- Maths
- Classes 9-10
- 5 min read
Polynomials and factorisation
Learn degree and zeroes of a polynomial, use the identities for squares, and factorise quadratics by splitting the middle term.
Let us begin
The big idea
A polynomial is an expression made from a variable, numbers, plus signs and whole-number powers, like 3x² + 5x − 2. Factorising means writing it as a product of simpler pieces, which makes solving equations easy.
The degree of a polynomial is its highest power. 4x³ − 2x + 7 has degree 3. A degree-1 polynomial is linear, degree 2 is quadratic and degree 3 is cubic. The numbers in front of the powers are the coefficients.
A zero of a polynomial p(x) is a value of x that makes p(x) equal to 0. For p(x) = x² − 3x + 2, p(1) = 1 − 3 + 2 = 0, so 1 is a zero. The factor theorem says: if p(a) = 0 then (x − a) is a factor of p(x).
Three identities you should know by heart: (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and (a + b)(a − b) = a² − b². They work both ways: to expand and to factorise.
To factorise x² + bx + c, find two numbers whose product is c and whose sum is b, then split the middle term and group. For ax² + bx + c the two numbers must multiply to a × c.
Worked examples
See how it works
Split the middle term: x² + 5x + 6
We need two numbers with product 6 and sum 5: they are 2 and 3. Write x² + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 2)(x + 3). Check by expanding: x² + 3x + 2x + 6 = x² + 5x + 6.
Factorise 6x² + 11x + 3
Product a × c = 18, sum 11: the numbers are 9 and 2. So 6x² + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (3x + 1)(2x + 3).
Use the factor theorem on x³ − 6x² + 11x − 6
Try x = 1: 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor. Dividing gives x² − 5x + 6, which factorises as (x − 2)(x − 3). So the cubic is (x − 1)(x − 2)(x − 3).
A quick square with an identity
98² = (100 − 2)² = 100² − 2 × 100 × 2 + 2² = 10000 − 400 + 4 = 9604. No long multiplication needed.
Try it at home
Cut-out proof of (a + b)²
You will need
- A sheet of paper
- A pencil and ruler
- Scissors
What to do
- Draw a square of side 10 cm. Mark 7 cm along the top edge and 7 cm down the left edge.
- Draw a vertical and a horizontal line through those marks. The big square is now cut into four pieces.
- Label the pieces: a 7 by 7 square (a²), two 7 by 3 rectangles (ab each) and a 3 by 3 square (b²).
- Cut them out and add up the areas: 49 + 21 + 21 + 9 = 100, which is the area of the whole 10 by 10 square.
- That is the identity (a + b)² = a² + 2ab + b² with a = 7 and b = 3.
Ask a grown-up: Please take care with the scissors.
Practice sheet
Check what you learned
Practice sheet: Polynomials and factorisation (Classes 9-10, Maths)
- What is the degree of 4x³ − 2x + 7?
- Is 1 a zero of p(x) = x² − 3x + 2? Is 2?
- Factorise x² + 7x + 12.
- Expand (x + 5)².
- Factorise 49 − y².
- Is (x − 1) a factor of x³ − 1?
- Find the zeroes of x² − 9.
- Expand (2x − 3)(x + 4).
- Use an identity to find 103².
- Factorise x² − 5x + 6.
Answer key
Answers: Polynomials and factorisation
- 3, because the highest power is x³.
- Yes to both: p(1) = 1 − 3 + 2 = 0 and p(2) = 4 − 6 + 2 = 0.
- (x + 3)(x + 4), since 3 × 4 = 12 and 3 + 4 = 7.
- x² + 10x + 25.
- (7 − y)(7 + y), using a² − b² = (a − b)(a + b).
- Yes, because putting x = 1 gives 1 − 1 = 0.
- 3 and −3, because x² − 9 = (x − 3)(x + 3).
- 2x² + 8x − 3x − 12 = 2x² + 5x − 12.
- (100 + 3)² = 10000 + 600 + 9 = 10609.
- (x − 2)(x − 3), since (−2) × (−3) = 6 and (−2) + (−3) = −5.