What a Polynomial Is
Expressions such as \( 3x^2 - 5x + 2 \) and \( x^4 - 3x^2 + 5x - 4 \) are built from a variable, its powers and some numbers. In each of them, every power of \( x \) is a whole number: 4, 3, 2, 1 or 0. A constant term such as 2 counts as \( 2x^0 \), so it has power 0. An algebraic expression of this kind is called a polynomial.
Polynomial: an algebraic expression in which the power (index) of the variable in every term is a whole number, that is, zero or a positive integer.
So \( 2x^{-1} + 3 \) is not a polynomial, because the power \( -1 \) is negative. Neither is \( \sqrt{x} + 1 \), because \( \sqrt{x} = x^{\frac{1}{2}} \) has a fractional power.
The Greek mathematician Diophantus of Alexandria did so much early work on these expressions that he is called the father of polynomials. After him, Babylonian, Egyptian and Chinese mathematicians, and the Islamic mathematicians Al-Karaji and Al-Samawal, developed the idea further. In the sixteenth century François Viète, and in the seventeenth century René Descartes, also made contributions to this field.
A polynomial is a function, so it turns up wherever one quantity depends on another. Outside mathematics it is used in daily work on demand, supply, speed, cost and production. By its highest power a polynomial may be linear, quadratic, cubic, biquadratic (degree four) and so on. A polynomial of degree \( n \) is written as a function in this form:
\[ \begin{aligned} f(x) &= a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} \\ &\quad + \dots + a_2x^2 + a_1x + a_0 \end{aligned} \]
Here the coefficients \( a_n, a_{n-1}, a_{n-2}, \dots, a_0 \) are real numbers, and the power \( n \) is a whole number. In a polynomial the power of the variable is only ever zero or a positive integer.
Degree of a polynomial: the highest power of the variable that appears in it with a non-zero coefficient. For example, the degree of \( x^4 - 3x^2 + 5x - 4 \) is 4.
Dividing a Polynomial the Ordinary Way
Polynomials can be divided just as numbers can. When 23 is divided by 5, the answer is 4 with 3 left over, and \( 23 = 5 \times 4 + 3 \). The same four parts appear when one polynomial is divided by another.
Dividend \( f(x) \): the polynomial being divided.
Divisor \( d(x) \): the polynomial we divide by.
Quotient \( q(x) \): the result of the division.
Remainder \( R \): what is left when no further division is possible.
Take \( f(x) = x^4 - 3x^2 + 5x - 4 \) and \( d(x) = x - 2 \). Long division finds the quotient one term at a time. At each stage the leading term is divided by \( x \), the divisor is multiplied by that result, and the product is subtracted.
This dividend has no \( x^3 \) term. Leave a gap for it, or write it as \( 0x^3 \), so that like terms stay in the same column. Writing \( -3x^2 \) under \( -2x^3 \) is the slip that spoils most long divisions.
Step 1: Dividing \( x^4 \) by \( x \) gives \( x^3 \), the first term of the quotient. Multiplying the divisor by \( x^3 \):
\[ x^3(x - 2) = x^4 - 2x^3 \]
Subtracting this from the dividend leaves:
\[ 2x^3 - 3x^2 + 5x - 4 \]
Step 2: Dividing \( 2x^3 \) by \( x \) gives \( 2x^2 \). Multiplying the divisor by \( 2x^2 \):
\[ 2x^2(x - 2) = 2x^3 - 4x^2 \]
Subtracting leaves:
\[ x^2 + 5x - 4 \]
Step 3: Dividing \( x^2 \) by \( x \) gives \( x \). Multiplying the divisor by \( x \):
\[ x(x - 2) = x^2 - 2x \]
Subtracting leaves:
\[ 7x - 4 \]
Step 4: Dividing \( 7x \) by \( x \) gives 7. Multiplying the divisor by 7:
\[ 7(x - 2) = 7x - 14 \]
Subtracting leaves:
\[ 10 \]
Nothing more can be divided, because 10 contains no \( x \). So the quotient is \( q(x) = x^3 + 2x^2 + x + 7 \) and the remainder is \( R = 10 \).
