When a Value Makes a Polynomial Zero
Take the cubic equation \( 5x^3 - 6x^2 + 7x - 6 = 0 \). We can check whether \( x = 1 \) satisfies it by putting 1 in place of every \( x \) on the left side.
Step 1: Replacing every \( x \) by 1:
\[ 5(1)^3 - 6(1)^2 + 7(1) - 6 \]
Step 2: Working out each term and adding from the left:
\[ 5 - 6 + 7 - 6 = 0 \]
The left side becomes 0, so \( x = 1 \) satisfies the equation. From the factor theorem you already know what this means: \( (x - 1) \) is a factor of \( 5x^3 - 6x^2 + 7x - 6 \). Dividing the cubic by \( (x - 1) \) leaves a quadratic, and a quadratic is something you can already factorise by splitting the middle term.
So factorising a cubic comes down to one question. How do we find that first value of \( x \) which makes the polynomial zero? Trying numbers at random could take a very long time, and the root might even be a fraction. The rational root theorem answers this question. It gives a short list of numbers, and if the equation has a rational root at all, that root is somewhere on the list.
Leading Coefficient and Constant Term
A polynomial equation of degree \( n \) can be written in this general form, with the powers of \( x \) in descending order:
\[ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0 \]
Two numbers in this form matter for the theorem: the number at the very front and the number at the very end.
Leading coefficient: the coefficient of the highest power of the variable, written \( a_n \). In \( 2x^3 - 3x^2 - 11x + 6 \), the leading coefficient is 2.
Constant term: the term that contains no variable at all, written \( a_0 \). In \( 2x^3 - 3x^2 - 11x + 6 \), the constant term is 6.
The sign belongs to the number. In \( x^3 - x^2 - 10x - 8 \) the leading coefficient is 1, because \( x^3 \) means \( 1x^3 \), and the constant term is \( -8 \). When we list factors for the theorem we use 8, because the \( \pm \) sign in front of the list already covers both signs.
Arrange the polynomial in descending powers of \( x \) before you read off \( a_n \) and \( a_0 \). In \( 6 + 4x^2 - x^3 \) the first number written is 6, but the leading coefficient is \( -1 \), the coefficient of \( x^3 \).
The Rational Root Theorem
Rational root theorem: if the polynomial equation \( a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0 \) has integer coefficients, its possible rational roots are \( \pm \) (factors of \( a_0 \)) divided by (factors of \( a_n \)), that is \( \pm \frac{p}{q} \), where \( p \) is a factor of the constant term, \( q \) is a factor of the leading coefficient, and \( \frac{p}{q} \) is in its lowest terms.
\[ x = \pm \frac{\text{factors of } a_0}{\text{factors of } a_n} = \pm \frac{p}{q} \]
Notice the word possible. The theorem does not say that every number on the list is a root. It gives the candidates, and each candidate still has to be tested by putting it into the polynomial. The first one that gives 0 is a root. The theorem is still very powerful, because it also works the other way round: a rational number that is not on the list can never be a root, so we never waste time on it.
Here is how the list is made for \( 2x^3 - 3x^2 - 11x + 6 = 0 \). The factors of the constant term 6 are 1, 2, 3 and 6. The factors of the leading coefficient 2 are 1 and 2. Dividing each factor of 6 by each factor of 2:
\[ \frac{1}{1}, \frac{2}{1}, \frac{3}{1}, \frac{6}{1}, \frac{1}{2}, \frac{2}{2}, \frac{3}{2}, \frac{6}{2} \]
Since \( \frac{2}{2} = 1 \) and \( \frac{6}{2} = 3 \) are already in the list, the possible rational roots in lowest terms are:
\[ x = \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2} \]
When the leading coefficient is 1, the only factor below the line is 1, so every candidate is a whole number. The possible roots are then simply plus or minus the factors of the constant term.
Roots or Zeros and the Factor They Give
Roots or zeros: the values of the variable that satisfy the polynomial equation \( f(x) = 0 \). If \( f(a) = 0 \), then \( a \) is a root (or zero) of \( f(x) \).
To factorise a polynomial or to solve a polynomial equation, the first job is always to find one root. Once a root \( a \) is found, the factor theorem tells us that \( (x - a) \) is a factor. Dividing \( f(x) \) by this divisor \( d(x) = x - a \) with synthetic division gives a quotient \( q(x) \) and a remainder of 0, so the polynomial splits as:
\[ f(x) = d(x) \times q(x) \]
For a cubic, \( q(x) \) has degree 2. That is why one root is enough to get started: the rest of the work is factorising a quadratic.
The factor has the opposite sign to the root. If \( f(-1) = 0 \), the factor is \( x - (-1) = x + 1 \), not \( x - 1 \). If \( f(2) = 0 \), the factor is \( x - 2 \). In synthetic division you divide by the root itself, so for the factor \( x + 1 \) you write \( -1 \) in the corner.
