11.1 Polynomials, Identities and Division
AHL 2.12
- what a polynomial is, and what its degree is
- how to add, subtract and multiply polynomials, and how to find one coefficient quickly
- what an identity is, and two ways to find unknown constants in an identity
- how to divide one polynomial by another (long division), and the division algorithm \(P(x) = D(x)\,Q(x) + R(x)\)
Polynomials
A polynomial in \(x\) is a sum of terms of the form \(a x^n\), where
- the power \(n\) is a non-negative integer (\(0, 1, 2, 3, \ldots\)), and
- the number \(a\) is a constant, called the coefficient of \(x^n\).
The degree of a polynomial is the highest power of \(x\) that appears (with a non-zero coefficient). We often name a polynomial with a letter, such as \(P(x)\).
| Polynomial | Degree | Name | Leading coefficient | Constant term |
|---|---|---|---|---|
| \(P(x) = 7x - 2\) | \(1\) | linear | \(7\) | \(-2\) |
| \(P(x) = 3 - x^2\) | \(2\) | quadratic | \(-1\) | \(3\) |
| \(P(x) = 2x^3 - 5x^2 + 4\) | \(3\) | cubic | \(2\) | \(4\) |
| \(P(x) = x^4 + x\) | \(4\) | quartic | \(1\) | \(0\) |
Expressions such as \(\sqrt{x}\), \(\dfrac{1}{x}\) or \(x^{-2}\) are not polynomials, because the powers are not non-negative integers.
Addition and subtraction of polynomials
To add or subtract polynomials, collect like terms (terms with the same power of \(x\)).
Example 1 Given that \(P(x) = 2x^2 - 3x + 4\) and \(Q(x) = x^2 + 5x - 2\), find
(a) \(P(x) + Q(x)\), \(\qquad\) (b) \(3P(x) - 2Q(x)\).
(a)
\[ P(x) + Q(x) = (2x^2 + x^2) + (-3x + 5x) + (4 - 2) = 3x^2 + 2x + 2 \]
(b) Multiply first, then collect. Be careful with the signs of \(-2Q(x)\).
\[ \begin{aligned} 3P(x) - 2Q(x) &= (6x^2 - 9x + 12) - (2x^2 + 10x - 4) \\ &= 6x^2 - 9x + 12 - 2x^2 - 10x + 4 \\ &= 4x^2 - 19x + 16 \end{aligned} \]
Multiplication of polynomials
To multiply two polynomials, multiply every term of the first by every term of the second (the distributive law), then collect like terms.
Example 2 Expand and simplify
(a) \((3x + 4)(x^2 - 2x + 5)\),
(b) \((x^3 + 2x^2 - 3)(2x^2 - x + 4)\).
(a)
\[ \begin{aligned} (3x + 4)(x^2 - 2x + 5) &= 3x^3 - 6x^2 + 15x + 4x^2 - 8x + 20 \\ &= 3x^3 - 2x^2 + 7x + 20 \end{aligned} \]
(b) Multiply each of \(x^3\), \(2x^2\) and \(-3\) by the second bracket:
\[ \begin{aligned} x^3(2x^2 - x + 4) &= 2x^5 - x^4 + 4x^3 \\ 2x^2(2x^2 - x + 4) &= 4x^4 - 2x^3 + 8x^2 \\ -3(2x^2 - x + 4) &= -6x^2 + 3x - 12 \end{aligned} \]
Add the three lines, collecting like terms:
\[ 2x^5 + 3x^4 + 2x^3 + 2x^2 + 3x - 12 \]
Check: the degree is \(3 + 2 = 5\), as expected.
If you need only one coefficient, you do not have to expand everything. Find all the pairs of terms whose powers add up to the power you want.
Example 3 Find the coefficient of \(x^2\) in the expansion of \((3x^2 + x - 2)(x^2 - 4x + 6)\).
The pairs that give \(x^2\) are
\[ 3x^2 \times 6, \qquad x \times (-4x), \qquad -2 \times x^2 . \]
\[ 18x^2 - 4x^2 - 2x^2 = 12x^2 \]
The coefficient of \(x^2\) is \(12\).
Identities
An identity is an equation that is true for every value of \(x\). We use the symbol \(\equiv\) (“is identically equal to”) to show an identity.
- \(x^2 - 9 \equiv (x - 3)(x + 3)\) is an identity: it is true for all \(x\).
- \(x^2 - 9 = 8x\) is not an identity. It is true only for \(x = 9\) and \(x = -1\) (the solutions of the equation). For example, it is false when \(x = 0\).
When an identity contains unknown constants, we can find them in two ways.
