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Factoring polynomials: a step-by-step guide

The steps for factoring polynomials — common factor first, then two, three or four terms — with worked examples, each checked by multiplying back out.

3 منٹ کا مطالعہتازہ کاری 2 نومبر، 2026از Sumzo teamجائزہ: Sumzo team

Factoring polynomials means writing a polynomial as a product of simpler polynomials — the reverse of expanding brackets. Expanding (x+2)(x+3)(x + 2)(x + 3) gives x2+5x+6x^2 + 5x + 6; factoring x2+5x+6x^2 + 5x + 6 takes you back to (x+2)(x+3)(x + 2)(x + 3). The product is called the factored form.

Factored form is useful: it shows where a polynomial equals zero (each factor gives a root), it simplifies fractions, and it makes graphs easier to sketch. In the UK the same skill is called factorising expressions.

Factoring steps: a checklist

  1. Take out the greatest common factor (GCF) of all the terms.
  2. Count the terms that are left.
  3. Two terms: look for a difference of two squares, a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b), or a sum or difference of two cubes.
  4. Three terms (a trinomial): find two numbers that multiply to a×ca \times c and add to bb.
  5. Four terms: try grouping them in pairs.
  6. Check each factor — can it be factored again? Then multiply out to check.

Common monomial factor: examples with answers

A common monomial factor is a single term that divides every term of the polynomial. Find the GCF of the numbers and the lowest power of each letter:

  • 6x2+9x=3x(2x+3)6x^2 + 9x = 3x(2x + 3)
  • 4x3−8x2=4x2(x−2)4x^3 - 8x^2 = 4x^2(x - 2)
  • 5a2b+10ab2=5ab(a+2b)5a^2b + 10ab^2 = 5ab(a + 2b)
  • 7y−21=7(y−3)7y - 21 = 7(y - 3)

حل شدہ مثال

Easy: take out the greatest common factor

Factor 12x3+18x212x^3 + 18x^2.

حل دکھائیں
  1. The GCF of 12 and 18 is 6, and the lowest power of xx is x2x^2, so the GCF is 6x26x^2.
    12x3+18x2=6x2⋅2x+6x2⋅312x^3 + 18x^2 = 6x^2 \cdot 2x + 6x^2 \cdot 3
  2. Divide each term by 6x26x^2.
    6x2(2x+3)6x^2(2x + 3)
  3. Check by expanding.
    6x2⋅2x+6x2⋅3=12x3+18x2✓6x^2 \cdot 2x + 6x^2 \cdot 3 = 12x^3 + 18x^2 \checkmark

جواب: 6x2(2x+3)6x^2(2x + 3)

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حل شدہ مثال

Medium: a trinomial with leading coefficient 1

Factor x2+5x+6x^2 + 5x + 6.

حل دکھائیں
  1. Find two numbers that multiply to 6 and add to 5: 2 and 3.
    2×3=6,2+3=52 \times 3 = 6,\quad 2 + 3 = 5
  2. Write the factors.
    (x+3)(x+2)(x + 3)(x + 2)
  3. Check.
    (x+3)(x+2)=x2+2x+3x+6=x2+5x+6✓(x + 3)(x + 2) = x^2 + 2x + 3x + 6 = x^2 + 5x + 6 \checkmark

جواب: (x+3)(x+2)(x + 3)(x + 2)

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حل شدہ مثال

Medium: the difference of two squares

Factor 9x2−259x^2 - 25.

حل دکھائیں
  1. Both terms are perfect squares: 9x2=(3x)29x^2 = (3x)^2 and 25=5225 = 5^2.
    a=3x,b=5a = 3x,\quad b = 5
  2. Use a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b).
    (3x+5)(3x−5)(3x + 5)(3x - 5)

جواب: (3x+5)(3x−5)(3x + 5)(3x - 5)

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حل شدہ مثال

Hard: a trinomial with leading coefficient other than 1

Factor 6x2+7x−36x^2 + 7x - 3.

حل دکھائیں
  1. Multiply a×c=6×(−3)=−18a \times c = 6 \times (-3) = -18. Two numbers that multiply to −18-18 and add to 7 are 9 and −2-2.
    9×(−2)=−18,9+(−2)=79 \times (-2) = -18,\quad 9 + (-2) = 7
  2. Split the middle term and group.
    6x2+9x−2x−3=3x(2x+3)−1(2x+3)6x^2 + 9x - 2x - 3 = 3x(2x + 3) - 1(2x + 3)
  3. Take out the common bracket.
    (2x+3)(3x−1)(2x + 3)(3x - 1)
  4. Check.
    6x2−2x+9x−3=6x2+7x−3✓6x^2 - 2x + 9x - 3 = 6x^2 + 7x - 3 \checkmark

جواب: (2x+3)(3x−1)(2x + 3)(3x - 1)

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حل شدہ مثال

Hard: four terms, factored by grouping

Factor x3+2x2−9x−18x^3 + 2x^2 - 9x - 18.

