Class 9 Maths Chapter 4: Exploring Algebraic Identities — Notes, Examples & Practice

An equation such as $2x + 1 = 7$ is true only for some values ($x = 3$). An identity such as $(a + b)^2 = a^2 + 2ab + b^2$ is true for all values. You can test a claimed identity with numbers, but one success does not prove it — proving needs algebra or a general picture.

Area picture: a square of side $a + b$ splits into a square $a \times a$, a square $b \times b$ and two rectangles $a \times b$. So $(a + b)^2 = a^2 + 2ab + b^2$. The middle term $2ab$ is the two rectangles — the part people forget.

Consecutive squares differ by odd numbers: $(n + 1)^2 - n^2 = 2n + 1$. For example $13^2 - 12^2 = 25$.

A square of side $a + b + c$ splits into three squares and six rectangles (two each of $ab$, $bc$, $ca$), which gives $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.

The difference of squares $a^2 - b^2 = (a + b)(a - b)$: cut a $b \times b$ square from the corner of an $a \times a$ square; the L-shaped leftover can be rearranged into a rectangle $(a + b)$ by $(a - b)$.

Spot it even when disguised: $49x^2 - 1 = (7x)^2 - 1^2$; $x^4 - 16 = (x^2)^2 - 4^2$; and $103 \times 97 = (100 + 3)(100 - 3)$.

Factorising reverses expanding: write an expression as a product. Always start by taking out any common factor, then look for a pattern.

Three terms with two perfect squares and a middle term equal to twice their product → a perfect square: $9x^2 + 30x + 25 = (3x)^2 + 2(3x)(5) + 5^2 = (3x + 5)^2$. Two squares with a minus between them → difference of squares.

Algebra tiles make this visible: $x^2$ tiles, $x$ strips and unit squares can be arranged into a rectangle exactly when the expression factorises, and the sides of the rectangle are the factors.

To factorise $ax^2 + bx + c$, find two numbers whose **product is $ac$ and whose sum is $b$**. Use them to split $bx$ into two terms, then group in pairs.

Why it works: $(px + q)(rx + s) = prx^2 + (ps + qr)x + qs$. The two pieces $ps$ and $qr$ add to $b$ and multiply to $(pr)(qs) = ac$.

Signs guide the search: if $ac > 0$ both numbers have the sign of $b$; if $ac < 0$ they have opposite signs.

Multiplying $(a + b)^2$ by $(a + b)$ once more gives $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, which can be written compactly as $a^3 + b^3 + 3ab(a + b)$. Picture a cube of side $a + b$ cut into 8 blocks: one $a^3$, one $b^3$, three $a^2b$ slabs and three $ab^2$ rods.

Rearranging gives the sum and difference of cubes, very useful for factorising and for simplifying fractions.

A special result: if $a + b + c = 0$ then $a^3 + b^3 + c^3 = 3abc$. It comes from the identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.

A rational expression is a fraction of polynomials, like $\dfrac{x^2 - 9}{x^2 + 5x + 6}$. To simplify, factorise top and bottom fully, then cancel factors that are common to the whole numerator and the whole denominator.

You may cancel only factors (things multiplied), never terms (things added). In $\dfrac{x + 3}{x + 5}$ nothing cancels.

Note the values that make the denominator zero — the expression is not defined there, even after cancelling.