Given the polynomial in factored form, it is possible to calculate $x$ intercepts, $y$ intercept and degree. With that information we can draw a reasonably accurate sketch of the graph.

Curve Sketching using the Factored Form

Consider the polynomial $p(x)=(x+5)(x+1)(x-1)^2$ .

The $x$ intercepts occur when $y=0$ . We solve $0=(x+5)(x+1)(x-1)^2$ . There are three distinct solutions, $x=-5, x=-1$ and $x=1$ . Note that $x=1$ is a double solution.

The $y$ intercept occurs when $x=0$ . We calculate $p(0)=(0+5)(0+1)(0-1)^2=(5)(1)(-1)^2=5$ .

Plotting the intercepts:

Multiplying the $x$ term in each bracket, we see that the degree of the polynomial is 4, and the leading coefficient is $+1$ . The first and last terms of the expansion of the regular factored form give us degree and y-intercept:

$p(x)=(x+5)(x+1)(x-1)(x-1)=x^4 + \dots + 5$

Using the intercepts we can sit a ‘positive quartic’ shape onto the points calculated, bearing in mind the various forms a positive quartic (anywhere between a symmetric ‘w’ to asymmetric ‘w’ to simple ‘u’ shape) can take. Note that there is a ‘double zero’ at $x=1$ . Also note that for the curve to be smooth, larger gaps between zeros produce larger ‘valleys’ or ‘mountains’. You may wish to ‘spot check’ a value of $x$ in a large gap to get an idea of how far to dip.

The Factor Theorem

The factor theorem allows us to find the factors of a polynomial if the factors are integers.

Given polynomial $p(x)$ , if $p(c)=0$ then $(x-c)$ is a factor.

Remember that the final term in an expansion of factors is the product of all the zeros ( $\pm$ ). Therefore, we need only consider factors of the constant as possible zeros of the polynomial.

Example

$p(x)=x^4-x^3-7x^2+x+6$

We need only consider the factors of $6$ which are $\pm1, \pm2,\pm 3, \pm6$ .

$p(1)=1^4-1^3-7(1)^2+1+6=0\checkmark$

Therefore $(x-1)$ is a factor.

$p(x)=x^4-x^3-7x^2+x+6=(x-1)(???)$

To determine what goes in the second bracket, we may use one of the following methods:

polynomial long division,
equating coefficients
juggling method.
repeat the factor theorem until all values $c$ where $p(c)=0$ have been found.

Equating Coefficients

$p(x)=x^4-x^3-7x^2+x+6=(x-1)(???)$

Since $(x-1)$ is a factor of $p(x)$ , and $p(x)$ is a polynomial of degree 4, we know that the unknown polynomial is a cubic.

$x^4-x^3-7x^2+x+6=(x-1)(ax^3+bx^2+cx+d)$

Multiplying out and gathering like terms gives us:

$\begin{align*}x^4-x^3-7x^2+x+6=&(x-1)(ax^3+bx^2+cx+d)\\[10 pt]=&ax^4+bx^3+cx^2+dx-ax^3-bx^2-cx-d\\[10 pt]=&ax^4+(b-a)x^3+(c-b)x^2+(d-c)x-d\end{align}$

We now equate coefficients of each term with the corresponding term in $p(x)$

$\begin{align*}x^4: \quad &a=1\\[10 pt]x^3: \quad &b-a=-1\quad\Rightarrow b-1=-1 \quad\Rightarrow b=0\\[10 pt] x^2: \quad &c-b=-7\quad\Rightarrow c-0 =-7 \quad\Rightarrow c=-7\\[10 pt]x: \quad &d-c=1\quad\Rightarrow d-(-7)=1 \quad\Rightarrow d+7=1 \quad\Rightarrow d=-6\end{align}$

We now have

$x^4-x^3-7x^2+x+6=(x-1)(x^3-7x-6)$

We begin the process again, considering the factors of $-6$ in the smaller polynomial $q(x)=x^3-7x-6$ .

Try $q(1)$ : $q(1)=1^3-7(1)-6=-12$ . Therefore $(x-1)$ is not a factor.

Try $q(-1)$ : $q(-1)=(-1)^3-7(-1)-6=-1+7-6=0$ . Therefore $(x+1)$ is a factor.