Finding unknowns in an identity
Method 1 — substitute values of \(x\). The identity is true for every \(x\), so choose values that make brackets equal to zero.
Method 2 — equate coefficients. If two polynomials are identical, the coefficients of \(x^n\) on both sides are equal, for every power \(n\) (including the constant terms).
Often the quickest way is to mix the two methods.
Example 4 Given that \[ x^3 + 4x^2 - 3x + 1 \equiv ax(x + 2)^2 + b(x + 2) + c , \] find the values of \(a\), \(b\) and \(c\).
Method 1. Put \(x = -2\) (this makes every bracket zero):
\[ -8 + 16 + 6 + 1 = 0 + 0 + c \quad\Longrightarrow\quad c = 15 . \]
Method 2. Expand the right-hand side:
\[ ax(x^2 + 4x + 4) + bx + 2b + c = ax^3 + 4ax^2 + (4a + b)x + (2b + c) . \]
Coefficient of \(x^3\): \(\;a = 1\).
Coefficient of \(x\): \(\;4a + b = -3\), so \(b = -7\).
Check with the constant term: \(2b + c = -14 + 15 = 1\) ✓, and the \(x^2\) term: \(4a = 4\) ✓.
So \(a = 1\), \(b = -7\), \(c = 15\).
Example 5 Given that \(9x^3 + cx + 2 \equiv (3x - 2)(x + 1)(ax + b)\), find the values of \(a\), \(b\) and \(c\).
Compare the \(x^3\) terms: on the right, \(3x \times x \times ax = 3ax^3\). So \(3a = 9\) and \(a = 3\).
Compare the constant terms: \((-2) \times 1 \times b = -2b\). So \(-2b = 2\) and \(b = -1\).
Now expand to find \(c\):
\[ (3x - 2)(x + 1)(3x - 1) = (3x^2 + x - 2)(3x - 1) = 9x^3 - 7x + 2 . \]
So \(c = -7\). (The \(x^2\) terms cancel, as they must, since the left-hand side has no \(x^2\) term.)
Division of polynomials
First recall long division with numbers. Dividing \(23\) by \(4\):
\[ 23 = 4 \times 5 + 3 \]
| dividend | divisor | quotient | remainder | |||
|---|---|---|---|---|---|---|
| \(23\) | \(=\) | \(4\) | \(\times\) | \(5\) | \(+\) | \(3\) |
We divide polynomials in the same way. At each step:
- Divide the leading term of what is left by the leading term of the divisor. This gives the next term of the quotient.
- Multiply the divisor by this term, and write the result underneath.
- Subtract, and bring down the next term.
- Repeat until the degree of what is left is less than the degree of the divisor. What is left is the remainder.
Example 6 Find the quotient and the remainder when \(2x^2 + 5x - 4\) is divided by \(x - 3\).
\[ \begin{array}{r} 2x + 11\phantom{)} \\[-3pt] x - 3\,\enclose{longdiv}{\,2x^2 + 5x - 4} \\ \underline{2x^2 - 6x\phantom{{} - 4}} \\ 11x - 4 \\ \underline{11x - 33} \\ 29 \end{array} \]
- \(2x^2 \div x = 2x\). Multiply: \(2x(x - 3) = 2x^2 - 6x\). Subtract: \(5x - (-6x) = 11x\). Bring down \(-4\).
- \(11x \div x = 11\). Multiply: \(11(x - 3) = 11x - 33\). Subtract: \(-4 - (-33) = 29\).
The quotient is \(2x + 11\) and the remainder is \(29\):
\[ 2x^2 + 5x - 4 \equiv (x - 3)(2x + 11) + 29 . \]
If a power of \(x\) is missing from the dividend, write it with a zero coefficient (for example \(+\,0x^2\)). This keeps the columns in line.
Example 7 Find the quotient and the remainder when \(4x^3 + 5x - 2\) is divided by \(2x + 1\).
Write the dividend as \(4x^3 + 0x^2 + 5x - 2\).
\[ \begin{array}{r} 2x^2 - x + 3\phantom{)} \\[-3pt] 2x + 1\,\enclose{longdiv}{\,4x^3 + 0x^2 + 5x - 2} \\ \underline{4x^3 + 2x^2\phantom{{} + 5x - 2}} \\ -2x^2 + 5x\phantom{{} - 2} \\ \underline{-2x^2 - x\phantom{{} - 2}} \\ 6x - 2 \\ \underline{6x + 3} \\ -5 \end{array} \]
The quotient is \(2x^2 - x + 3\) and the remainder is \(-5\).