The Division Relation
Just as \( 23 = 5 \times 4 + 3 \), the dividend can be rebuilt from the other three parts. Multiplying the divisor by the quotient and adding the remainder gives back the polynomial we started with:
\[ \begin{aligned} x^4 - 3x^2 + 5x - 4 &= (x - 2)(x^3 + 2x^2 + x + 7) \\ &\quad + 10 \end{aligned} \]
This happens in every division, so it is written as a general rule.
Division relation: when a polynomial \( f(x) \) is divided by a divisor \( d(x) \), giving quotient \( q(x) \) and remainder \( R \), the four polynomials are related by \( f(x) = d(x) \times q(x) + R \).
\[ f(x) = d(x) \times q(x) + R \]
Dividing both sides by \( d(x) \) writes the same relation in fraction form:
\[ \frac{f(x)}{d(x)} = \frac{d(x) \times q(x)}{d(x)} + \frac{R}{d(x)} \]
Cancelling \( d(x) \) in the first fraction:
\[ \frac{f(x)}{d(x)} = q(x) + \frac{R}{d(x)} \]
If \( R = 0 \), the last fraction disappears and \( \frac{f(x)}{d(x)} = q(x) \). The division is then exact: \( d(x) \) divides \( f(x) \) with nothing left over.
When the divisor is linear, that is of degree 1, division stops as soon as what is left has no \( x \) in it. So the remainder on dividing by a linear divisor is always a plain number.
A Pattern Hidden in the Values
Go back to the first division and put \( x = 2 \) into \( f(x) = x^4 - 3x^2 + 5x - 4 \):
\[ f(2) = 16 - 12 + 10 - 4 = 10 \]
That is exactly the remainder long division gave. To see whether this is luck, take another polynomial, \( f(x) = x^3 + 2x^2 + 3x - 6 \), and work out its value for several values of \( x \).
| Value of x | Working | f(x) |
| \( -1 \) | \( -1 + 2 - 3 - 6 \) | \( -8 \) |
| \( 0 \) | \( 0 + 0 + 0 - 6 \) | \( -6 \) |
| \( 1 \) | \( 1 + 2 + 3 - 6 \) | \( 0 \) |
| \( -2 \) | \( -8 + 8 - 6 - 6 \) | \( -12 \) |
| \( 2 \) | \( 8 + 8 + 6 - 6 \) | \( 16 \) |
| \( -3 \) | \( -27 + 18 - 9 - 6 \) | \( -24 \) |
| \( 3 \) | \( 27 + 18 + 9 - 6 \) | \( 48 \) |
The value \( f(1) \) is 0. Now divide the same polynomial by \( x - 1 \). Long division gives:
\[ \begin{aligned} x^3 + 2x^2 + 3x - 6 &= (x - 1)(x^2 + 3x + 6) \\ &\quad + 0 \end{aligned} \]
The remainder is 0, the same as \( f(1) \). Dividing the same polynomial by \( x - 2 \) instead leaves a remainder of 16, which is \( f(2) \). Every time, putting \( x = a \) into the polynomial gives the same number as the remainder on dividing by \( x - a \).
The value of \( f(x) \) at \( x = a \) is the remainder when \( f(x) \) is divided by the linear polynomial \( x - a \). So \( f(a) \) gives the remainder without any division.
The Remainder Theorem
Remainder theorem: when a polynomial \( f(x) \) of degree \( n \) is divided by \( x - a \), the remainder is \( f(a) \) and the quotient has degree \( n - 1 \). That is, \( R = f(a) \).
The theorem turns a long division into a single substitution. To find the remainder there is no need to find the quotient at all. The quotient has degree \( n - 1 \) because its first term comes from dividing \( a_nx^n \) by \( x \), which gives \( a_nx^{n-1} \).
Proof of the Remainder Theorem
Let the polynomial \( f(x) \) be divided by \( x - a \), giving quotient \( q(x) \) and remainder \( R \).
