3. Polynomials#
Coming up:
We will define what is meant by a polynomial, as well as associated terminology
Long division of polynomials
Factorising polynomials (factor & remainder theorems)
3.1. Introduction#
A polynomial (in \(x\), of degree \(n\)) is an expression of the form
where \(a_n,\) \(a_{n-1},\dots,\) \(a_1,\) and \(a_0\) are real numbers and \(a_n \neq 0.\)
For each \(i=0,1,\ldots,n\) we say that \(a_i\) is the coefficient of \(x^i\). We say that \(a_n\) is the leading coefficient and \(a_0\) is the constant coefficient while \(a_nx^n\) is known as the leading term and \(a_0\) as the constant term of the polynomial.
Note that, from the definition, the degree of the polynomial is the highest power of \(x\) occurring. We call a polynomial linear if it has degree 1. So, for example,
Example 3.1
(i) \(x+7\) and \(3x-2\) are both linear polynomials.
(ii) All quadratics are polynomials of degree 2.
(iii) \(x^5+10x +1\) is a polynomial in \(x\) of degree 5. In the notation of (3.1), \(a_0=1\), \(a_1=10\), \(a_3=a_4=0\) and \(a_5=1\).
(iv) \(t^3 -t\) is a polynomial in \(t\) of degree \(3\). Its leading coefficient is \(1\), its linear coefficient is \(-1\), and all other coefficients equal zero.
We say that two polynomials are equal or identically equal if they have the same coefficients. Note that a consequence of this definition is that equal polynomials must have the same degree.
Example 3.2
If \(ax^3 + bx^2 + cx + d = x^3 -3x^2 +1\) (as polynomials) then \(a = 1,\) \(b = -3,\) \(c = 0\) and \(d =1.\)
The above process is called comparing or equating the coefficients.
Polynomials are fairly easy to work with and so they are often used as approximations to more difficult functions. You will no doubt meet these ideas further in your studies, but not in this course.
3.2. Algebraic Examples#
Polynomials can be added and multiplied by a real number or by another polynomial.
Example 3.3
Let \(p(x) = 3x -1\) and \(q(x) = 2x^3 +3x +1.\) Calculate \(p(x)+q(x)\), \(3p(x) - q(x)\) and \(p(x)q(x)\).
Solution.
We have \(p(x) + q(x) = (3x -1 +2x^3) + (3x +1) = 2x^3 +6x\),
and
Sometimes we are only interested in one coefficient of a product of two polynomials. For this, we can of course multiply the two polynomials together and collect like terms. However, there is a shortcut.
Example 3.4
Let \(f(x) = x^2 + 2x -3\) and \(g(x) = x^2 +3x -2.\) What is the coefficient of \(x^3\) in \(f(x)g(x)\)?
Solution.
We are only interested in the terms in the product \((x^2+2x-3)(x^2+3x-2)\) which give \(x^3\) terms. We write a table as shown below, where each entry in the bottom row is obtained by multiplying the two terms above it.
terms which make \(x^3\) |
|
|---|---|
\(f(x)\) forwards |
\(0x^3\) \(+x^2\) \(+2x\) |
\(g(x)\) backwards |
\(-2\) \(+3x\) \(+x^2\) |
\(x^3\) term in \(f(x)g(x)\) |
\(0\) \(+3x^3\) \(2x^3\) |
Now we add up the numbers in the bottom row, getting \(3x^3+2x^3 =5x^3\). Hence the coefficient of \(x^3\) in \(f(x)g(x)\) is \(5\).