From Example 7 we have the identity
\[ \underbrace{4x^3 + 5x - 2}_{\text{dividend, degree 3}} \equiv \underbrace{(2x + 1)}_{\text{divisor, degree 1}} \underbrace{(2x^2 - x + 3)}_{\text{quotient, degree 2}} \;+\; \underbrace{(-5)}_{\text{remainder, degree 0}} . \]
Notice that
- degree of divisor \(+\) degree of quotient \(=\) degree of dividend (\(1 + 2 = 3\));
- when the divisor is linear, the remainder is a constant.
The division algorithm Not in the formula booklet
When a polynomial \(P(x)\) is divided by a polynomial \(D(x)\) of degree at least 1,
\[ P(x) \equiv D(x)\,Q(x) + R(x), \]
where the quotient \(Q(x)\) and the remainder \(R(x)\) are polynomials, and
\[ \text{degree of } R(x) < \text{degree of } D(x) . \]
So when you divide by a quadratic, the remainder has the form \(px + q\) (degree 1 or less).
Example 8 Express \(6x^5 - x^4 + 4x^3 + 7x^2 - 3x - 2\) in the form \((2x^2 - x + 3)\,Q(x) + R(x)\), where \(Q(x)\) and \(R(x)\) are polynomials and the degree of \(R(x)\) is less than \(2\).
\[ \begin{array}{r} 3x^3 + x^2 - 2x + 1\phantom{)} \\[-3pt] 2x^2 - x + 3\,\enclose{longdiv}{\,6x^5 - x^4 + 4x^3 + 7x^2 - 3x - 2} \\ \underline{6x^5 - 3x^4 + 9x^3\phantom{{} + 7x^2 - 3x - 2}} \\ 2x^4 - 5x^3 + 7x^2\phantom{{} - 3x - 2} \\ \underline{2x^4 - x^3 + 3x^2\phantom{{} - 3x - 2}} \\ -4x^3 + 4x^2 - 3x\phantom{{} - 2} \\ \underline{-4x^3 + 2x^2 - 6x\phantom{{} - 2}} \\ 2x^2 + 3x - 2 \\ \underline{2x^2 - x + 3} \\ 4x - 5 \end{array} \]
The remainder \(4x - 5\) has degree \(1 < 2\), so we stop.
\[ 6x^5 - x^4 + 4x^3 + 7x^2 - 3x - 2 \equiv (2x^2 - x + 3)(3x^3 + x^2 - 2x + 1) + (4x - 5) \]
When the divisor is \(x - k\), you can work with the coefficients only. Here is Example 6 again (\(2x^2 + 5x - 4\) divided by \(x - 3\), so \(k = 3\)):
\[ \begin{array}{r|rrr} 3 & 2 & 5 & -4 \\ & & 6 & 33 \\ \hline & 2 & 11 & 29 \end{array} \]
- Bring down the first coefficient, \(2\).
- Multiply by \(k = 3\) and write the result in the next column: \(6\). Add: \(5 + 6 = 11\).
- Multiply by \(3\): \(33\). Add: \(-4 + 33 = 29\).
The last number, \(29\), is the remainder. The others, \(2\) and \(11\), are the coefficients of the quotient \(2x + 11\). Remember to include a \(0\) for any missing power. For a divisor \(x + 2\), use \(k = -2\).
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \(3P - 2Q = 6x^2 - 9x + 12 - 2x^2 + 10x - 4\) | \(\ldots - 2x^2 - 10x + 4\) | \(-2\) multiplies every term of \(Q(x)\). |
| Dividing \(4x^3 + 5x - 2\) with the \(x^2\) term left out | Write \(4x^3 + 0x^2 + 5x - 2\) | A missing power needs a \(0\) placeholder. |
| Subtracting \(5x - 6x = -x\) in Example 6 | \(5x - (-6x) = 11x\) | You subtract the whole line, including its signs. |
| Stopping with remainder \(2x^2 + 3x - 2\) when dividing by a quadratic | Divide once more | The remainder must have lower degree than the divisor. |
| Using only \(x = -2\) to find \(a\), \(b\), \(c\) | Also equate coefficients | One substitution usually gives only one unknown. |
Exercises
1 Given \(P(x) = x^3 - 2x + 5\) and \(Q(x) = 2x^2 + 3x - 1\), find \(P(x) - 2Q(x)\).
\[ x^3 - 2x + 5 - 4x^2 - 6x + 2 = x^3 - 4x^2 - 8x + 7 \]
2 Expand and simplify \((2x - 3)(x^2 + 4x - 1)\).
\[ 2x^3 + 8x^2 - 2x - 3x^2 - 12x + 3 = 2x^3 + 5x^2 - 14x + 3 \]
3 Find the coefficient of \(x^3\) in the expansion of \((x^3 - 2x^2 + 4)(3x^2 + x - 5)\).
Pairs giving \(x^3\): \(\;x^3 \times (-5) = -5x^3\) and \(-2x^2 \times x = -2x^3\). (\(4\) would need an \(x^3\) term in the second bracket; there is none.)