Step 1: Writing the division relation for this divisor:
\[ f(x) = q(x) \times (x - a) + R \quad \dots (i) \]
Step 2: Equation (i) is true for every value of \( x \), so putting \( x = a \) in it:
\[ f(a) = q(a) \times (a - a) + R \]
Step 3: Replacing \( a - a \) by 0:
\[ f(a) = q(a) \times 0 + R \]
Step 4: Multiplying by 0, which makes the whole product 0:
\[ f(a) = 0 + R \]
Step 5: Adding 0 to \( R \) leaves \( R \), which completes the proof:
\[ R = f(a) \]
The whole proof rests on one idea. The value \( x = a \) makes the divisor zero, so the part that holds the quotient vanishes and only the remainder is left.
Remainders for Other Linear Divisors
The theorem is stated for a divisor of the form \( x - a \). Any other linear divisor can first be written in that form, and then the same idea gives its remainder.
| Polynomial | Divisor | Remainder R |
| \( f(x) \) | \( x - a \) | \( f(a) \) |
| \( f(x) \) | \( x + a = x - (-a) \) | \( f(-a) \) |
| \( f(x) \) | \( ax - b = a\left(x - \frac{b}{a}\right) \) | \( f\left(\frac{b}{a}\right) \) |
| \( f(x) \) | \( ax + b = a\left\{x - \left(-\frac{b}{a}\right)\right\} \) | \( f\left(-\frac{b}{a}\right) \) |
Here is why the divisor \( ax - b \) gives the remainder \( f\left(\frac{b}{a}\right) \). The division relation says \( f(x) = (ax - b) \times q(x) + R \). The divisor is zero when \( ax - b = 0 \), that is, when \( x = \frac{b}{a} \). Putting this value in makes the first part zero and leaves \( f\left(\frac{b}{a}\right) = R \). The divisors \( x + a \) and \( ax + b \) work the same way, with \( x = -a \) and \( x = -\frac{b}{a} \).
One way to remember all four rows: set the divisor equal to zero, solve for \( x \), and put that value into \( f(x) \). The number that comes out is the remainder.
For the divisor \( x + 3 \), put \( x = -3 \), not \( x = 3 \). For the divisor \( 2x - 3 \), put \( x = \frac{3}{2} \), not \( x = 3 \). The wrong value still produces a number, so this mistake is easy to miss.
Using the Theorem Step by Step
Every remainder question below follows the same short method.
Step 1: Write down the polynomial \( f(x) \) and the divisor.
Step 2: The divisor is set equal to zero and solved for \( x \). The value it gives is \( a \):
\[ x = a \]
Step 3: That value is put into \( f(x) \) in place of \( x \). By the remainder theorem, the value it gives is the remainder:
\[ f(a) \]
Step 4: Simplify carefully: powers first, then multiplication, then addition and subtraction.
Step 5: If the remainder is given and a letter such as \( k \) is unknown, set \( f(a) \) equal to the given remainder:
\[ f(a) = \text{given remainder} \]
Solving this equation gives the value of the unknown letter.
Example 1: Dividing by a Simple Linear Divisor
Using the remainder theorem, find the remainder when \( f(x) = 4x^3 + 3x^2 - 2x + 1 \) is divided by \( d(x) = x - 2 \).
Step 1: Comparing \( x - 2 \) with \( x - a \) to find the value to substitute:
\[ a = 2 \]
Step 2: By the remainder theorem the remainder is \( f(2) \), so substituting \( x = 2 \):
\[ R = 4(2)^3 + 3(2)^2 - 2(2) + 1 \]
Step 3: Working out the powers:
\[ R = 4 \times 8 + 3 \times 4 - 2 \times 2 + 1 \]
Step 4: Multiplying:
\[ R = 32 + 12 - 4 + 1 \]
Step 5: Adding and subtracting from the left:
\[ R = 41 \]
So the remainder is 41. Long division would reach the same number, but only after three rounds of divide, multiply and subtract.
Example 2: When x Has a Coefficient
Using the remainder theorem, find the remainder when \( f(x) = 2x^4 - 7x^3 + 12x^2 - 17x + 7 \) is divided by \( d(x) = 2x - 3 \).