Exercise
Let \(p(x) = 2 -3x,\) \(q(x) = 2x^3 +x^2 -2x +1\) and \(r(x) = x^2 -3.\)
(i) Simplify \(2q(x) - p(x).\)
(ii) Expand \(2p(x)q(x).\)
(iii) Find the coefficient of \(x^2\) in \(r(x)q(x).\)
Solution
(i) We have
(ii) We have
(iii) We draw a table.
terms which make \(x^2\) |
|
|---|---|
\(r(x)\) |
\(x^2\) \(+0x\) \(-3\) |
\(q(x)\) |
\(1\) \(-2x\) \(+x^2\) |
\(x^2\) term in \(r(x)q(x)\) |
\(x^2\) \(0\) \(-3x^2\) |
So the coefficient of \(x^2\) in \(r(x)q(x)\) is \(1 -3 = -2.\)
3.3. Long Division of Polynomials#
A key problem for polynomials is factorisation. That is, given a polynomial of degree \(n\) can we write it as a product of brackets of the form \((ax+b)\)? We’ve seen techniques for approaching quadratics, but so far have no way of factorising higher degree polynomials. As factorisation is the reverse process of multiplication (of brackets), it may not be too surprising that one of our approaches uses long division.
3.3.1. Long division of numbers#
We recall the technique of long division of numbers which we will generalise to polynomials. Remember that, when using long division, one ends up with a whole number (called the quotient}) and a remainder.
Example 3.5
Find \(3702 \div 23\) using long division.
Solution.
We approach this by asking how many \(23\)s can be subtracted from \(3702\). The whole calculation is usually written as follows.
Here we have subtracted first 100 lots of 23, followed by 60 more and we have ended up with a remainder of 22 - too small to subtract any further multiples from. That is, we have found that \(3702\div 23 = 160\) remainder \(22\). Rephrasing, we have \(3702=160\times 23 + 22\). Here, \(160\) is called the quotient and 22 the remainder.
3.3.2. Long division of polynomials#
We generalise the above ideas to polynomials.
Example 3.6
Find \((2x^4+x^2+x-3)\div (x^2-1)\) .
Solution.
The idea is to subtract as many copies of \((x^2-1)\) away from \(2x^4+x^2+x-3\) as possible. We work with a
standard layout, which grows larger as we carry out the calculation. Below is how the layout looks at the
various points in the calculation.
Step 1: The Starting Layout.
Notice that we’ve added \(0x\) and \(0x^3\) terms. The reason is that on the right hand side each vertical column corresponds to a different power of \(x\).
Step 2: The Multiplication Question.
Note the two terms in boxes.
Question: What do we multiply \(x^2\) by to get \(2x^4\)?
The answer is \(2x^2\). Write the answer above the top line, above the \(x^2\) term.
Step 3: The Multiplication Action.
Multiply the answer in Step 2 by the polynomial on the left to get, in our case, \(2x^2 \times (x^2+0x-1)=2x^4+0x^3-2x^2\). Write the answer to this on the bottom row.
Step 4: The Subtraction.
Subtract the polynomial you added to the layout in Step 3 from
the polynomial immediately above it. Write your answer at the bottom, keeping the columns intact.
To indicate the subtraction, put the polynomial you added in Step 3 in brackets, put a \(-\) sign in front of it and a line underneath it.
Note: Be careful with the signs! For example, \(x^2-(-2x^2)\) is equal to \(3x^2\), not \(-x^2\).
Step 5: The Multiplication Question.
This is essentially the same as Step 2.
Note the two terms in boxes. What do we multiply \(x^2\) by to get \(3x^2\)? The answer is \(+3\). We write the answer on the top line, in the constant terms column. As there is no \(x\) term we can insert \(0x\) in the \(x\) terms column.
Step 6: The Multiplication Action.
This is essentially the same as Step 3. Multiply the answer in Step 5 by the polynomial on the left to get \(3 \times (x^2+0x-1)=3x^2+0x-3\). Write the answer to this at the bottom.
Step 7: The Subtraction.
This is essentially the same as Step 4. Subtract the polynomial you added to the layout in Step 6, from
the polynomial immediately above it. Write your answer at the bottom. As before, include a \(-\) sign, brackets and a horizontal line.
Step 8: The Multiplication Question.
Note the two terms in boxes. What do we multiply \(x^2\) by to get \(x\)? Here we cannot find an answer because the degree of \(x\) is \(1\) which is less than the degree of \(x^2\) which is \(2\). This means we have subtracted as many copies of (\(x^2-1\)) away from \(2x^4+x^2+x-3\) as possible so we stop.