The coefficient is \(-5 - 2 = -7\).
4 Find \(a\), \(b\) and \(c\) such that \(2x^2 - 5x + 7 \equiv a(x - 1)^2 + b(x - 1) + c\).
\(x = 1\): \(\;2 - 5 + 7 = c\), so \(c = 4\).
Coefficient of \(x^2\): \(a = 2\).
Coefficient of \(x\): the right-hand side gives \(-2a + b\), so \(-4 + b = -5\) and \(b = -1\).
Check (constant): \(a - b + c = 2 + 1 + 4 = 7\) ✓.
5 Given that \(x^3 + ax^2 + bx - 6 \equiv (x - 2)(x + 3)(x + c)\), find \(a\), \(b\) and \(c\).
\((x - 2)(x + 3) = x^2 + x - 6\), so the right-hand side is
\[ (x^2 + x - 6)(x + c) = x^3 + (c + 1)x^2 + (c - 6)x - 6c . \]
Constant: \(-6c = -6\), so \(c = 1\). Then \(a = c + 1 = 2\) and \(b = c - 6 = -5\).
6 Find the quotient and remainder when \(x^3 + 2x^2 - 5x + 1\) is divided by \(x - 2\).
Synthetic division with \(k = 2\):
\[ \begin{array}{r|rrrr} 2 & 1 & 2 & -5 & 1 \\ & & 2 & 8 & 6 \\ \hline & 1 & 4 & 3 & 7 \end{array} \]
Quotient \(x^2 + 4x + 3\), remainder \(7\).
7 Find the quotient and remainder when \(2x^3 - 3x + 4\) is divided by \(x + 2\).
Write \(2x^3 + 0x^2 - 3x + 4\) and use \(k = -2\):
\[ \begin{array}{r|rrrr} -2 & 2 & 0 & -3 & 4 \\ & & -4 & 8 & -10 \\ \hline & 2 & -4 & 5 & -6 \end{array} \]
Quotient \(2x^2 - 4x + 5\), remainder \(-6\).
8 Find the quotient and remainder when \(6x^3 + x^2 - 4x + 5\) is divided by \(3x - 1\).
\[ \begin{array}{r} 2x^2 + x - 1\phantom{)} \\[-3pt] 3x - 1\,\enclose{longdiv}{\,6x^3 + x^2 - 4x + 5} \\ \underline{6x^3 - 2x^2\phantom{{} - 4x + 5}} \\ 3x^2 - 4x\phantom{{} + 5} \\ \underline{3x^2 - x\phantom{{} + 5}} \\ -3x + 5 \\ \underline{-3x + 1} \\ 4 \end{array} \]
Quotient \(2x^2 + x - 1\), remainder \(4\).
9 Find the quotient and remainder when \(x^4 - 3x^3 + 2x^2 + x - 4\) is divided by \(x^2 + x - 1\).
\[ \begin{array}{r} x^2 - 4x + 7\phantom{)} \\[-3pt] x^2 + x - 1\,\enclose{longdiv}{\,x^4 - 3x^3 + 2x^2 + x - 4} \\ \underline{x^4 + x^3 - x^2\phantom{{} + x - 4}} \\ -4x^3 + 3x^2 + x\phantom{{} - 4} \\ \underline{-4x^3 - 4x^2 + 4x\phantom{{} - 4}} \\ 7x^2 - 3x - 4 \\ \underline{7x^2 + 7x - 7} \\ -10x + 3 \end{array} \]
Quotient \(x^2 - 4x + 7\), remainder \(-10x + 3\).
10 When \(x^3 + px^2 + qx + 3\) is divided by \(x^2 - x + 1\), the remainder is \(2x + 1\). Find \(p\) and \(q\).
The quotient is linear with leading term \(x\), so write it as \(x + k\):
\[ x^3 + px^2 + qx + 3 \equiv (x^2 - x + 1)(x + k) + 2x + 1 . \]
The right-hand side is \(x^3 + (k - 1)x^2 + (3 - k)x + (k + 1)\).
Constant: \(k + 1 = 3\), so \(k = 2\). Then \(p = k - 1 = 1\) and \(q = 3 - k = 1\).