Factorisation of polynomials | ICSE Class 10 Maths Notes

This note covers polynomial terms and degree, substitution and zeroes, the Remainder Theorem, the Factor Theorem, unknown coefficients, polynomial division, quadratic factorisation, complete factorisation of cubic polynomials, and checks on factors and signs.

What do polynomial terms, degree and factorisation mean?

A variable, written here as x, is a symbol whose value can change. Real numbers are numbers represented on the number line. A polynomial in x combines variable powers with real numerical multipliers, called coefficients. Its variable powers are non-negative whole numbers: zero or positive integers.

A term is a part of an expression separated by addition or subtraction, with its sign retained. Its coefficient is its numerical multiplier. A constant term has no variable factor. An exponent, or power, specifies repeated multiplication: x² means x × x, and x³ means x × x × x. A non-zero number raised to power zero equals 1.

The notation p(x) names a polynomial in x; it does not mean p multiplied by x. For example, p(x) = x² − 3x − 4 has terms x², −3x and −4. The coefficient of x is −3, and the constant term is −4.

How is degree identified?

The degree of a non-zero polynomial is its highest power with a non-zero coefficient. A linear polynomial has degree 1, a quadratic polynomial has degree 2, and a cubic polynomial has degree 3. The zero polynomial has every coefficient zero; its degree is not defined.

Here a, b, c and d denote real coefficients; ≠ means “is not equal to”.

The coefficient of the highest power is the leading coefficient. Its non-zero condition ensures that the stated degree is correct.

Factorisation means writing an expression as a product of factors. A factor divides the polynomial with zero remainder. Multiplying the factors together reverses factorisation and recovers the original expression. For instance, x² − 3x − 4 = (x + 1)(x − 4).

How do substitution and zeroes prepare us to find factors?

Substitution means replacing the variable by a specified number everywhere in the expression. If k denotes a real number, p(k) is the value obtained by replacing every x in p(x) by k. A zero of p(x) is a number k for which p(k) = 0.

Keep the distinction between an expression and its value. The expression p(x) can take different values as x changes. The notation p(2) requests one particular value. It does not request the coefficient of x² or the result of multiplying the polynomial by 2.

How should negative substitutions be written?

Place a negative value inside brackets before taking a power. In particular, (−1)² = 1, while (−1)³ = −1. Calculate powers first, then multiplication, then addition and subtraction. This keeps the sign belonging to the substituted number separate from a coefficient's sign.

Worked example 1. For p(x) = x² − 3x − 4, find p(2), p(−1) and p(4). Identify which of the tested values are zeroes.

Answer: p(2) = 2² − 3 × 2 − 4 = −6.

p(−1) = (−1)² − 3 × (−1) − 4 = 1 + 3 − 4 = 0.

p(4) = 4² − 3 × 4 − 4 = 16 − 12 − 4 = 0. Therefore −1 and 4 are zeroes, whereas 2 is not a zero.

The zeroes are numbers, while the corresponding factors are expressions. Here the zero −1 corresponds to x + 1, and the zero 4 corresponds to x − 4. The Factor Theorem will justify this connection.

A non-zero value is useful too: it rules out that particular candidate as a zero. It does not say that the polynomial has no zeroes. In the example, p(2) is non-zero even though two other tested values make the same polynomial zero.

How does the Remainder Theorem replace long division?

In division, the dividend is the polynomial being divided, the divisor is the expression dividing it, the quotient is the resulting polynomial, and the remainder is what remains. For division by a linear polynomial, the remainder is a constant, possibly zero.

Theorem: Remainder Theorem

Definition: If p(x) is divided by x − k, where k is a real number, the remainder is p(k).

The theorem provides the remainder without requiring the quotient. First find the value that makes the divisor zero, then substitute that value into the dividend. For x + 1, that value is −1; for x − 3, it is 3.

Identity: Dividend equals divisor times quotient plus remainder

An identity is an equality true for every permitted value of the variable. Write the quotient as q(x) and the constant remainder as r. Division by x − k gives the identity p(x) = (x − k)q(x) + r.

Substitute x = k. The product (k − k)q(k) is zero, leaving p(k) = r. This explains both the substitution value and why calculating the quotient is unnecessary when only the remainder is required.

