Factorise completely x3 - 7x + 6.

The coefficients add to 1 - 7 + 6 = 0, so x - 1 is a factor. Dividing gives x2 + x - 6 = (x + 3)(x - 2). So x3 - 7x + 6 = (x - 1)(x - 2)(x + 3).

Factorisation of Polynomials

ICSE Class 10 · Mathematics · Chapter 8

What this chapter covers

Polynomials, their degree and their value
Dividing a polynomial by a linear polynomial
The remainder theorem
Finding unknown coefficients with the remainder theorem
The factor theorem
Finding the first factor of a cubic
Factorising a polynomial completely
Problems with two unknown coefficients
Common mistakes and how to avoid them

Key points

A polynomial's degree is its highest power; this chapter works only with polynomials of degree at most 3.
Division algorithm: dividend = divisor × quotient + remainder, with the remainder of lower degree than the divisor.
Remainder theorem: the remainder when f(x) is divided by x - a is f(a); for ax + b it is f(-(b)/(a)).
Factor theorem: x - a is a factor of f(x) if and only if f(a) = 0.
To find unknown coefficients, turn each given remainder or factor into one equation and solve.
Try divisors of the constant term to find a first zero; a zero sum of coefficients means x - 1 is a factor.
Equal sums of even-power and odd-power coefficients mean x + 1 is a factor.
To factorise completely: find one factor, divide to get a quadratic, split the quadratic, and check by multiplying back.
If the quadratic quotient has a negative discriminant, it cannot be split further.

Questions answered

State the remainder theorem.

When a polynomial f(x) is divided by x - a, the remainder is f(a).

State the factor theorem.

x - a is a factor of f(x) if and only if f(a) = 0.

Why is the remainder on dividing a polynomial by a linear polynomial always a constant?

The remainder must have a lower degree than the divisor. A linear divisor has degree 1, so the remainder has degree 0, which means it is a constant.

Find the remainder when x³ + 2x² - 3x + 4 is divided by x - 1.

f(1) = 1 + 2 - 3 + 4 = 4.

Find the remainder when 2x³ - x² + 3 is divided by 2x + 1.

Put x = -(1)/(2): 2(-(1)/(8)) - (1)/(4) + 3 = -(1)/(4) - (1)/(4) + 3 = (5)/(2).

Show that x + 2 is a factor of x³ + 8.

f(-2) = -8 + 8 = 0, so by the factor theorem x + 2 is a factor.

Find k if x - 1 is a factor of 3x³ + kx² - 2x + 5.

f(1) = 3 + k - 2 + 5 = k + 6 = 0, so k = -6.

Factorise completely x³ - 7x + 6.

The coefficients add to 1 - 7 + 6 = 0, so x - 1 is a factor. Dividing gives x² + x - 6 = (x + 3)(x - 2). So x³ - 7x + 6 = (x - 1)(x - 2)(x + 3).

Past board exam questions

ICSE 2027 sample paper · 1 mark

(x + 2) and (x + 3) are the factors of x³ + 4x² + x - 6. The third factor of the given polynomial is:

(x - 1)
(x - 4)
(x + 1)
(x + 4)

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(x - 1)

ICSE 2027 sample paper · 3 marks

Using Remainder and Factor Theorem, factorise the following polynomial: 2x³ + 5x² - 28x - 15.

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Let f(x) = 2x³ + 5x² - 28x - 15. f(3) = 2(27) + 5(9) - 28(3) - 15 = 54 + 45 - 84 - 15 = 0, so (x - 3) is a factor. Dividing: f(x) = (x - 3)(2x² + 11x + 5) 2x² + 11x + 5 = 2x² + 10x + x + 5 = (x + 5)(2x + 1) f(x) = (x - 3)(x + 5)(2x + 1)

ICSE 2026 sample paper · 1 mark

(x - 2) and (x + 2) are the factors of x³ + x² - 4x - 4. The third factor of the given polynomial is:

(x - 1)
(x - 4)
(x + 1)
(x + 4)

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(x + 1)

ICSE 2026 board paper · 1 mark

What must be added to x³ + 7x² + 3x + 2 so that the result is completely divisible by (x + 2)?

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-16

ICSE 2026 board paper · 4 marks

The polynomial kx³ + 3x² - 11x - 6 when divided by (x + 1), leaves a remainder of 6. (a) Find the value of **k**. (b) Using the value of **k** factorise completely the polynomial kx³ + 3x² - 11x - 6

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(a) By the remainder theorem, f(-1) = 6: -k + 3 + 11 - 6 = 6 ⇒ 8 - k = 6 ⇒ k = 2. (b) f(x) = 2x³ + 3x² - 11x - 6. f(2) = 16 + 12 - 22 - 6 = 0, so (x - 2) is a factor. Dividing: 2x³ + 3x² - 11x - 6 = (x - 2)(2x² + 7x + 3) 2x² + 7x + 3 = 2x² + 6x + x + 3 = (2x + 1)(x + 3) ∴ 2x³ + 3x² - 11x - 6 = (x - 2)(2x + 1)(x + 3).

ICSE 2025 compartment paper · 1 mark

When polynomial x³ - 3x² - 6x + 8 is divided by (x + 2), the remainder is zero. The probability of (x + 2) to be one of the factors of the given polynomial is:

0
(1)/(3)
(2)/(3)
1

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ICSE 2025 sample paper · 1 mark

A polynomial in ‘x’ is divided by (x - a) and for (x - a) to be a factor of this polynomial, the remainder should be:

-a
0
a
2a

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ICSE 2025 compartment paper · 1 mark

The value of p for which (x - p) is a factor of x³ - px² + x + 5 is:

– 5
– 4
5
p + 5

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– 5