When the Remainder Is Zero
Every whole number has factors. The number 3 is a factor of 12 because 12 divided by 3 gives 4 with nothing left over, so we can write \( 12 = 3 \times 4 \). The number 5 is not a factor of 12, because 12 divided by 5 leaves a remainder of 2. A factor is something that divides exactly, with remainder zero. Polynomials follow the same idea, and the factor theorem turns it into a rule we can test in one line.
Take the polynomial \( f(x) = x^2 - 5x + 6 \). Putting \( x = 2 \) gives \( f(2) = 4 - 10 + 6 = 0 \), and putting \( x = 3 \) gives \( f(3) = 9 - 15 + 6 = 0 \). So this polynomial becomes zero at two values of x, namely 2 and 3. Now factorise it by splitting the middle term:
\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]
The factors are \( x - 2 \) and \( x - 3 \), and the numbers inside them are the same 2 and 3 that made the polynomial zero. This is not a coincidence. Because \( x - 2 \) and \( x - 3 \) are factors, each of them divides \( x^2 - 5x + 6 \) exactly.
Now try a cubic. Divide \( f(x) = x^3 + 7x^2 - 12x + 4 \) by the divisor \( d(x) = x - 1 \) using ordinary long division.
Step 1: The first term of the quotient is \( x^3 \div x = x^2 \). Multiplying \( x - 1 \) by \( x^2 \) and subtracting it from the first two terms:
\[ (x^3 + 7x^2) - (x^3 - x^2) = 8x^2 \]
Step 2: Bringing down \( -12x \), the next term of the quotient is \( 8x^2 \div x = 8x \). Multiplying \( x - 1 \) by \( 8x \) and subtracting:
\[ (8x^2 - 12x) - (8x^2 - 8x) = -4x \]
Step 3: Bringing down \( +4 \), the last term of the quotient is \( -4x \div x = -4 \). Multiplying \( x - 1 \) by \( -4 \) and subtracting:
\[ (-4x + 4) - (-4x + 4) = 0 \]
Step 4: Reading off the quotient, with remainder \( R = 0 \):
\[ q(x) = x^2 + 8x - 4 \]
In every division the dividend, divisor, quotient and remainder are related by \( f(x) = d(x) \times q(x) + R \). Here \( R = 0 \), so the relation becomes a plain product:
\[ x^3 + 7x^2 - 12x + 4 = (x - 1)(x^2 + 8x - 4) \]
Notice also that \( f(1) = 1 + 7 - 12 + 4 = 0 \). Once again the polynomial is zero at the very number that sits inside the exact divisor.
When a polynomial f(x) is divided by d(x) and the quotient is q(x) with remainder 0, the three are related by \( f(x) = d(x) \times q(x) \). So if f(x) is exactly divisible by \( x - a \), then \( x - a \) is a factor of f(x).
The Factor Theorem
What we saw with the two examples above holds for every polynomial. Stated as a theorem, it reads:
Factor theorem: If a polynomial f(x) of degree n, where \( n \ge 1 \), is exactly divisible by \( x - a \), then \( x - a \) is a factor of f(x). That is, \( f(x) = q(x) \times (x - a) \), where q(x) is the quotient.
Proof. The proof uses only the relation between the parts of a division, the same relation you use to check any long division.
Step 1: Let the quotient be q(x) and the remainder be R when f(x) is divided by \( (x - a) \). By the rule of division, these are related as:
\[ f(x) = q(x) \times (x - a) + R \quad \ldots (i) \]
Step 2: Since f(x) is exactly divisible by \( x - a \), the remainder is zero. Putting \( R = 0 \) in equation (i):
\[ f(x) = q(x) \times (x - a) + 0 \]
Step 3: Adding zero changes nothing, so dropping it:
\[ f(x) = q(x) \times (x - a) \]
The polynomial f(x) is now written as \( (x - a) \) multiplied by another polynomial q(x). That is exactly what it means for \( x - a \) to be a factor. Hence \( (x - a) \) is a factor of f(x). Proved.
We do not want to carry out a long division every time we test a factor. The remainder theorem saves us the work: when f(x) is divided by \( x - a \), the remainder is \( f(a) \). So saying that f(x) is exactly divisible by \( x - a \) is the same as saying \( f(a) = 0 \). The factor theorem therefore gives a test that needs only one substitution.
To test whether \( x - a \) is a factor of f(x), find \( f(a) \). If \( f(a) = 0 \), then \( x - a \) is a factor. If \( f(a) \) is any other number, \( x - a \) is not a factor, and that number is the remainder.
The Converse of the Factor Theorem
The converse of a statement runs it backwards. The factor theorem starts from exact division and ends with a factor. Its converse starts from a factor and tells us the value of the remainder.
Converse of the factor theorem: If \( x - a \) is a factor of a polynomial f(x) of degree n, where \( n \ge 1 \), then the remainder \( f(a) = 0 \).
Proof. This time we begin with the product form and put \( x = a \) into it.