Worked example 2. Find the remainder when p(x) = x³ − 2x² − 4x − 1 is divided by x + 1.

Answer: x + 1 = 0 gives x = −1. Hence the remainder is p(−1) = −1 − 2 + 4 − 1 = 0. The divisor therefore divides this polynomial exactly.

Worked example 3. Find the remainder when p(x) = x³ − 3x² + 4x + 50 is divided by x − 3.

Answer: The remainder is p(3) = 27 − 27 + 12 + 50 = 62. Since this is non-zero, x − 3 is not a factor.

The two examples use the same rule but reach different conclusions. A remainder can be zero or non-zero; exact divisibility means specifically that it is zero.

What changes when the linear divisor has a coefficient before x?

A divisor need not begin with x alone. For the general linear divisor ax + b, with a ≠ 0, solve ax + b = 0 to obtain x = −b/a. The remainder is the dividend evaluated at this value.

The division identity explains this extension. With quotient q(x) and remainder r, write p(x) = (ax + b)q(x) + r. At x = −b/a, the entire divisor becomes zero. Its product with the quotient vanishes, leaving r = p(−b/a).

Why must the entire divisor be set equal to zero?

For 2x − 1, substitution of 1 would leave the divisor equal to 1, not zero. Solving 2x − 1 = 0 gives x = 1/2. Use that value in every term of the dividend, including its square and cube.

Worked example 4. Find the remainder when p(x) = 4x³ − 12x² + 14x − 3 is divided by 2x − 1.

Answer: Solve 2x − 1 = 0, giving x = 1/2.

p(1/2) = 4 × (1/2)³ − 12 × (1/2)² + 14 × (1/2) − 3 = 1/2 − 3 + 7 − 3 = 3/2.

The remainder is 3/2. Since it is not zero, 2x − 1 is not a factor of p(x).

The same approach handles a negative coefficient of x. For the divisor 2 − 3x, the required substitution is x = 2/3. The order in which the constant and variable terms are written does not change the equation to solve.

Note: Evaluate the dividend at the zero of the divisor. Do not substitute the divisor itself, and do not multiply or divide the resulting remainder by the divisor's leading coefficient.

Fractions in the substitution are legitimate. Calculate each power of the fraction before multiplying by the corresponding coefficient. Keep exact fractions through the working so that a zero remainder, when it occurs, is recognised exactly.

How does the Factor Theorem turn a zero into a factor?

Theorem: Factor Theorem

Definition: For a polynomial p(x) of degree at least 1, x − k is a factor if and only if p(k) = 0, where k is a real number.

The phrase if and only if states both directions. If x − k is a factor, division leaves no remainder, so p(k) = 0. Conversely, if p(k) = 0, the Remainder Theorem gives remainder zero, so x − k divides p(x) exactly.

This distinction matters when reading a question. “Show that an expression is a factor” asks for a zero remainder calculation. “Find a factor” requires a successful candidate first. “Factorise completely” requires further work after that first factor has been obtained.

How is a proposed factor tested?

1. Identify the proposed linear factor, keeping its coefficient and constant sign.
2. Set that factor equal to zero and solve for the substitution value.
3. Evaluate the full polynomial at this value, showing the arithmetic.
4. Use the zero or non-zero result to state whether the expression is a factor.

Worked example 5. Show that x + 3 is a factor of p(x) = 69 + 11x − x² + x³.

Answer: Set x + 3 = 0, so x = −3. Then p(−3) = 69 + 11 × (−3) − (−3)² + (−3)³ = 69 − 33 − 9 − 27 = 0.

Therefore x + 3 is a factor of p(x), by the Factor Theorem.

Notice that the polynomial need not be written with its highest power first to perform this test. Substitution works on the expression as given. Rearranging terms into descending powers can nevertheless make the signs easier to track.

A multiple of a polynomial is an expression obtained by multiplying it by another polynomial. Thus saying that x + 3 is a factor of p(x) also says that p(x) is a multiple of x + 3. Both statements describe the same exact divisibility.

How can a given factor determine an unknown coefficient?

A polynomial question may include an unknown coefficient, a fixed number that must be found from an additional condition. A given factor supplies that condition: substitute its zero into the polynomial and set the result equal to zero.

