Lesson illustration

Factoring Algebraic Expressions Using Common Factors and Identities

Hello. In the previous lesson, you expanded brackets and established three key identities: the squares of a sum and difference, and the product of conjugates. Factorisation reverses that process: instead of multiplying factors to produce an expression, you recognise an expression’s structure and write it as a product.

This is a high-return skill for entrance exams. A factored form often exposes cancellation, makes substitution quicker, and later lets you solve equations efficiently. In this lesson, you will learn a disciplined sequence: first extract any common factor, then look for an exact standard-identity pattern, and finally check whether further factorisation is possible.


Factorisation is reverse distribution

To factorise an expression means to rewrite it as multiplication. For example,

6x2+18x6x^2+18x

has 6x6x in both terms. Since

6x(x+3)=6x2+18x,6x(x+3)=6x^2+18x,

we can factorise it as

6x2+18x=6x(x+3)\boxed{6x^2+18x=6x(x+3)}

The number or expression placed outside the bracket is a common factor: it divides every term exactly.

A useful way to avoid mistakes is to divide each original term by the factor you are taking out:

12x320x212x^3-20x^2

The greatest common factor of the coefficients 1212 and 2020 is 44. Both terms also contain x2x^2, so the greatest common factor is 4x24x^2.

12x3÷4x2=3x12x^3 \div 4x^2=3x 20x2÷4x2=5-20x^2 \div 4x^2=-5

Therefore,

12x320x2=4x2(3x5)\boxed{12x^3-20x^2=4x^2(3x-5)}

The signs inside the bracket come directly from division. Never change a sign merely because it “looks more natural.”


Finding the greatest common factor

For a polynomial with several terms, construct the GCF from two parts:

  1. The greatest common numerical divisor of all coefficients.
  2. Every variable common to all terms, taken to its lowest exponent.

Consider:

18x3y224x2y+6xy318x^3y^2-24x^2y+6xy^3

The coefficient GCF is 66. For the variables:

  • the lowest power of xx is x1x^1;
  • the lowest power of yy is y1y^1.

So the GCF is 6xy6xy:

18x3y224x2y+6xy318x^3y^2-24x^2y+6xy^3 =6xy(3x2y4x+y2)=6xy(3x^2y-4x+y^2)

Check it by multiplying back:

6xy(3x2y4x+y2)6xy(3x^2y-4x+y^2) =18x3y224x2y+6xy3=18x^3y^2-24x^2y+6xy^3

A variable missing from even one term cannot belong to the common factor. For example, in

8x2+12x+4,8x^2+12x+4,

the GCF is 44, not 4x4x, because the final term has no xx.

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A repeated bracket can also be a common factor

The common factor need not be a number or a single variable. Sometimes an entire bracket repeats:

x(3x+1)5(3x+1)x(3x+1)-5(3x+1)

Both terms contain 3x+13x+1. Treat that bracket as one object:

x(3x+1)5(3x+1)=(3x+1)(x5)\boxed{x(3x+1)-5(3x+1)=(3x+1)(x-5)}

This is still reverse distribution:

(3x+1)(x5)=x(3x+1)5(3x+1)(3x+1)(x-5)=x(3x+1)-5(3x+1)

The key visual habit is to scan for a repeated complete bracket before doing any expansion.

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Factor out the GCF first

A standard exam rule is:

Extract the GCF before attempting an identity.\boxed{\text{Extract the GCF before attempting an identity.}}

Why? A hidden common factor can conceal an identity.

For example:

8x2+24x+188x^2+24x+18

All terms are divisible by 22:

8x2+24x+18=2(4x2+12x+9)8x^2+24x+18=2(4x^2+12x+9)

Now the bracket is a perfect-square trinomial:

4x2=(2x)24x^2=(2x)^2 9=329=3^2 12x=2(2x)(3)12x=2(2x)(3)

Therefore,

8x2+24x+18=2(2x+3)2\boxed{8x^2+24x+18=2(2x+3)^2}

Stopping at 2(4x2+12x+9)2(4x^2+12x+9) is partially correct, but not fully factorised.

Sometimes extracting a negative factor makes the remaining structure clearer:

8x2+18-8x^2+18 =2(4x29)=-2(4x^2-9)

The bracket is now a difference of squares:

4x29=(2x)2324x^2-9=(2x)^2-3^2

So,

8x2+18=2(2x+3)(2x3)\boxed{-8x^2+18=-2(2x+3)(2x-3)}

Taking out a negative GCF is not compulsory, but it is often the cleanest choice when the leading term is negative.