Steps to Factorise a Polynomial of Degree 3
Every cubic in this topic is factorised by the same five steps.
- (a) Arrange the terms of \( f(x) \) in descending order of the powers of the variable.
- (b) Use the constant term \( a_0 \) and the leading coefficient \( a_n \) to list the possible roots \( \pm \frac{p}{q} \) in lowest terms. Then put values such as \( x = \pm 1, \pm 2, \pm 3, \pm \frac{1}{2}, \pm \frac{2}{3} \) from the list into \( f(x) \) until one of them makes the value of the polynomial zero.
- (c) Once a value of \( x \) makes the polynomial zero, use synthetic division with that value to find the quotient \( q(x) \).
- (d) The quotient \( q(x) \) has degree 2. Factorise it, usually by splitting the middle term.
- (e) Write \( f(x) \) as the product of all three factors.
Start testing with the smallest candidates, \( 1 \) and \( -1 \), then \( 2 \) and \( -2 \). Small numbers are quick to substitute, and very often one of them is a root.
Worked Example 1: Factorising a Cubic
Factorise \( x^3 - x^2 - 10x - 8 \).
Step 1: The polynomial, already in descending powers, is:
\[ f(x) = x^3 - x^2 - 10x - 8 \]
Step 2: Listing the possible roots with the rational root theorem. Here \( a_0 = -8 \) and \( a_n = 1 \):
\[ x = \pm \frac{\text{factors of } 8}{\text{factors of } 1} = \pm \frac{1, 2, 4, 8}{1} \]
So the possible rational roots are \( x = \pm 1, \pm 2, \pm 4, \pm 8 \).
Step 3: Testing \( x = 1 \):
\[ f(1) = 1 - 1 - 10 - 8 = -18 \]
The value is not zero, so 1 is not a root. Step 4: Testing \( x = -1 \):
\[ f(-1) = (-1)^3 - (-1)^2 - 10(-1) - 8 \]
\[ = -1 - 1 + 10 - 8 = 0 \]
Since \( f(-1) = 0 \), one factor of \( f(x) \) is \( x - (-1) = x + 1 \).
Step 5: Dividing by \( x + 1 \) with synthetic division. We write \( -1 \) in the corner and the coefficients 1, \( -1 \), \( -10 \), \( -8 \) along the top. Bring down 1, multiply by \( -1 \), add to the next coefficient, and repeat:
\[ \begin{array}{r|rrrr} -1 & 1 & -1 & -10 & -8 \\ & & -1 & 2 & 8 \\ \hline & 1 & -2 & -8 & 0 \end{array} \]
Step 6: Reading the bottom row. The last number 0 is the remainder, and 1, \( -2 \), \( -8 \) are the coefficients of the quotient, one degree lower than the cubic:
\[ q(x) = x^2 - 2x - 8 \]
Step 7: Factorising the quotient by splitting the middle term. We need two numbers whose product is \( -8 \) and whose sum is \( -2 \), which are \( -4 \) and 2:
\[ x^2 - 2x - 8 = x^2 - (4 - 2)x - 8 \]
\[ = x^2 - 4x + 2x - 8 \]
\[ = x(x - 4) + 2(x - 4) \]
\[ = (x - 4)(x + 2) \]
Step 8: Writing all three factors together:
\[ x^3 - x^2 - 10x - 8 = (x + 1)(x + 2)(x - 4) \]
A quick check: put \( x = 0 \) on both sides. The left side gives \( -8 \). The right side gives \( (1)(2)(-4) = -8 \). Both sides agree, so the factors are very likely right.
Worked Example 2: Solving a Cubic Equation
Solve \( (x - 1)(x^2 - 2x - 12) + 12 = 0 \). The equation is not yet in the general form, so the brackets have to be removed first.