Step 1: Taking the coefficient 2 outside so that the divisor shows the form \( x - a \):
\[ 2x - 3 = 2\left(x - \frac{3}{2}\right) \]
Comparing \( x - \frac{3}{2} \) with \( x - a \) gives \( a = \frac{3}{2} \).
Step 2: By the remainder theorem \( R = f\left(\frac{3}{2}\right) \), so substituting \( x = \frac{3}{2} \):
\[ \begin{aligned} R &= 2\left(\tfrac{3}{2}\right)^4 - 7\left(\tfrac{3}{2}\right)^3 \\ &\quad + 12\left(\tfrac{3}{2}\right)^2 - 17\left(\tfrac{3}{2}\right) + 7 \end{aligned} \]
Step 3: Working out the powers, using \( \left(\frac{3}{2}\right)^4 = \frac{81}{16} \), \( \left(\frac{3}{2}\right)^3 = \frac{27}{8} \) and \( \left(\frac{3}{2}\right)^2 = \frac{9}{4} \):
\[ \begin{aligned} R &= 2 \times \tfrac{81}{16} - 7 \times \tfrac{27}{8} \\ &\quad + 12 \times \tfrac{9}{4} - 17 \times \tfrac{3}{2} + 7 \end{aligned} \]
Step 4: Multiplying each term:
\[ R = \frac{81}{8} - \frac{189}{8} + 27 - \frac{51}{2} + 7 \]
Step 5: Writing every term over the common denominator 8:
\[ R = \frac{81 - 189 + 216 - 204 + 56}{8} \]
Step 6: Simplifying the numerator from the left and dividing by 8:
\[ R = \frac{-40}{8} = -5 \]
So the remainder is \( -5 \). A remainder can be negative, and a fraction for \( a \) is nothing to fear: the work is the same, only the arithmetic is longer.
Example 3: Finding k From a Given Remainder
When \( f(x) = 4x^3 - 3x^2 + 3x - k \) is divided by \( x - 2 \), the remainder is 12. Find the value of \( k \).
Step 1: The divisor \( x - 2 \) gives \( a = 2 \), and by the remainder theorem the remainder is \( f(2) \). Setting it equal to the given remainder:
\[ f(2) = 12 \]
Step 2: Substituting \( x = 2 \) into \( f(x) \):
\[ 4(2)^3 - 3(2)^2 + 3(2) - k = 12 \]
Step 3: Working out the powers and products:
\[ 32 - 12 + 6 - k = 12 \]
Step 4: Collecting the numbers on the left:
\[ 26 - k = 12 \]
Step 5: Solving for \( k \):
\[ k = 26 - 12 = 14 \]
Check: with \( k = 14 \), \( f(2) = 32 - 12 + 6 - 14 = 12 \), which is the given remainder.
Example 4: Finding k When Two Remainders Are Equal
The polynomials \( f(x) = 2x^2 - 5x + k \) and \( g(x) = x^3 - x^2 + kx + 5 \) leave the same remainder when each is divided by \( x - 2 \). Find the value of \( k \).
Here \( a = 2 \) for both polynomials. Neither remainder is given as a number, but the two are equal. So each remainder is found in terms of \( k \), and then the two are set equal.
Step 1: Finding the remainder for \( f(x) \) by substituting \( x = 2 \):
\[ \begin{aligned} f(2) &= 2(2)^2 - 5(2) + k \\ &= 8 - 10 + k \\ &= -2 + k \end{aligned} \]
Step 2: Finding the remainder for \( g(x) \) by substituting \( x = 2 \):
\[ \begin{aligned} g(2) &= (2)^3 - (2)^2 + k(2) + 5 \\ &= 8 - 4 + 2k + 5 \\ &= 9 + 2k \end{aligned} \]
Step 3: Setting the two remainders equal, since \( f(2) = g(2) \):
\[ -2 + k = 9 + 2k \]
Step 4: Moving the numbers to the left side and the \( k \) terms to the right side:
\[ -2 - 9 = 2k - k \]
Step 5: Simplifying both sides:
\[ k = -11 \]
Check: with \( k = -11 \), \( f(2) = -2 - 11 = -13 \) and \( g(2) = 9 - 22 = -13 \). The two remainders are equal, as the question requires.
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