This produces an equation for the unknown coefficient. Solve that equation and then check the value in the original factor condition. Treat the variable and the unknown coefficient as different roles, even when both are represented by letters.

What happens when the factor is already numerical?

Worked example 6. Find the real coefficient m for which p(x) = x³ − 2mx² + 16 is divisible by x + 2.

Answer: The factor condition gives p(−2) = 0. Hence (−2)³ − 2m(−2)² + 16 = 0, so −8 − 8m + 16 = 0.

Thus 8 − 8m = 0 and m = 1. Checking gives −8 − 8 + 16 = 0, as required.

The square applies to −2 before multiplication by −2m. Missing those brackets can reverse a sign and produce the wrong coefficient. The condition concerns the entire polynomial, so its constant term must also be included.

What if the unknown also appears in the factor?

Worked example 7. Find the real number a if x − a is a factor of p(x) = x³ − ax² + 2x + a − 1.

Answer: Use p(a) = 0. This gives a³ − a × a² + 2a + a − 1 = 0. The cubic terms cancel, leaving 3a − 1 = 0.

Therefore a = 1/3. The reduced factor condition becomes 3 × (1/3) − 1 = 0, confirming the value.

Here a plays two connected roles: it is a coefficient in the polynomial and the number substituted for x. Replacing x by a does not mean removing the existing coefficient a. Write the substitution fully before simplifying the resulting powers.

How is the remaining quadratic found after a cubic factor is known?

When a cubic polynomial is divided exactly by a linear factor, the quotient has degree 2. This quadratic quotient contains the remaining factorisation work. Finding one factor alone does not normally complete the factorisation of a cubic.

Write the dividend and divisor in descending powers, meaning highest power first. If a power is absent, a zero coefficient can hold its place in written division. At each stage divide the leading term of the current expression by the leading term of the divisor.

How does repeated multiplication and subtraction work?

1. Divide the leading terms to obtain the next quotient term.
2. Multiply the whole divisor by this term.
3. Subtract that entire product, changing every sign within the subtracted expression.
4. Continue with the remaining terms until the remainder has lower degree than the divisor.

Worked example 8. Obtain one factor of p(x) = x³ − 6x² + 11x − 6 by the Factor Theorem, then find the quotient.

Answer: p(1) = 1 − 6 + 11 − 6 = 0. Therefore x − 1 is a factor.

The first quotient term is x². Subtract x²(x − 1) = x³ − x² from the dividend, leaving −5x² + 11x − 6.

The next quotient term is −5x. Subtract −5x(x − 1) = −5x² + 5x, leaving 6x − 6.

The final quotient term is 6. Subtract 6(x − 1) = 6x − 6, leaving zero. Thus p(x) = (x − 1)(x² − 5x + 6).

The quotient is x² − 5x + 6. Its leading term x² is consistent with dividing a cubic leading term by a linear leading term. The remainder zero agrees with the earlier factor test, giving a useful internal check on the division.

Two errors deserve particular attention: subtracting only the first term of a product, and dropping the constant from the dividend. Keep the whole subtraction visible. If division gives a non-zero remainder after a verified factor test, recheck substitution and subtraction before proceeding.

How is the quadratic quotient split into linear factors?

For a quadratic expression ax² + bx + c, where a ≠ 0, splitting the middle term means writing bx as the sum of two terms whose coefficients add to b and multiply to ac. Grouping those terms can reveal a common binomial factor.

A binomial is a polynomial with two non-zero terms. A common factor divides each of the grouped terms. After taking common factors from the two groups, look for the same binomial appearing in both.

How is the previous cubic completed?

The quotient x² − 5x + 6 needs two numbers with sum −5 and product 6. They are −2 and −3. Thus x² − 5x + 6 = x² − 2x − 3x + 6 = x(x − 2) − 3(x − 2).

Taking out the common factor x − 2 gives (x − 2)(x − 3). Retain the cubic's first factor to obtain x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3).

Worked example 9. Factorise p(x) = 3x² + 5x − 2 by splitting the middle term.

Answer: The required sum is 5 and the product is 3 × (−2) = −6. Use 6 and −1.

3x² + 5x − 2 = 3x² + 6x − x − 2 = 3x(x + 2) − 1(x + 2) = (3x − 1)(x + 2).