Step 1: Multiplying out the brackets, first by \( x \) and then by \( -1 \):
\[ x^3 - 2x^2 - 12x - x^2 + 2x + 12 + 12 = 0 \]
Step 2: Collecting like terms:
\[ x^3 - 3x^2 - 10x + 24 = 0 \]
Step 3: Listing the possible roots. Here \( a_0 = 24 \) and \( a_n = 1 \), so the candidates are plus or minus the factors of 24:
\[ x = \pm 1, \pm 2, \pm 3, \pm 4, \]
\[ \pm 6, \pm 8, \pm 12, \pm 24 \]
Step 4: Testing \( x = 1 \) gives \( 1 - 3 - 10 + 24 = 12 \), which is not zero. Testing \( x = 2 \):
\[ f(2) = (2)^3 - 3(2)^2 - 10(2) + 24 \]
\[ = 8 - 12 - 20 + 24 = 0 \]
So \( (x - 2) \) is a factor. Step 5: Dividing by \( x - 2 \) with synthetic division, writing 2 in the corner:
\[ \begin{array}{r|rrrr} 2 & 1 & -3 & -10 & 24 \\ & & 2 & -2 & -24 \\ \hline & 1 & -1 & -12 & 0 \end{array} \]
Step 6: Factorising the quotient \( q(x) = x^2 - x - 12 \). The two numbers with product \( -12 \) and sum \( -1 \) are \( -4 \) and 3:
\[ x^2 - x - 12 = x^2 - 4x + 3x - 12 \]
\[ = x(x - 4) + 3(x - 4) = (x - 4)(x + 3) \]
Step 7: Using \( f(x) = d(x) \times q(x) \), the equation becomes:
\[ (x - 2)(x - 4)(x + 3) = 0 \]
Step 8: A product is zero only when one of its factors is zero. So \( x - 2 = 0 \), or \( x - 4 = 0 \), or \( x + 3 = 0 \). This gives:
\[ x = 2, \; 4, \; -3 \]
When a question says factorise, the answer is the product of factors. When it says solve, the answer is the values of \( x \). Stopping at \( (x - 2)(x - 4)(x + 3) = 0 \) does not answer a solve question.
Worked Example 3: When the Leading Coefficient Is Not 1
Factorise \( 2x^3 - 3x^2 - 11x + 6 \). This time the leading coefficient is 2, so fractions appear among the candidates.
Step 1: Listing the possible roots. The factors of 6 are 1, 2, 3, 6 and the factors of 2 are 1, 2. As worked out earlier, the candidates in lowest terms are:
\[ x = \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2} \]
Step 2: Testing the small candidates one by one. The results are:
| Value of \( x \) | Value of \( f(x) \) | Is it a root? |
| \( 1 \) | \( 2 - 3 - 11 + 6 = -6 \) | No |
| \( -1 \) | \( -2 - 3 + 11 + 6 = 12 \) | No |
| \( 2 \) | \( 16 - 12 - 22 + 6 = -12 \) | No |
| \( -2 \) | \( -16 - 12 + 22 + 6 = 0 \) | Yes |
Since \( f(-2) = 0 \), \( (x + 2) \) is a factor. Step 3: Dividing by \( x + 2 \) with synthetic division, writing \( -2 \) in the corner:
\[ \begin{array}{r|rrrr} -2 & 2 & -3 & -11 & 6 \\ & & -4 & 14 & -6 \\ \hline & 2 & -7 & 3 & 0 \end{array} \]
Step 4: Factorising the quotient \( q(x) = 2x^2 - 7x + 3 \). We need two numbers with product \( 2 \times 3 = 6 \) and sum \( -7 \), which are \( -6 \) and \( -1 \):
\[ 2x^2 - 7x + 3 = 2x^2 - 6x - x + 3 \]
\[ = 2x(x - 3) - 1(x - 3) = (x - 3)(2x - 1) \]
Step 5: Writing all three factors together:
\[ 2x^3 - 3x^2 - 11x + 6 = (x + 2)(2x - 1)(x - 3) \]
Look at the factor \( (2x - 1) \). It is zero when \( x = \frac{1}{2} \), and \( \frac{1}{2} \) was on our list of candidates. This is exactly why the theorem divides by the factors of the leading coefficient: a root like \( \frac{1}{2} \) could never have been found by trying whole numbers only.
When a Power of x Is Missing
Some cubics skip a power, such as \( x^3 - 7x + 6 \), which has no \( x^2 \) term. The coefficient of the missing power is 0, and that 0 must be written in its place in the synthetic division. Here the candidates are \( \pm 1, \pm 2, \pm 3, \pm 6 \), and \( f(1) = 1 - 7 + 6 = 0 \), so \( (x - 1) \) is a factor.
Step 1: Dividing by \( x - 1 \) with the coefficients 1, 0, \( -7 \), 6:
\[ \begin{array}{r|rrrr} 1 & 1 & 0 & -7 & 6 \\ & & 1 & 1 & -6 \\ \hline & 1 & 1 & -6 & 0 \end{array} \]
Step 2: Factorising the quotient \( x^2 + x - 6 \), using 3 and \( -2 \):
\[ x^2 + x - 6 = (x + 3)(x - 2) \]
Step 3: Writing all three factors together:
\[ x^3 - 7x + 6 = (x - 1)(x - 2)(x + 3) \]
Leaving out the 0 for a missing power shifts every coefficient into the wrong column. Written as 1, \( -7 \), 6, the division treats the polynomial as the quadratic \( x^2 - 7x + 6 \), and every step after that is wrong.
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