Step 9: The Result.
From the final layout we can read off the result, namely that \((2x^4+x^2+x-3) \div (x^2-1) = 2x^2+3\)
remainder \(x\). We usually state the result in the following form:
That is, when \(2x^4+x^2+x-3\) is divided by \(x^2-1\), the quotient is \(2x^2+3\) and the remainder is \(x\).
Example 3.7
Find the quotient and remainder when \(6x^3 - 19x^2 +13x +2\) is divided by \(2x -5\).
Solution.
Using the method above, we have the following calculation.
Hence \(6x^3 - 19x^2 +13x +2=(2x -5) \times (3x^2 -2x + \frac 32) + \frac {19}{2}\). That is, when \(6x^3 - 19x^2 +13x +2\) is divided by \(2x-5\) the quotient is \(3x^2 -2x + \frac 32\) and the remainder is \(\frac {19}{2}\).
Exercise
(i) Find the quotient and the remainder when \(x^3 -2x^2 +x -1\) is divided by \(x-1.\)
(ii) Find \((2x^4 - x + 1) \div (x^2 +1)\).
(iii) Find the quotient and the remainder on dividing \(3x^3 -2x^2 +1\) by \(2x +1.\)
Solution
(i) We have
So when \(x^3 -2x^2 +x -1\) is divided by \(x-1,\) the quotient is \(x^2 -x\) and the remainder is \(-1.\)
(ii) We have
So \((2x^4-x+1) \div (x^2 +1)=2x^2-2\) remainder \((-x + 3)\); that is, the quotient is \(2x^2-2\) and the remainder is \(-x+3=3-x\).
(iii) We have
So when \(3x^3 -2x^2 +1\) is divided by \(2x+1\) the quotient is \(\frac 32x^2 -\frac 74x + \frac 78\) and the remainder is \(\frac {1}{8}.\)
3.3.3. Factorising polynomials#
The following theorem generalizes a result that we have used with already with quadratics.
Theorem 3.1 (The Factor Theorem)
For any polynomial \(f(x)\) and any real number \(a\)
\((x-a)\) is a factor of \(f(x)\) if and only if \(f(a) = 0.\)
(In other words, if \((x-a)\) is a factor of \(f(x)\), then \(f(a)=0\) (easy to see) and, conversely, if \(f(a)=0\) then \((x-a)\) is a factor of \(f(x)\) (harder!).)
Proof: By using long division, we can divide \(f(x)\) by \((x-a)\) to get
for some polynomials \(q(x)\) (the quotient) and \(r(x)\) (the remainder). But since \((x-a)\) has degree 1, \(r(x)\) must just be a constant (or else we would have carried on with the long division). Hence \(f(x)=(x-a)q(x)+k\) for some real number \(k\). Now
\(f(a)= 0\),
so \((a-a)\times q(a)+ k=0\)
so \(0\times q(a)+k=0\)
so \(k=0\)
so \(f(x)=(x-a)\times q(x)\)
so \((x-a)\) is a factor of \(f(x)\). \(\square\)
For example, the polynomial \(f(x) = x^2 - 7x +10\) has \(f(2)=0\) and \(f(5)=0\) and factorises as \((x-2)(x-5)\); that is, \((x-2)\) and \((x-5)\) are both factors.\
Example 3.8
Factorise \(f(x) = x^4 - 8x^3 +17x^2 +2x - 24.\)
Solution.
The idea is to try some values of \(x\) to see if we can find some that give \(f(x) = 0\) and then, using the Factor Theorem, do some long division to reduce the degree of the polynomial to 2 or less (that is, a quadratic).
Step 1: Try to find \(x\) with \(f(x) = 0\).
So, by the Factor Theorem, \((x-2)\) is a factor of \(f(x) = x^4 - 8x^3 +17x^2 +2x - 24\). We try further numbers.