When the leading coefficient is not 1, the product to use is ac, not c alone. Also keep numerical factors: 2x² − 8x + 6 = 2(x − 1)(x − 3). Omitting the initial 2 changes the polynomial even though the zeroes remain the same.

Expansion checks the result independently. Multiplying (3x − 1)(x + 2) gives 3x² + 6x − x − 2, which combines to the original quadratic. This checks both the middle coefficient and the sign of the constant term.

How do we factorise different cubic polynomials completely?

The central sequence is test a candidate, obtain a linear factor, find the quadratic quotient, then factorise the quotient. Trial values are candidates rather than established zeroes. A calculation giving zero is the evidence that makes the first factor valid.

Different cubics may leave different kinds of quadratic quotient. Some require splitting the middle term; others match an algebraic identity. Keep the first factor visible throughout so that the final answer is a product equal to the entire original cubic.

Worked example 10. Factorise p(x) = 2x³ − 3x² − 17x + 30 completely after obtaining a factor by the Factor Theorem.

Answer: p(2) = 16 − 12 − 34 + 30 = 0, so x − 2 is a factor.

Division by x − 2 gives 2x² + x − 15. Split its middle term: 2x² + 6x − 5x − 15 = 2x(x + 3) − 5(x + 3).

Therefore p(x) = (x − 2)(2x − 5)(x + 3). All three displayed polynomial factors are linear.

Identity: Difference of two squares

For real numbers or algebraic expressions a and b, a² − b² = (a − b)(a + b). The two factors differ only in the sign between their terms. Multiplication cancels the two middle terms and leaves the difference of the squares.

Worked example 11. Factorise p(x) = x³ + x² − 4x − 4 after finding a factor by the Factor Theorem.

Answer: p(−1) = −1 + 1 + 4 − 4 = 0, so x + 1 is a factor. The quotient is x² − 4.

Since x² − 4 = x² − 2² = (x − 2)(x + 2), the complete factorisation is p(x) = (x + 1)(x − 2)(x + 2).

The factor x + 1 could also be spotted by grouping x³ + x² and −4x − 4. However, when a question requests the Factor Theorem, include the evaluation p(−1) = 0 as the justification for that first factor.

How can the final factorisation be checked without losing signs or factors?

Expansion gives the most direct check: multiply the factors and compare the resulting coefficients with the original polynomial. A correct product reproduces every term, including its sign, coefficient and constant. Checking only one substituted value is not a complete comparison of two polynomials.

What checks are useful before full expansion?

Compare the leading coefficient with the product of the leading coefficients of the factors. Compare the constant term with the product of their constants. In (x − 2)(2x − 5)(x + 3), these checks give leading coefficient 2 and constant (−2)(−5)(3) = 30.

Those checks agree with 2x³ − 3x² − 17x + 30, but they do not check the middle terms. Complete the expansion: (2x − 5)(x + 3) = 2x² + x − 15. Multiplying by x − 2 then gives 2x³ − 3x² − 17x + 30.

Can a factor appear more than once?

A repeated factor occurs more than once in a product. For example, x³ − x² = x²(x − 1) = x × x × (x − 1). Its zeroes are 0 and 1. The factor x occurs twice, but that does not make 0 two different numbers.

A cubic polynomial can have at most three zeroes. “At most” must not be changed to “exactly”: the example x³ − x² has two distinct zeroes, while x³ has only the zero 0. Distinct means different in value.

Finally, distinguish factorising an expression from solving an equation. Factorising gives a product. If that product is set equal to zero, its factors give equations for the zeroes. Do not replace a requested product with only a list of numbers.

Read the requested endpoint before finishing: a remainder is a number, a factor test needs a conclusion, an unknown-coefficient question needs its value, and complete factorisation needs the full product. These outputs use related methods but answer different questions.