حل دکھائیں
  1. Group in pairs and take out a factor from each.
    x2(x+2)−9(x+2)x^2(x + 2) - 9(x + 2)
  2. Take out the common bracket (x+2)(x + 2).
    (x+2)(x2−9)(x + 2)(x^2 - 9)
  3. x2−9x^2 - 9 is a difference of squares, so factor it too.
    (x+2)(x+3)(x−3)(x + 2)(x + 3)(x - 3)

جواب: (x+3)(x+2)(x−3)(x + 3)(x + 2)(x - 3)

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Factoring completely

Factoring completely means continuing until no factor can be broken down any further. The most common slip is stopping after the first step. In the example below, taking out 2x2x leaves x2−9x^2 - 9, which still factors.

حل شدہ مثال

Factor this polynomial completely

Factor 2x3−18x2x^3 - 18x completely.

حل دکھائیں
  1. Take out the GCF, 2x2x.
    2x(x2−9)2x(x^2 - 9)
  2. Factor the difference of squares.
    2x(x+3)(x−3)2x(x + 3)(x - 3)

جواب: 2x(x+3)(x−3)2x(x + 3)(x - 3)

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Special patterns to recognise

  • Perfect square trinomial: a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2 and a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a - b)^2. Example: x2−10x+25=(x−5)2x^2 - 10x + 25 = (x - 5)^2.
  • Difference of two squares: a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b).
  • Difference of two cubes: a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2).
  • Sum of two cubes: a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2).

حل شدہ مثال

A perfect square trinomial

Factor x2+6x+9x^2 + 6x + 9.

حل دکھائیں
  1. The first and last terms are squares (x2x^2 and 323^2), and the middle term is 2×x×32 \times x \times 3.
    x2+2(x)(3)+32x^2 + 2(x)(3) + 3^2
  2. So it is a perfect square.
    (x+3)2(x + 3)^2

جواب: (x+3)2(x + 3)^2

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حل شدہ مثال

A difference of two cubes

Factor x3−8x^3 - 8.

حل دکھائیں
  1. Both terms are cubes: x3x^3 and 8=238 = 2^3, so a=xa = x and b=2b = 2.
    a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2)
  2. Substitute.
    (x−2)(x2+2x+4)(x - 2)(x^2 + 2x + 4)
  3. The quadratic factor has no real roots, so this is fully factored.
    ✓\checkmark

جواب: (x−2)(x2+2x+4)(x - 2)(x^2 + 2x + 4)

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Why factored form is useful

Factored form turns hard equations into easy ones. To solve x2+5x+6=0x^2 + 5x + 6 = 0, factor it as (x+3)(x+2)=0(x + 3)(x + 2) = 0. A product is zero only when a factor is zero, so x=−3x = -3 or x=−2x = -2. The same idea finds where a graph crosses the x-axis and simplifies algebraic fractions such as x2−9x+3=x−3\frac{x^2 - 9}{x + 3} = x - 3 (for x≠−3x \neq -3).

Common mistakes

  • Skipping the common factor. Without it, 2x3−18x2x^3 - 18x looks much harder than it is.
  • Sign errors in the two numbers: for 6x2+7x−36x^2 + 7x - 3 the product is −18-18, so one number must be negative.
  • Stopping too early — always check whether each factor can be factored again.
  • Treating a sum of squares like a difference: x2+25x^2 + 25 does not factor as (x+5)(x−5)(x + 5)(x - 5), which expands to x2−25x^2 - 25.

اکثر پوچھے گئے سوالات

What is factored form?

A polynomial written as a product of factors, such as (x+3)(x+2)(x + 3)(x + 2) instead of x2+5x+6x^2 + 5x + 6. Setting each factor to zero gives the roots.

What are the steps for factoring polynomials?

Take out the greatest common factor, then count the terms: two terms — difference of squares; three terms — find two numbers for the trinomial; four terms — grouping. Repeat on each factor, then check by expanding.

What does factoring completely mean?

Factoring until none of the factors can be factored again. 2x3−18x=2x(x2−9)2x^3 - 18x = 2x(x^2 - 9) is not complete; 2x(x+3)(x−3)2x(x + 3)(x - 3) is.

How do I factorise expressions with a common factor?

Find the biggest number that divides every term and the lowest power of each letter that appears in every term, write them outside a bracket as one factor, and divide each term by that factor to fill the bracket: 10x2+15x=5x(2x+3)10x^2 + 15x = 5x(2x + 3).

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