Skill: Solve polynomial equations

To solve a polynomial equation, first write it in the form

$$P(x)=0.$$

Then try to factor $P(x)$. Once the polynomial is written as a product, use the zero-product property: if

$$A(x)B(x)=0,$$

then $A(x)=0$ or $B(x)=0$.

Worked example

Solve

$$x^4-4x^3+5x^2-4x+4=0.$$

1. Find a possible rational root

The possible rational roots are factors of the constant term divided by factors of the leading coefficient:

$$\pm1,\ \pm2,\ \pm4.$$

Test $x=2$:

$$2^4-4(2^3)+5(2^2)-4(2)+4=16-32+20-8+4=0.$$

Since the result is $0$, $x=2$ is a root, so $(x-2)$ is a factor.

2. Divide by the factor

Using synthetic division,

Therefore,

$$x^4-4x^3+5x^2-4x+4=(x-2)(x^3-2x^2+x-2).$$

3. Factor the quotient

Group the terms in the cubic:

$$\begin{aligned} x^3-2x^2+x-2 &=(x^3-2x^2)+(x-2)\\ &=x^2(x-2)+1(x-2)\\ &=(x-2)(x^2+1). \end{aligned}$$

Thus the original equation becomes

$$(x-2)^2(x^2+1)=0.$$

4. Solve each factor

Using the zero-product property:

$$(x-2)^2=0 \quad\Rightarrow\quad x=2,$$

and

$$x^2+1=0 \quad\Rightarrow\quad x^2=-1 \quad\Rightarrow\quad x=\pm i.$$

Therefore, the solutions are

$$\boxed{x=2,\quad x=i,\quad x=-i}.$$

The root $x=2$ has multiplicity $2$ because the factor $(x-2)$ appears twice. The polynomial has degree $4$, so it has four complex solutions when counted with multiplicity:

$$2,\ 2,\ i,\ -i.$$

Only $x=2$ is real, so the graph has one real $x$-intercept, at $(2,0)$.