Year 9 Algebraic techniques: revision mind map

1Expanding binomial products

Multiply every term in the first bracket by every term in the second.

$(x + 3)(x + 5) = x^2 + 8x + 15$

$(2x - 1)(x + 4) = 2x^2 + 7x - 4$

2Perfect squares and DOTS

$(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)(a + b) = a^2 - b^2$

$(x + 4)^2 = x^2 + 8x + 16$

$(x - 5)(x + 5) = x^2 - 25$

Watch out: $(x + 3)^2 \ne x^2 + 9$: don't forget the middle term.

3Factorising expressions

Take out the highest common factor (HCF).

$6x + 9 = 3(2x + 3)$   $x^2 - 7x = x(x - 7)$

$-4a - 8 = -4(a + 2)$

Check by expanding: you should get back to the start.

4Factorising DOTS

Difference of two squares: $a^2 - b^2 = (a - b)(a + b)$

$x^2 - 49 = (x - 7)(x + 7)$

$4x^2 - 9 = (2x - 3)(2x + 3)$

$3x^2 - 12 = 3(x - 2)(x + 2)$

5Factorising by grouping

Group into pairs, take out a factor from each, then the common bracket.

$x^2 + 3x + 2x + 6 = x(x + 3) + 2(x + 3)$

$= (x + 3)(x + 2)$

6Monic quadratic trinomials

$x^2 + bx + c$: find two numbers that multiply to $c$ and add to $b$.

$x^2 + 7x + 12 = (x + 3)(x + 4)$

$x^2 - x - 6 = (x - 3)(x + 2)$

7Trinomials $ax^2 + bx + c$

Find two numbers that multiply to $a \times c$ and add to $b$. Split the middle term, then group.

$2x^2 + 7x + 3$: $2 \times 3 = 6$, use 6 and 1

$2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3)$

$= (2x + 1)(x + 3)$

8Algebraic fractions: × and ÷

Factorise, cancel common factors, then multiply. To divide, flip the second fraction.

$\dfrac{x + 2}{6} \times \dfrac{9}{x + 2} = \dfrac{9}{6} = \dfrac32$

$\dfrac{4x}{5} \div \dfrac{2x}{15} = \dfrac{4x}{5} \times \dfrac{15}{2x} = 6$

9Algebraic fractions: + and −

Use a common denominator, then add or subtract the numerators.

$\dfrac{x}{3} + \dfrac{x}{4} = \dfrac{4x}{12} + \dfrac{3x}{12} = \dfrac{7x}{12}$

$\dfrac{2}{x} - \dfrac{1}{2x} = \dfrac{4}{2x} - \dfrac{1}{2x} = \dfrac{3}{2x}$

10Further simplification

Factorise first, then cancel whole factors.

$\dfrac{x^2 - 9}{x + 3} = x - 3$

$\dfrac{x^2 + 5x + 6}{x + 2} = x + 3$

Watch out: cancel factors, never single terms: $\dfrac{x + 3}{3} \ne x$

11Equations with algebraic fractions

Multiply both sides by the lowest common denominator to clear the fractions.

$\dfrac{x + 1}{3} = \dfrac{x - 2}{2} \Rightarrow 2(x + 1) = 3(x - 2) \Rightarrow x = 8$

$\dfrac{x}{2} + \dfrac{x}{3} = 10$: multiply by 6, $3x + 2x = 60$, so $x = 12$