So, by the Factor Theorem, \((x-(-1))=(x+1)\) is also a factor of \(f(x) = x^4 - 8x^3 +17x^2 +2x - 24\). Now, \(f\) has degree \(4\) so it has at most four factors. We have found two so there are at most two left. But these remaining two factors would form a quadratic and we can deal with quadratics easily so we stop looking for further factors.
Step 2: Use long division to find the remaining quadratic.\ By our work above we know that \(f(x) = (x-2)(x+1)g(x)\) where \(g(x)\) is a quadratic. We use long division to find \(g(x)\). \
First, we multiply out the factors to get \((x-2)(x+1) = x^2 -x -2\). Next, we divide:
(Note that the remainder of \(0\) confirms that \(x^2-x-2\) was a factor of \(f(x)\)). So
Step 3: Factorise the remaining quadratic.
By our usual methods, we have \(x^2 - 7x + 12 = (x-3)(x-4)\). Thus we get
and \(f(x)\) is fully factorised.
Note
Of course, in the above example, you could try to find all the factors by testing numbers but often you will encounter examples where the first few factors are easy numbers (like \(2\), \(1\), \(0\), \(-1\), \(-2\)) and the remaining factors are large or even surds. The key point is to find enough to reduce down to a quadratic.
It can happen that a factor \((x-a)\) appears twice. In such a case, \((x-a)\) is called a repeated factor.
Example 3.9
Let \(f(x) = x^4 - 8x^3 +17x^2 +2x - 24\). Solve \(f(x) = 0.\)
Solution.
By the previous example we know that \(f(x)\) can be factorised as \((x-2)(x+1)(x-3)(x-4)\). Hence
so \(x-2 = 0\) or \(x+1 = 0\) or \(x-3 = 0\) or \(x-4 =0\). Thus the solutions are \(x=2\), \(x=-1\), \(x =3\) and \(x =4\).
Example 3.10
Factorise \(f(x) = x^3 -2x^2 -5x +6\). Hence, or otherwise, find all the solutions to \(f(x) = 0\).
Solution.
Step 1: Try some numbers.
First time lucky! Hence, by the Factor Theorem, \((x-1)\) is a factor of \(f(x)\).
Step 2: Divide by to get a quadratic.
We know that \(f(x) = (x-1)g(x)\) where \(g(x)\) is a quadratic and use long division to find \(g(x)\).
So \(f(x) = (x-1)(x^2 -x -6).\)
Step 3: Factorise the quadratic \(x^2 - x -6 = (x+2)(x-3).\)
Thus we get \(f(x) = (x-1)(x+2)(x-3).\) So the solutions to \(f(x) = 0\) are \(x = 1\), \(x = -2\) and \(x = 3.\)
Exercise
Factorise \(g(x) = x^4 -2x^3 +2x^2 -2x +1\). Hence, or otherwise, find all the solutions to \(g(x) = 0.\)
Solution
Step 1: Try some numbers:
The standard numbers do not work except \(1.\) So maybe \((x-1)\) is a repeated factor…
Step 2: Divide \(x^4 - 2x^3 +2x^2 -2x +1\) by \((x-1)\):
So \(g(x) = (x-1)(x^3 -x^2 + x-1).\)
Repeat Step 1: We check if \(x-1\) is a factor of \(x^3 -x^2 +x -1\). Putting \(x=1\) we get
So \((x-1)\) is a factor of \(x^3 -x^2 +x -1\) and (hence is a repeated factor of \(g(x)\)).
Repeat Step 2: Divide \(x^3 -x^2 +x-1\) by \(x-1\):
So \(x^3 -x^2 +x -1=(x-1)(x^2+1)\) and hence \(g(x) = (x-1)^2(x^2+1)\).
Note that \((x^2 +1)\) has no real solutions (see this by looking at the discriminant or the quadratic formula) and so we get
and \(g(x)\) is fully factorised. Hence \(g(x) = 0\) if and only if \(x = 1\).
Exercise
Show that \(2x^3 +x^2 -13x +6\) is divisible by \(x-2\) and find the factors of the expression.