Glossary

• Polynomial — An expression formed from terms with non-negative whole-number variable powers and real coefficients.
• Coefficient — The numerical multiplier of a variable power in a polynomial term, including its sign.
• Constant term — The term independent of the variable, unchanged when a value is substituted for that variable.
• Degree — The highest variable power with a non-zero coefficient in a non-zero polynomial.
• Linear polynomial — A polynomial of degree one, with a non-zero coefficient of its variable.
• Quadratic polynomial — A polynomial of degree two, whose squared-variable term has a non-zero coefficient.
• Cubic polynomial — A polynomial of degree three, whose cubed-variable term has a non-zero coefficient.
• Zero — A number that makes a polynomial's value equal to zero when substituted for its variable.
• Factor — A polynomial expression that divides another polynomial exactly, leaving a zero remainder.
• Quotient — The polynomial obtained in division before any separate remainder is added to the division identity.
• Remainder — The part left after division; division by a linear polynomial leaves a constant remainder.
• Factorisation — Rewriting a polynomial as a product of factors whose multiplication recovers the original expression.
• Identity — An equality between algebraic expressions that holds for every permitted value of their variables.
• Repeated factor — A factor occurring more than once in the product representing a polynomial.

Common errors and misconceptions

• Misconception: For a divisor x + 1, the substitution is x = 1. Correct: Solve x + 1 = 0 and substitute −1 into the dividend.
• Misconception: The zero −1 gives the factor x − 1. Correct: The factor corresponding to a zero k is x − k, so −1 gives x + 1.
• Misconception: A non-zero remainder still establishes a factor. Correct: A factor requires exact divisibility, meaning that the remainder is zero.
• Misconception: For 2x − 1, substitute 1. Correct: The divisor becomes zero at x = 1/2, which is the value required by the remainder test.
• Misconception: One linear factor completes a cubic factorisation. Correct: Find the quadratic quotient and factorise it further when possible, retaining the first factor.
• Misconception: Squaring a negative substitution keeps it negative. Correct: Brackets give (−1)² = 1; powers must be evaluated before multiplication by coefficients.
• Misconception: Numerical factors may be dropped once the zeroes are known. Correct: Retain them to preserve the original polynomial's leading coefficient and values.
• Misconception: Every cubic has exactly three different zeroes. Correct: A cubic has at most three zeroes; repeated factors can correspond to the same zero.

Exam-style questions with model answers

Q1. Use the Remainder Theorem to find the remainder when p(x) = x³ − 3x² + 4x + 50 is divided by x − 3. [2 marks]

1. The divisor x − 3 becomes zero at x = 3. By the Remainder Theorem, the required remainder is p(3).
2. Substituting gives p(3) = 27 − 27 + 12 + 50 = 62. Therefore the remainder is 62.

Q2. Show by the Factor Theorem that x + 3 is a factor of p(x) = 69 + 11x − x² + x³. [3 marks]

1. Set the proposed factor equal to zero: x + 3 = 0 gives x = −3. This is the value that must be substituted into the polynomial.
2. Evaluate all four terms: p(−3) = 69 + 11 × (−3) − (−3)² + (−3)³ = 69 − 33 − 9 − 27 = 0.
3. Since p(−3) = 0, the Factor Theorem establishes that x + 3 divides p(x) exactly and is therefore a factor.

Q3. Find the real coefficient m if x³ − 2mx² + 16 is divisible by x + 2. Verify the factor condition. [3 marks]

1. Name the polynomial p(x). Since x + 2 is a factor, its zero −2 must satisfy p(−2) = 0 by the Factor Theorem.
2. Substitution gives −8 − 8m + 16 = 0, so 8 − 8m = 0 and m = 1. The square of −2 is positive.
3. With m = 1, p(−2) = −8 − 8 + 16 = 0. This verifies the required exact divisibility by x + 2.

Q4. Find the remainder when p(x) = 4x³ − 12x² + 14x − 3 is divided by 2x − 1. State whether the divisor is a factor. [3 marks]

1. Solve the divisor equation 2x − 1 = 0, giving x = 1/2. The required remainder is p(1/2), because this substitution makes the divisor zero.
2. Evaluate p(1/2) = 4 × (1/2)³ − 12 × (1/2)² + 14 × (1/2) − 3 = 1/2 − 3 + 7 − 3 = 3/2.
3. The remainder is 3/2, which is non-zero. Therefore 2x − 1 does not divide p(x) exactly and is not a factor.