Perfect-square trinomials

From the previous lesson:

(a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 (ab)2=a22ab+b2(a-b)^2=a^2-2ab+b^2

Reversing these identities gives the factorisation forms:

a2+2ab+b2=(a+b)2\boxed{a^2+2ab+b^2=(a+b)^2} a22ab+b2=(ab)2\boxed{a^2-2ab+b^2=(a-b)^2}

A perfect-square trinomial must meet all three conditions:

  • the first term is a perfect square;
  • the last term is a perfect square;
  • the middle term is exactly twice the product of their square roots.

Consider:

25p230pq+9q225p^2-30pq+9q^2

The first and last terms are squares:

25p2=(5p)225p^2=(5p)^2 9q2=(3q)29q^2=(3q)^2

Now test the middle term:

2(5p)(3q)=30pq-2(5p)(3q)=-30pq

It matches exactly, so:

25p230pq+9q2=(5p3q)2\boxed{25p^2-30pq+9q^2=(5p-3q)^2}

The sign of the middle term decides the sign inside the bracket:

Expression patternFactorised form
a2+2ab+b2a^2+2ab+b^2(a+b)2(a+b)^2
a22ab+b2a^2-2ab+b^2(ab)2(a-b)^2

Do not factor an expression as a perfect square merely because its first and last terms are squares. For instance,

x2+6x+16x^2+6x+16

has square end terms, but the relevant middle term for (x+4)2(x+4)^2 would be

2(x)(4)=8x,2(x)(4)=8x,

not 6x6x. So this is not a perfect-square trinomial.

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Difference of two squares

The third essential identity from the previous lesson was:

(a+b)(ab)=a2b2(a+b)(a-b)=a^2-b^2

Reversing it gives:

a2b2=(a+b)(ab)\boxed{a^2-b^2=(a+b)(a-b)}

Use this identity only when all of the following are true:

  1. There are exactly two main terms.
  2. The terms are separated by subtraction.
  3. Both terms are perfect squares.

For example:

9y425x29y^4-25x^2

can be written as

(3y2)2(5x)2(3y^2)^2-(5x)^2

Therefore,

9y425x2=(3y2+5x)(3y25x)\boxed{9y^4-25x^2=(3y^2+5x)(3y^2-5x)}

The two factors are conjugates: the terms are the same, but the internal signs differ.

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The image also explains why this identity is geometrically sound: the leftover L-shaped region from the larger square can be rearranged into a rectangle. Its area has not changed, only its shape has.

Be alert to what the identity does not say:

a2+b2a^2+b^2

is a sum of squares, not a difference of squares. It does not factor using this identity over integers.


Combining methods: factor completely

Entrance-exam expressions are often designed in layers. After extracting a GCF, inspect the remaining bracket again.

Consider:

12x33x12x^3-3x

First take out the GCF 3x3x:

12x33x=3x(4x21)12x^3-3x=3x(4x^2-1)

The bracket is a difference of squares:

4x21=(2x)2124x^2-1=(2x)^2-1^2

Factor it again:

12x33x=3x(2x+1)(2x1)\boxed{12x^3-3x=3x(2x+1)(2x-1)}

This is the fully factorised form.

The following recognition sequence is reliable under time pressure:

  1. Count terms and scan for a GCF. Look for common numbers, variables, or complete brackets.
  2. Extract the GCF. Divide every term carefully.
  3. Inspect what remains. Check for a perfect-square trinomial or a difference of squares.
  4. Repeat if necessary. A factored bracket may itself factor further.
  5. Verify by expansion. This is especially useful when signs or powers are involved.

For the final check of the previous example:

(2x+1)(2x1)=4x21(2x+1)(2x-1)=4x^2-1

and then

3x(4x21)=12x33x3x(4x^2-1)=12x^3-3x

The original expression is recovered.

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Key takeaways

Factorisation is reverse expansion: you rewrite a sum or difference as a product.

Start with the GCF. For monomials, it consists of the numerical GCF and the lowest shared exponent of each common variable. A complete bracket can also be a common factor.

The three core identity patterns are:

a2+2ab+b2=(a+b)2a^2+2ab+b^2=(a+b)^2 a22ab+b2=(ab)2a^2-2ab+b^2=(a-b)^2 a2b2=(a+b)(ab)a^2-b^2=(a+b)(a-b)

The most important habit is structural recognition: do not force an identity because an expression merely resembles one. Confirm the exact middle term in a perfect-square trinomial, and confirm two perfect squares separated by subtraction for a difference of squares.

Next, you will use algebraic manipulation in a more operational setting: solving one-variable linear equations with integer and fractional coefficients.

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