Show that \(12x^3 + 16x^2 -5x -3\) is divisible by \(2x-1\) and find the factors of the expression.
Factorise and solve \(f(x) = 0\) for the following:
(i) \(f(x) = x^3 -4x^2 +x +6\),
(ii) \(f(x) = 2x^3 +x^2 -8x -4\),
(iii) \(f(x) = 2x^3 +5x^2 +x -2\),
(iv) \(f(x) = 2x^3 +11x^2 +17x +6\),
(v) \(f(x) = 2x^3- x^2 +2x -1\).
3.3.4. The Remainder Theorem#
Example 3.11
(i) Find the remainder when \(x^4 -7x^3 +x^2 - 6\) is divided by \((x-3)\).
(ii) Calculate \(f(3)\) and compare your answer.
Solution.
(i) The long division gives
Thus the remainder when \(x^4 -7x^3 +x^2 - 6\) is divided by \((x-3)\) is \(-105\).
(ii) \(f(3)=3^4 -7\times 3^3 +3^2 - 6 = 81 - 189 + 9 - 6 = -105\), which is the same!
This is no coincidence! We have the following theorem.
Theorem 3.2 (The Remainder Theorem)
Let \(f(x)\) be a polynomial and \(a\in\mathbb{R}\). Then,
when \(f(x)\) is divided by \((x - a)\) the remainder is \(f(a).\)
Proof: Similarly to the proof of the Factor Theorem, dividing \(f(x)\) by \((x-a)\) we get \(f(x)=(x-a)q(x)+r(x)\) where \(r(x)\) is the remainder and has degree 0; that is, it is just a number, say \(r(x)=k\). Hence \(f(x)=(x-a)q(x)+k\) and, putting \(x=a\), we get
Thus the remainder is \(f(a)\), as claimed. \(\square\)
Exercise
Find the remainder when \(f(x) = -2x^3 +3x^2 -x +1\) is divided by \(x+2.\)
Solution
Note that \(x + 2 = x- (-2)\) and so, by the Remainder Theorem, we get that the remainder is \(f(-2) = -2\times (-2)^3 + 3\times(-2)^2 - (-2) + 1 = 16 +12 + 2 +1 = 31.\)
Note
If you give it some thought, you’ll see that the Factor Theorem is really just a special case of the Remainder Theorem.
Exercise
Use the remainder theorem to find the remainders when
\(x^3 +3x^2 -4x +2\) is divided by \(x-1;\)
\(x^3 -2x^2 +5x +8\) is divided by \(x-2;\)
\(x^5 +x -9\) is divided by \(x+1;\)
\(x^3 +3x^2+3x+1\) is divided by \(x+2;\)
Solution
\(2\), 2. \(18\), 3. \(-11\), 4. \(-1\).
Exercise
Find the values of \(a\) in the expression below so that the following conditions are satisfied.
(i) \(x^3 + ax^2 +3x -5\) has remainder \(-3\) when divided by \(x-2.\)
(ii) \(x^3 +x^2 +ax + 8\) is divisible by \(x-1.\)
(iii) \(x^3 +x^2 -2ax +a^2\) has remainder \(8\) when divided by \(x-2.\)
(iv) \(x^4 -3x^2 +2x +a\) is divisible by \(x+1.\)
(v) \(x^3 -3x^2 +ax +5\) has remainder \(17\) when divided by \(x-3.\)
(vi) \(x^5 +4x^4 -6x^2 +ax +2\) has remainder \(6\) when divided by \(x+2.\)
Solution
(i) By the remainder theorem, we require \(-3=(2)^3+a(2)^2+3(2)-5\), so
so \(a=-3\).
(ii) The same method yields \(0=1^3+1^2+a+8\), so \(a=-10\).
(iii) We get \(8=2^3+2^2-4a+8\), so \(a^2-4a+4=0\), so \((a-2)^2=0\), hence \(a=2\).
(iv) \(a=4\).
(v) \(a=4\).
(vi) \(a=2\).
There are many more exercises for you to practice in the problem booklet.