Polynomial Algebra and Theorems

Algebraic manipulation of polynomials

Algebraic manipulation of polynomials, including:

- **(a)** expanding brackets and collecting like terms
- **(b)** factorisation and simple algebraic division (by a linear polynomial, including those of the form $ax + b$, and by quadratics, including those of the form $ax^2 + bx + c$)
- **(c)** use of the Factor Theorem and the Remainder Theorem

In the TMUA/ESAT we expect you to be able to multiply our brackets and collect like terms; and recall collecting "like terms" means collecting all the constants together, and separately collecting all the $x$ terms together, and separately the $x^2$ terms, and separately the $x^3$ terms, and so on.

You should also be able to factorise simple algebraic expression – certainly quadratics and other expressions with common factors. You should also be able to factorise cubics using the factor theorem [see below].

In addition, make sure you can perform simple algebraic [long] division – you should be able to perform long division by linear and quadratic expressions. Here is an example:

Example

What is $x^4 + 2x^2 + 3x - 4$ divided by $x + 3$?

We write this out in a grid [below this green box] to make things easier to follow – and notice we have a column for each power of $x$ and we have also included $0x^3$ in our grid – this is a good idea as otherwise it is easy to make errors as you work through the long division. We also keep the same powers of $x$ in the same vertical columns throughout [we do this for the top line [the $x^3 - 3x^2 + 11x - 30$ line] for consistency but it is not so important there and you might prefer to shift that line to the left so that the $x^3$ appears above the $x^4$ etc.]

		$x^3$	$-3x^2$	$+11x$	$-30$
$x + 3$	$x^4$	$+0x^3$	$+2x^2$	$+3x$	$-4$
	$x^4$	$+3x^3$	$\downarrow$	$\downarrow$	$\downarrow$
	$0x^4$	$-3x^3$	$+2x^2$	$+3x$	$-4$
		$-3x^3$	$-9x^2$	$\downarrow$	$\downarrow$
		$0x^3$	$+11x^2$	$+3x$	$-4$
			$+11x^2$	$+33x$	$\downarrow$
			$0x^2$	$-30x$	$-4$
				$-30x$	$-90$
				$0x$	$86$

Notice that division involves just first terms at each stage, but when we multiply back to get what we need to subtract at each stage, then we use all the terms. We stop the long division when we get $86$ as $x + 3$ does not divide into $86$. In fact, $86$ is our remainder.

This long division tells us that when $x^4 + 2x^2 + 3x - 4$ is divided by $x + 3$ the answer is $x^3 - 3x^2 + 11x - 30$ with a remainder of $86$. We can write this as follows:

$x^4 + 2x^2 + 3x - 4 = (x^3 - 3x^2 + 11x - 30)(x + 3) + 86$

Study this carefully and make sure you can see exactly how it fits with our grid above and how it fits with your normal understanding of division [e.g., 11 divided by 4 is 2 remainder 3 so that we can write $11 = 2 \times 4 + 3$].

Factor and remainder theorem

The factor theorem appears in many advanced school mathematics specifications, but the remainder theorem appears in fewer, so you might not have met both theorems in your maths classes. We have decided to keep both theorems in the TMUA/ESAT specification as they are closely related to each other and easy to understand.

We will start by looking at the factor theorem. For this we will use the notation $f(x)$ and you will need to know what is meant by a factor in algebra. Let's start by refreshing our understanding of these two things.

First, what does it mean to be a factor of an algebraic expression? Here are a couple for examples:

• Consider $(x + 2)(x - 3)$ then both $(x + 2)$ and $(x - 3)$ are factors but, for instance, $(2x + 5)$ isn't.
• Consider $(2x - 7)(x^2 + 3x + 7)$ then both $(2x - 7)$ and $(x^2 + 3x + 7)$ are factors.

In general, one algebraic expression is a factor of another of it divides into it exactly - that is, without any remainder. This is just like the idea of factors for numbers.

