Given the polynomial in factored form, it is possible to calculate intercepts, 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 .
The intercepts occur when . We solve . There are three distinct solutions, and . Note that is a double solution.
The intercept occurs when . We calculate .
Plotting the intercepts:
Multiplying the term in each bracket, we see that the degree of the polynomial is 4, and the leading coefficient is . The first and last terms of the expansion of the regular factored form give us degree and y-intercept:
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 . 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 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.
Remember that the final term in an expansion of factors is the product of all the zeros (). Therefore, we need only consider factors of the constant as possible zeros of the polynomial.
We need only consider the factors of which are .
Therefore is a factor.
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 where have been found.
Since is a factor of , and is a polynomial of degree 4, we know that the unknown polynomial is a cubic.
Multiplying out and gathering like terms gives us:
We now equate coefficients of each term with the corresponding term in
We now have
We begin the process again, considering the factors of in the smaller polynomial .
Try : . Therefore is not a factor.
Try : . Therefore is a factor.
(Here we use and again for convenience, however don’t mix these up with the and we already used.) Multiply out, gather like terms and equate coefficients gives us:
We are now down to a quadratic which is straightforward to factor:
And our factoring is done.
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:
Using the factor theorem, let’s try factors of until we have a value where .
We find that . Therefore is a factor.
The first term in the second bracket must be in order to multiply to .
Multiplying with we have
The first term is correct, we calculated that term. However we need a total of , which is more than we have. Let’s make the next term in the bracket and multiply out:
We now have the correct coefficient for and , but we need to ‘correct’ the term. Currently, we have but we need which is more. Let’s make the next term .
Now, all terms are correct except the term. We currently have but we need which is more.
Now because is a factor, the last term in the bracket works out perfectly.
Now we repeat the process with the cubic . By some examination we find that , so is a factor.
We need another so let’s add the term , which will multiply out to .
We need another so let’s add the term which will multiply out to .
We are now down to a quadratic:
After a little investigation, we see that doesn’t factor. When graphed, this quadratic does not cross the axis, so there are no real zeros. This is called an ‘irreducible quadratic’. Therefore,
is the fully factored form of this quartic. There are only two real zeros which occur at and . 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.
Using technology, we find that there is one turning point. The graph:
All of the polynomials in this applet have at least one value where and .