Step 1: Since \( x - a \) is a factor of f(x), there is a quotient q(x) such that:
\[ f(x) = q(x) \times (x - a) \quad \ldots (i) \]
Step 2: Putting \( x = a \) in equation (i):
\[ f(a) = q(a) \times (a - a) \]
Step 3: Since \( a - a = 0 \), the bracket becomes zero:
\[ f(a) = q(a) \times 0 \]
Step 4: Any number multiplied by zero is zero, so:
\[ f(a) = 0 \]
Hence \( f(a) = 0 \). Proved. This explains why 2 and 3 made \( x^2 - 5x + 6 \) zero. They are the numbers that make the factors \( x - 2 \) and \( x - 3 \) equal to zero, and zero multiplied by anything is zero.
| Statement | What we start from | What we conclude |
| Factor theorem | f(x) is exactly divisible by \( x - a \) | \( x - a \) is a factor, so \( f(x) = q(x) \times (x - a) \) |
| Converse | \( x - a \) is a factor of f(x) | The remainder \( f(a) = 0 \) |
Testing Whether a Binomial Is a Factor
Together, the two theorems give a quick test. We never need to divide. We only need the value of x that makes the binomial zero, and then one substitution. The method is always the same.
- Step 1: Put the binomial equal to zero and solve it to find the value of x.
- Step 2: Substitute that value of x into the polynomial f(x).
- Step 3: Simplify carefully, working out one term at a time.
- Step 4: If the result is 0, the binomial is a factor. If the result is any other number, it is not a factor.
The value to substitute depends on the sign inside the binomial, and on the coefficient of x when x does not stand alone:
| Binomial | Solving it for x | Value to find |
| \( x - 1 \) | \( x = 1 \) | \( f(1) \) |
| \( x + 3 \) | \( x = -3 \) | \( f(-3) \) |
| \( x + k \) | \( x = -k \) | \( f(-k) \) |
| \( 2x - 1 \) | \( x = \frac{1}{2} \) | \( f\left(\frac{1}{2}\right) \) |
| \( 2x + 1 \) | \( x = -\frac{1}{2} \) | \( f\left(-\frac{1}{2}\right) \) |
| \( cx - d \) | \( x = \frac{d}{c} \) | \( f\left(\frac{d}{c}\right) \) |
For the factor \( x + 3 \), substitute \( x = -3 \), not \( x = 3 \). Writing \( x + 3 \) as \( x - (-3) \) shows that here \( a = -3 \). Copying the sign exactly as it appears in the bracket is the most frequent error in this topic.
Why does a binomial like \( cx - d \) follow the same rule? Because \( cx - d = c\left(x - \frac{d}{c}\right) \). If \( f(x) = q(x)\left(x - \frac{d}{c}\right) \), then \( f(x) = \frac{q(x)}{c}(cx - d) \), so \( cx - d \) is a factor as well. Therefore \( cx - d \) is a factor of f(x) exactly when \( f\left(\frac{d}{c}\right) = 0 \).
Worked example 1: Prove that \( x - 1 \) is a factor of the polynomial \( f(x) = 2x^3 - 5x^2 + 4x - 1 \).
Step 1: Solving \( x - 1 = 0 \) gives \( x = 1 \). Substituting it in f(x):
\[ f(1) = 2(1)^3 - 5(1)^2 + 4(1) - 1 \]
Step 2: Every power of 1 is 1, so each term becomes a plain number:
\[ f(1) = 2 - 5 + 4 - 1 \]
Step 3: Adding the positive terms together and the negative terms together:
\[ f(1) = 6 - 6 = 0 \]
Since \( f(1) = 0 \), by the factor theorem \( x - 1 \) is a factor of \( 2x^3 - 5x^2 + 4x - 1 \). Proved.
Worked example 2: Show that \( 3x - 2 \) is a factor of \( g(x) = 3x^3 - 2x^2 + 3x - 2 \).
Step 1: Solving \( 3x - 2 = 0 \) gives \( x = \frac{2}{3} \). Since \( \left(\frac{2}{3}\right)^3 = \frac{8}{27} \) and \( \left(\frac{2}{3}\right)^2 = \frac{4}{9} \), substituting gives:
\[ \begin{aligned} g\left(\tfrac{2}{3}\right) &= 3 \times \tfrac{8}{27} - 2 \times \tfrac{4}{9} \\ &\quad + 3 \times \tfrac{2}{3} - 2 \end{aligned} \]
Step 2: Multiplying out each term:
\[ g\left(\tfrac{2}{3}\right) = \tfrac{8}{9} - \tfrac{8}{9} + 2 - 2 \]
Step 3: The equal and opposite terms cancel:
\[ g\left(\tfrac{2}{3}\right) = 0 \]
Since \( g\left(\frac{2}{3}\right) = 0 \), the binomial \( 3x - 2 \) is a factor of g(x). Proved.
Finding an Unknown Coefficient
Sometimes a polynomial carries an unknown letter such as k, and we are told that a certain binomial is a factor. The converse of the factor theorem then hands us an equation: the polynomial must equal zero at the value of x that makes the binomial zero. Solving that equation finds the letter.