Next, we remind ourselves that $f(x)$ [or $g(x)$ or $h(x)$ etc] is just another notation for a function of $x$, or really just an algebraic expression. We often write $f(x)$ instead of $y$ when we are dealing with algebraic expression. So, for instance we could write $f(x) = x^2 + 3x + 2$. This notation is useful as we can then indicate the value of the expression for different $x$ values. For instance, $f(2)$ means the value of the expression that is represented by $f(x)$ when $x = 2$; so, when $f(x) = x^2 + 3x + 2$, then $f(2) = 2^2 + 3 \times 2 + 2 = 12$.

Now let's use our knowledge of factors and the notation $f(x)$ to set up and explain the factor theorem:

Let's start with $f(x) = (x - 2)(x + 7)$. We know that both $(x - 2)$ and $(x + 7)$ are factors of $f(x)$. Let's look at the values of $f(2)$ and $f(-7)$. In both cases, a quick calculation shows that each has a value 0. This leads to the factor theorem which states:

If $f(x)$ is a polynomial in $x$, then $f(a) = 0$ if and only if $x - a$ is a factor of $f(x)$

Before we look at how we can use the factor theorem, we will make a few comments. First, notice that the factor theorem is applicable only to polynomials [and in TMUA/ESAT that means polynomials with real coefficients]. And also notice it is an "if and only if" statement. That means we can use it two ways:

1. We can find that $f(a) = 0$ and then we know that $(x - a)$ is a factor of $f(x)$.
2. We can start with $(x - a)$ as a factor of $f(x)$ and then we know that $f(a) = 0$.

You should realise that you know the factor theorem in a slightly different form: the factor theorem links the roots of the equation [when the graph crosses the $x$-axis – that is the solution to $f(x) = 0$] with the factors you get when you factorize an expression into brackets. So, for instance, if you are asked to find the roots of $f(x) = (x + 4)(x - 7)$ [in other words find where $y = (x + 4)(x - 7)$ crosses the $x$-axis] you know you have to solve $(x + 4)(x - 7) = 0$. So you are asking what $x$ values make $f(x) = 0$ and they must be the $x$ values that make $(x + 4) = 0$ or $(x - 7) = 0$. Read the factor theorem again to make sure you see the connections.

The factor theorem itself should be fairly "obvious". It is useful in form 1 above as it helps us to factorise polynomial expression [mostly quadratics and cubics]. Here are some examples of using the factor theorem:

Example 1

Factorise $f(x) = x^2 + 3x + 2$.

We will look at this in a little more detail than usual. First, we note that if the factors are $(x - a)$ and $(x - b)$ then $ab = 2$ so we only consider $a = \pm 1$ and $a = \pm 2$ [once we know $a$ we can immediately work out $b$; of course, we are assuming it CAN be factorised into nice brackets with integers!!].

Next, we use the factor theorem to work out what $a$ might be by a sort of "trial and error" process. We want to find an $a$ such that $f(a) = 0$ and we see that as all the bits of the quadratic are positive, then it is best to try negative values for $a$. We start with $a = -1$ and work out $f(-1) = 0$ so we know that $(x - -1)$ is a factor, ie $(x + 1)$ is a factor. This immediately gives $b = -2$ and so the other factor is $(x + 2)$.

Example 2

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

Here, using the same idea as in Example 1, we can see that this probably factorises to $(x + a)(x + b)(x + c)$ with $abc = 3$ [the constant term in the cubic]. So, we should start by using the factor theorem using factors of 3 [i.e., $1, -1, 3$ and $-3$].

We start with looking at $f(1) = 1^3 + 1^2 - 5 \times 1 + 3 = 0$ so we know that $(x - 1)$ is a factor.

Now we could continue to check $-1, 3$ and $-3$ and if you were to do so, you would find that $(x + 3)$ is a factor because $f(-3) = 0$. As there are no other factors of 3 and we have found only two factors then it looks like something has gone wrong. Actually, it hasn't, one factor is repeated, and the answer is $(x - 1)(x - 1)(x + 3)$.

