$C \left(x\right) = \left(x - 3\right) \times \text{quadratic expression}$,

and indeed we get:

$\setminus \setminus \setminus \setminus C \left(x\right) = \left(x - 3\right) \left(2 {x}^{2} - 7 x + 3\right)$
$\therefore f \left(x\right) = \left(2 x - 1\right) \left(x - 3\right) \left(2 {x}^{2} - 7 x + 3\right)$

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

$\left(2 {x}^{2} - 7 x + 3\right) = \left(x - 3\right) \left(2 x - 1\right)$:

$\therefore f \left(x\right) = \left(2 x - 1\right) \left(x - 3\right) \left(x - 3\right) \left(2 x - 1\right)$
$\therefore f \left(x\right) = {\left(2 x - 1\right)}^{2} {\left(x - 3\right)}^{2}$

And so the roots are $x = \frac{1}{2}$ (double root), $x = 3$ (double root).

We can confirm this graphically:

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