Worked example 3: If \( x + 1 \) is a factor of the polynomial \( f(x) = x^3 - kx^2 + 3x + 6 \), find the value of k.
Step 1: Solving \( x + 1 = 0 \) gives \( x = -1 \). Since \( x + 1 \) is a factor, \( f(-1) = 0 \). Substituting \( x = -1 \):
\[ f(-1) = (-1)^3 - k(-1)^2 + 3(-1) + 6 \]
Step 2: Putting \( f(-1) = 0 \) and simplifying each term, using \( (-1)^3 = -1 \) and \( (-1)^2 = 1 \):
\[ 0 = -1 - k - 3 + 6 \]
Step 3: Collecting the numbers:
\[ 0 = 2 - k \]
Step 4: Solving for k:
\[ k = 2 \]
Check: with \( k = 2 \), \( f(-1) = -1 - 2 - 3 + 6 = 0 \), so \( x + 1 \) really is a factor.
The unknown may also sit in a power. Then the laws of indices do the work.
Worked example 4: If \( x + 3 \) is a factor of the polynomial \( f(x) = 2x^k + 3x^{k-1} - 5x + 12 \), find the value of k.
Step 1: Solving \( x + 3 = 0 \) gives \( x = -3 \), so \( f(-3) = 0 \). Substituting:
\[ 0 = 2(-3)^k + 3(-3)^{k-1} - 5(-3) + 12 \]
Step 2: By the law of indices \( (-3)^{k-1} = (-3)^k \times (-3)^{-1} \), and \( -5(-3) = 15 \). Rewriting:
\[ 0 = 2(-3)^k + 3(-3)^k (-3)^{-1} + 15 + 12 \]
Step 3: Since \( 3 \times (-3)^{-1} = \frac{3}{-3} = -1 \), taking \( (-3)^k \) common:
\[ 0 = (-3)^k \{2 - 1\} + 27 \]
Step 4: Simplifying the bracket:
\[ 0 = (-3)^k + 27 \]
Step 5: Moving 27 to the other side and writing \( -27 \) as a power of \( -3 \):
\[ (-3)^k = -27 = (-3)^3 \]
Step 6: The bases are equal, so equating the indices:
\[ k = 3 \]
Check: with \( k = 3 \) the polynomial is \( 2x^3 + 3x^2 - 5x + 12 \), and \( f(-3) = -54 + 27 + 15 + 12 = 0 \).
Finding the Number to Subtract or Add
A polynomial may not have the factor we want, but it can be given that factor by changing its constant term. If a number k is subtracted from f(x), the new polynomial is \( f(x) - k \). For \( x - a \) to be a factor of the new polynomial, the new polynomial must be zero at \( x = a \). This gives an equation in k.
Worked example 5: What number must be subtracted from \( f(x) = 2x^3 - 7x^2 + 7x - 9 \) so that \( x - 3 \) becomes a factor of it?
Step 1: Let the number to be subtracted be k. The new polynomial is \( 2x^3 - 7x^2 + 7x - 9 - k \), and since \( x - 3 \) must be its factor, it is zero at \( x = 3 \):
\[ 2(3)^3 - 7(3)^2 + 7(3) - 9 - k = 0 \]
Step 2: Working out each term, with \( 3^3 = 27 \) and \( 3^2 = 9 \):
\[ 54 - 63 + 21 - 9 - k = 0 \]
Step 3: Collecting the numbers:
\[ 3 - k = 0 \]
Step 4: Solving for k:
\[ k = 3 \]
So 3 must be subtracted. Check: the new polynomial \( 2x^3 - 7x^2 + 7x - 12 \) gives \( 54 - 63 + 21 - 12 = 0 \) at \( x = 3 \). In fact it factorises as:
\[ 2x^3 - 7x^2 + 7x - 12 = (x - 3)(2x^2 - x + 4) \]
Worked example 6: What number must be added to \( f(x) = x^3 - 2x^2 + 3x - 4 \) so that \( x - 1 \) becomes a factor of it?
Step 1: Let the number to be added be k. The new polynomial \( x^3 - 2x^2 + 3x - 4 + k \) must be zero at \( x = 1 \):
\[ (1)^3 - 2(1)^2 + 3(1) - 4 + k = 0 \]
Step 2: Working out each term:
\[ 1 - 2 + 3 - 4 + k = 0 \]
Step 3: Collecting the numbers:
\[ -2 + k = 0 \]
Step 4: Solving for k:
\[ k = 2 \]
So 2 must be added. The new polynomial \( x^3 - 2x^2 + 3x - 2 \) gives \( 1 - 2 + 3 - 2 = 0 \) at \( x = 1 \), so \( x - 1 \) is now a factor.
The number to subtract is \( f(a) \) itself, and the number to add is \( -f(a) \). In worked example 5, \( f(3) = 3 \), so 3 is subtracted. In worked example 6, \( f(1) = -2 \), so 2 is added.
Write the new polynomial with \( -k \) when the question says subtract and with \( +k \) when it says add. If k comes out negative, the opposite operation is needed: adding \( -5 \) is the same as subtracting 5.
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