$x^3-7x-6=(x+1)(ax^2+bx+c)$

(Here we use $a, b$ and $c$ again for convenience, however don’t mix these up with the $a, b$ and $c$ we already used.) Multiply out, gather like terms and equate coefficients gives us:

$\begin{align*}x^3-7x-6&=(x+1)(ax^2+bx+c)\\[10 pt]&=ax^3+bx^2+cx+ax^2+bx+c\\[10 pt]&=ax^3+(b+a)x^2+(c+b)x+c\\[20 pt]&x^3:\quad a=1\\[10 pt]&x^2: \quad b+a=0 \quad \Rightarrow b+1 = 0 \quad \Rightarrow b=-1\\[10 pt]&x: \quad c+b=-7 \quad \Rightarrow c-1 = -7 \quad \Rightarrow c=-6\end{align}$

That is,

$x^3-7x-6=(x+1)(x^2-x-6)$

and

$x^4-x^3-7x^2+x+6=(x-1)(x+1)(x^2-x-6)$

We are now down to a quadratic which is straightforward to factor:

$x^2-x-6=(x-3)(x+2)$

Gives us:

$x^4-x^3-7x^2+x+6=(x-1)(x+1)(x-3)(x+2)$

And our factoring is done.

Juggling

Juggling is a condensed version of the ‘equate coefficients’ method. Here, we calculate one term and ‘deal with the consequences’ ( or, correct the following term).

Let’s take a new example:

$p(x)=x^4+8x^3+26x^2+43x+30$

Using the factor theorem, let’s try factors of $30$ until we have a value $x$ where $p(x)=0$ .

We find that $p(-2)=0$ . Therefore $(x+2)$ is a factor.

$x^4+8x^3+26x^2+43x+30=(x+2)(x^3 + \dots )$

The first term in the second bracket must be $x^3$ in order to multiply to $x^4$ .

Multiplying $(x+2)$ with $x^3$ we have

$\begin{align*}x^4+8x^3+26x^2+43x+30&=(x+2)(x^3 +\dots)\\[10 pt]&=x^4+2x^3 +\dots\end{align}$

The first term is correct, we calculated that term. However we need a total of $8x^3$ , which is $6$ more than we have. Let’s make the next term in the bracket $6x^2$ and multiply out:

$\begin{align*}x^4+8x^3+26x^2+43x+30&=(x+2)(x^3 + 6x^2+\dots)\\[10 pt]&=x^4+2x^3 +6x^3+12x^2\dots\end{align}$

We now have the correct coefficient for $x^4$ and $x^3$ , but we need to ‘correct’ the $x^2$ term. Currently, we have $12x^2$ but we need $26$ which is $14$ more. Let’s make the next term $14x$ .

$\begin{align*}x^4+8x^3+26x^2+43x+30&=(x+2)(x^3 + 6x^2+14 x\dots)\\[10 pt]&=x^4+2x^3 +6x^3+12x^2+14x^2+28x\dots\\[10 pt]&=x^4+8x^3+26x^2+28x\dots\end{align}$

Now, all terms are correct except the $x$ term. We currently have $28x$ but we need $43x$ which is $15$ more.

$\begin{align*}x^4+8x^3+26x^2+43x+30&=(x+2)(x^3 + 6x^2+14 x+15)\\[10 pt]&=x^4+2x^3 +6x^3+12x^2+14x^2+28x+15x+30\dots\\[10 pt]&=x^4+8x^3+26x^2+43x+30\end{align}$

Now because $(x+2)$ is a factor, the last term in the bracket works out perfectly.

$x^4+8x^3+26x^2+43x+30=(x+2)(x^3 + 6x^2+14 x+15)$

Now we repeat the process with the cubic $x^3+6x^2+14x+15$ . By some examination we find that $(-3)^3+6(-3)^2+14(-3)+15=0$ , so $(x+3)$ is a factor.

$\begin{align*}x^3 + 6x^2+14 x+15&=(x+3)(x^2 \dots)\\[10 pt]&=x^3+3x^2 ...\end{align}$

We need another $3x^2$ so let’s add the term $3x$ , which will multiply out to $3x^2$ .

$\begin{align*}x^3 + 6x^2+14 x+15&=(x+3)(x^2 +3x\dots)\\[10 pt]&=x^3+3x^2 +3x^2+9x...\end{align}$

We need another $5x$ so let’s add the term $5$ which will multiply out to $5x$ .

$\begin{align*}x^3 + 6x^2+14 x+15&=(x+3)(x^2 +3x+5)\\[10 pt]&=x^3+3x^2 +3x^2+9x+5x+15\end{align}$

We are now down to a quadratic:

$\begin{align*}x^4+8x^3+26x^2+43x+30&=(x+2)(x+3)(x^2 +3x+5)\\[10 pt]&=(x+2)(x+3)(x+?)(x+?)\end{align}$

After a little investigation, we see that $x^2 +3x+5$ doesn’t factor. When graphed, this quadratic does not cross the $x$ axis, so there are no real zeros. This is called an ‘irreducible quadratic’. Therefore,

$x^4+8x^3+26x^2+43x+30=(x+2)(x+3)(x^2+3x+5)$

is the fully factored form of this quartic. There are only two real zeros which occur at $x=-2$ and $x=-3$ . We still know the end behaviour of this graph, however we don’t have enough information yet to determine if there is one or two turning points.