There is an alternative way to proceed. Once we know that $(x - 1)$ is a factor we can use long division [or factorising by inspection – we have not included that technique in this notes] to find a quadratic factor of $f(x) = x^3 + x^2 - 5x + 3$ and then factorise that quadratic.

Long division gives:

$f(x) = x^3 + x^2 - 5x + 3 = (x - 1)(x^2 + 2x - 3) = (x - 1)(x - 1)(x + 3)$

Now we have an idea of what the factor theorem says and how to use it, we will turn to look at the remainder theorem. Recall above when we undertook an algebraic long division, we obtained the following expression

$f(x) = x^4 + 2x^2 + 3x - 4 = (x^3 - 3x^2 + 11x - 30)(x + 3) + 86$

If we use this to calculate $f(-3)$ we see from the final expression that the answer is 86, the remainder we obtained when we divided $f(x)$ by $(x + 3)$. This, in essence, is that the remainder theorem says. In this case it says that the remainder when $f(x)$ is divided by $(x + 3)$ is $f(-3)$.

Now we have a rough idea of what the remainder theorem says, let's look in more general terms so we can build up a good grasp of the remainder theorem for a general polynomial. Consider the polynomial $f(x)$ divided by $(x + b)$ for some non-zero [integer; $b$ does not have to be an integer, but it is easier here to assume it is. Most cases of using the remainder theorem will involve $a$ as an integer or a fraction; fraction in the case, for instance, of diving by $(2x + 3)$] $b$, the remainder would have to be just a number [think about how algebraic long division works. If you get a $px + q$ during long division when dividing by something like $x - 3$ [for example] then you can always complete a further step in the long division. It is only when you reach a number alone in the long division that you can no longer divide by something like $x - 3$] so we can write a general expression for this division as follows:

$f(x) = g(x)(x - b) + R$

Where $g(x)$ is some unique polynomial [can you explain why it is unique?] and $R$ is the remainder [as an aside, you should be able to argue that the highest power of $x$ in $g(x)$ is one less than the highest power of $x$ in $f(x)$].

And from this expression we immediately see by putting $x = b$ that

$f(b) = g(b)(b - b) + R = g(b) \times 0 + R = R$

So, we can construct a version of the Remainder Theorem:

When a polynomial $f(x)$ is divided by $(x - b)$ the remainder is $f(b)$

This also works for division by $(px - q)$. What do you think the reminder is in this case and can you explain why? [When a polynomial $f(x)$ is divided by $(px - q)$ the remainder is $f(q/p)$ and this is because we can write $f(x) = g(x) (px - q) + R$ so we need to find what $x$ makes $px - q = 0$ and then substitute that value into $f(x)$.]

We can also look at general division here, for instance, when we divide a polynomial $f(x)$ by a quadratic $x^2 + bx + c$ [or even $(x - p)(x - q)$]. Before we explore this briefly, have a think about what you expect the general form of the remainder to be in this case...

Here is the general case for quadratics

$f(x) = g(x) (x^2 + bx + c) + mx + n$

And you should be able to explain that the highest power of $x$ in $g(x)$ is two less than the highest power of $x$ in $f(x)$; and you should also be able to explain [and extend the concept] that the highest power of $x$ in the remainder is [at most, since it could be that $m = 0$] one less than the highest power of what we are dividing $f(x)$ by. And if both $m$ and $n$ are zero then the quadratic is a factor of $f(x)$.

What is the relationship between the factor theorem and the remainder theorem? In simple terms, the factor theorem is a special case of the remainder theorem. Simply put: if the remainder on dividing $f(x)$ by $(x - b)$ is zero, then $(x - b)$ must be a factor of $f(x)$. Clever.

Exercise

The degree of a polynomial is the highest power of $x$ that appears there. So, the degree of a quadratic is two, and that of a cubic is three, and so on. Here is a general expression for dividing some polynomial $f(x)$ by a polynomial $h(x)$:

$$f(x) = g(x)h(x) + r(x)$$

What can you say about the degree of $h(x)$, $g(x)$ and $r(x)$ in this expression. Be very careful about how you express your answer.