PYQ Vault
Ebooks

SPI Mathematics · Formula sheet

Algebraic Identities and Simplification formulas

7 formulas and 12 common traps for SPI Mathematics Algebraic Identities and Simplification, grouped by subtopic.

Full notes with worked examples

Factorising with Common Factors and Identities

Learn this subtopic in the notes

Taking out a common factor and factorising by grouping

Common factor

ab+ac=a(b+c)ab + ac = a(b + c)

The identity for (x + a)(x + b)

Product of two binomials

(x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab

Perfect square identities

Perfect squares

(a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2(a + b)^2 = a^2 + 2ab + b^2 \qquad (a - b)^2 = a^2 - 2ab + b^2

Difference of two squares

Difference of squares

a2−b2=(a+b)(a−b)x2+1x2=(x−1x)2+2a^2 - b^2 = (a + b)(a - b) \qquad x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2

Common traps

Giving the common bracket alone

x(p+q)+y(p+q)x(p + q) + y(p + q) is (x+y)(p+q)(x + y)(p + q). The shared bracket is one factor and what is left is the other; neither one alone is the factorisation.

Pairing terms that share nothing

In ab+cd+ad+bcab + cd + ad + bc, pair ab with ad and cd with bc: a(b+d)+c(d+b)=(a+c)(b+d)a(b + d) + c(d + b) = (a + c)(b + d). Taking the terms in the printed order gives pairs with no common factor.

Getting the middle sign wrong

(x+4)(x−7)(x + 4)(x - 7) has middle term (4−7)x=−3x(4 - 7)x = -3x and constant −28-28. Add a and b with their signs, then multiply them with their signs.

Dropping the common factor

50x2−20x+2=2(5x−1)250x^2 - 20x + 2 = 2(5x - 1)^2. Leaving out the 2 gives (5x−1)2(5x - 1)^2, which is only 25x2−10x+125x^2 - 10x + 1.

Forgetting the middle term

(a+b)2(a + b)^2 is not a2+b2a^2 + b^2. (x+5)2=x2+10x+25(x + 5)^2 = x^2 + 10x + 25: the 10x10x is twice the product.

Factorising a sum of squares

a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b), but a2+b2a^2 + b^2 has no such factors. x2+9x^2 + 9 cannot be written as (x+3)(x−3)(x + 3)(x - 3), which is x2−9x^2 - 9.

Calling a hidden difference 'not possible'

x4+4x^4 + 4 looks like a sum, but x4+4=(x2+2)2−4x2=(x2+2x+2)(x2−2x+2)x^4 + 4 = (x^2 + 2)^2 - 4x^2 = (x^2 + 2x + 2)(x^2 - 2x + 2). Try completing the square before ruling out factors.

Finding Values with Identities

Learn this subtopic in the notes

Squares and cubes from the sum and the product

Sum and product identities

x2+y2=(x+y)2−2xyx3+y3=(x+y)3−3xy(x+y)x3−y3=(x−y)3+3xy(x−y)x^2 + y^2 = (x + y)^2 - 2xy \qquad x^3 + y^3 = (x + y)^3 - 3xy(x + y) \qquad x^3 - y^3 = (x - y)^3 + 3xy(x - y)

Three-term identities and the condition a + b + c = 0

Three-term identities

(a+b+c)2=a2+b2+c2+2(ab+bc+ca)a+b+c=0⇒a3+b3+c3=3abc(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \qquad a + b + c = 0 \Rightarrow a^3 + b^3 + c^3 = 3abc

Simplifying a chain of brackets that cancel

One bracket as one fraction

1+1n=n+1n1−1n=n−1n1 + \frac{1}{n} = \frac{n + 1}{n} \qquad 1 - \frac{1}{n} = \frac{n - 1}{n}

Common traps

Squaring the sum and stopping

(x+y)2(x + y)^2 is x2+y2+2xyx^2 + y^2 + 2xy, not x2+y2x^2 + y^2. Subtract 2xy2xy: with x+y=6x + y = 6 and xy=5xy = 5, x2+y2=36−10=26x^2 + y^2 = 36 - 10 = 26.

Leaving out the bracket in the cube identity

x3+y3=(x+y)3−3xy(x+y)x^3 + y^3 = (x + y)^3 - 3xy(x + y). The term taken away is 3xy3xy times the sum, not just 3xy3xy.

Forgetting the 2 in the square of three terms

(a+b+c)2=a2+b2+c2+2(ab+bc+ca)(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca). Without the 2, a2+b2+c2a^2 + b^2 + c^2 comes out wrong by ab+bc+caab + bc + ca.

Dropping the signs of the three terms

In p3−27q3−1p^3 - 27q^3 - 1 the three numbers being cubed are p, −3q-3q and −1-1, not p, 3q and 1. Their sum and their product must use those signs.

Multiplying the brackets out

Expanding (1+13)(1+14)\left(1 + \frac{1}{3}\right)\left(1 + \frac{1}{4}\right) term by term is slow and error-prone. As fractions it is 43×54=53\frac{4}{3} \times \frac{5}{4} = \frac{5}{3} at once.

More SPI Mathematics formula sheets