Q5. Factorise p(x) = x³ − 6x² + 11x − 6 completely, first obtaining a factor by the Factor Theorem. Show the quotient calculation. [5 marks]

1. Test x = 1: p(1) = 1 − 6 + 11 − 6 = 0. The Factor Theorem therefore gives x − 1 as a factor.
2. Begin division by x − 1 with quotient term x². Subtract x³ − x² from the dividend to leave −5x² + 11x − 6.
3. The next quotient term is −5x. Subtract −5x² + 5x, leaving 6x − 6. The final quotient term 6 leaves remainder zero.
4. The quotient is x² − 5x + 6. Split its middle term as −2x − 3x and group to obtain x(x − 2) − 3(x − 2).
5. Thus the quotient factors as (x − 2)(x − 3). Retaining the original factor gives p(x) = (x − 1)(x − 2)(x − 3), the complete factorisation.

Q6. Factorise p(x) = 2x³ − 3x² − 17x + 30 completely, first using the Factor Theorem. Check your answer by expansion. [5 marks]

1. Evaluate p(2) = 16 − 12 − 34 + 30 = 0. Therefore x − 2 is a factor by the Factor Theorem.
2. Dividing p(x) by x − 2 gives the quotient 2x² + x − 15 and zero remainder. Hence p(x) = (x − 2)(2x² + x − 15).
3. Split the quotient's middle term as 6x − 5x: 2x² + 6x − 5x − 15 = 2x(x + 3) − 5(x + 3).
4. Factor out the common binomial, obtaining (2x − 5)(x + 3). The complete product is therefore p(x) = (x − 2)(2x − 5)(x + 3).
5. For the check, (2x − 5)(x + 3) = 2x² + x − 15. Multiplication by x − 2 gives 2x³ − 3x² − 17x + 30, matching the dividend.

Q7. Given that x − a is a factor of x³ − ax² + 2x + a − 1, find the real number a and check the resulting condition. [3 marks]

1. Let p(x) be the given polynomial. Since x − a is a factor, the Factor Theorem requires p(a) = 0; replace every x by a.
2. This gives a³ − a × a² + 2a + a − 1 = 0. The cubic terms cancel, leaving 3a − 1 = 0, so a = 1/3.
3. Substitution into the reduced condition gives 3 × (1/3) − 1 = 0. Thus a = 1/3 satisfies the original factor condition.

Key takeaways

• A polynomial's degree is its highest variable power with a non-zero coefficient; linear, quadratic and cubic mean degrees one, two and three.
• To evaluate a polynomial, substitute the specified number everywhere, using brackets around negative values before calculating their powers.
• The remainder on division by x − k is p(k); obtain the substitution value by setting the divisor equal to zero.
• The Factor Theorem connects a zero k with the factor x − k, and requires the polynomial's value to be exactly zero.
• A known factor supplies an equation for an unknown coefficient: evaluate at the factor's zero and set the result equal to zero.
• Complete cubic factorisation continues beyond the first factor: obtain the quadratic quotient and factorise it, retaining all numerical multipliers.
• Expansion checks the full answer; leading-coefficient and constant-term checks are useful but do not verify every middle coefficient.
• A cubic has at most three zeroes, and a repeated factor can represent repeated occurrence of the same zero.

Test yourself

For p(x) = x² − 3x − 4, is 2 a zero?

No. Substitution gives p(2) = 4 − 6 − 4 = −6, which is not zero.

Which factor corresponds to the zero −1 of a polynomial?

The corresponding factor is x − (−1), which simplifies to x + 1.

What value should be substituted to find a remainder for the divisor 2x − 1?

Set 2x − 1 equal to zero and substitute x = 1/2 into the dividend.

What is the difference between a zero and a factor?

A zero is a number making the polynomial zero. Its corresponding linear factor is an expression dividing the polynomial exactly.

What remains to be done after writing x³ − 6x² + 11x − 6 = (x − 1)(x² − 5x + 6)?

Factorise the quadratic quotient as (x − 2)(x − 3), retaining x − 1 in the complete product.

What sum and product are used to split the middle term of 3x² + 5x − 2?

The required sum is 5 and the product is −6. The numbers 6 and −1 satisfy both conditions.

Why must the number 2 remain in 2x² − 8x + 6 = 2(x − 1)(x − 3)?

It preserves the polynomial's leading coefficient and values. Removing it changes the expression even though its zeroes stay the same.

Does x³ − x² have three distinct zeroes because it is cubic?

No. It equals x²(x − 1), so its distinct zeroes are 0 and 1; x is a repeated factor.