E2.2 - Special factorisations (Extended)

E2.2 - Special factorisations (Extended)

Why can x2−25x^2-25 be factorised immediately, while x2+25x^2+25 does not use the same pattern? This Extended E2.2 lesson recognises cancelling cross-products, repeated brackets and shared brackets hidden in four-term expressions.

Difference of squares

Expand (x−5)(x+5)(x-5)(x+5): the cross-products are +5x+5x and −5x-5x, so they cancel. What remains is x2−25x^2-25. The subtraction between the squares is essential.

Difference of two squares

A2−B2=(A−B)(A+B).A^2-B^2=(A-B)(A+B).

This identity holds for all real AA and BB. Each of AA and BB may itself be a term such as 3x3x or 2y2y.

Not on the supplied formula lists: learn it.

Identify the two whole terms being squared before writing the brackets. A plus between the squares does not give cancelling cross-products: (A−B)(A+B)(A-B)(A+B) always produces −B2-B^2.

Worked example 1: numerical and variable squares

Question. Factorise 9x2−169x^2-16.

Plan. Recognise 9x2=(3x)29x^2=(3x)^2 and 16=4216=4^2.

9x2−16=(3x)2−42=(3x−4)(3x+4)difference-of-squares pattern\begin{aligned} 9x^2-16&=(3x)^2-4^2\\ &=(3x-4)(3x+4) &&\text{difference-of-squares pattern} \end{aligned}

Answer. (3x−4)(3x+4)(3x-4)(3x+4).

Check. Expansion gives 9x2+12x−12x−16=9x2−169x^2+12x-12x-16=9x^2-16.

Replace the numerical square by a squared variable term. The pattern is unchanged, but the whole term 4y4y must appear in each bracket.

Worked example 2: two variable terms

Question. Factorise 9x2−16y29x^2-16y^2.

Plan. Use A=3xA=3x and B=4yB=4y.

9x2−16y2=(3x)2−(4y)2=(3x−4y)(3x+4y)\begin{aligned} 9x^2-16y^2&=(3x)^2-(4y)^2\\ &=(3x-4y)(3x+4y) \end{aligned}

Answer. (3x−4y)(3x+4y)(3x-4y)(3x+4y).

Check. The two mixed products are +12xy+12xy and −12xy-12xy. They cancel, leaving the required two squares.

A common factor may come first. For instance, 2x2−50=2(x2−25)=2(x−5)(x+5)2x^2-50=2(x^2-25)=2(x-5)(x+5). Keep inspecting the brackets until factorisation is complete.

Watch out

9x2+16y2≠(3x−4y)(3x+4y)9x^2+16y^2\ne(3x-4y)(3x+4y): the right side has a minus. To write 9x29x^2 as a square, use (3x)2(3x)^2, not (9x)2(9x)^2. This identifies a term whose square is 9x29x^2; it makes no assumption about the sign of xx.

Extract the common factor before applying the pattern.

Perfect squares

A squared bracket behaves differently from a difference of squares. In (A+B)(A+B)(A+B)(A+B), the cross-products are both ABAB, so they add to 2AB2AB. A three-term expression is a perfect square only if its middle term fits as well as its first and last terms.

Perfect-square factorisations

A2+2AB+B2=(A+B)2A2−2AB+B2=(A−B)2.\begin{aligned} A^2+2AB+B^2&=(A+B)^2\\ A^2-2AB+B^2&=(A-B)^2. \end{aligned}

These identities hold for all real AA and BB. In both, the last squared term is positive.

Not on the supplied formula lists: learn them.

For example, x2+6x+9x^2+6x+9 is a perfect square because 6x=2(x)(3)6x=2(x)(3). The expression x2+5x+9x^2+5x+9 has the same first and last terms, but the wrong middle term for this pattern.

Worked example 3: a squared sum

Question. Factorise 4a2+12ab+9b24a^2+12ab+9b^2.

Plan. Identify the squared terms, then check twice their product against the middle term.

4a2=(2a)29b2=(3b)22(2a)(3b)=12abmiddle term fits4a2+12ab+9b2=(2a+3b)2\begin{aligned} 4a^2&=(2a)^2\\ 9b^2&=(3b)^2\\ 2(2a)(3b)&=12ab &&\text{middle term fits}\\ 4a^2+12ab+9b^2&=(2a+3b)^2 \end{aligned}

Answer. (2a+3b)2(2a+3b)^2.

Check. (2a+3b)(2a+3b)=4a2+6ab+6ab+9b2(2a+3b)(2a+3b)=4a^2+6ab+6ab+9b^2, the original expression.

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