A4 - Expanding double brackets and factorising quadratics

Expanding two brackets

Is $51 \times 49$ equal to $50^2$? The numbers on either side of $50$ look balanced, but their product is slightly smaller. By the end you will explain the gap without long multiplication.

In $(x + 3)(x + 2)$, the first bracket is itself a whole factor. Multiply it by $x$, then by $2$. Equivalently, make a grid: each term from one bracket multiplies each term from the other. Two terms times two terms give four products before collecting.

[DIAGRAM: asset_name: Four areas in a product of two brackets; asset_slug: aqa8300_a04_binomial_area; caption: The four areas x squared, 3x, 2x and 6 add to x squared + 5x + 6; description: A rectangle has width x + 3 and height x + 2, split vertically into x and 3 and horizontally into x and 2. The large top-left region is pale blue and labelled x squared; the top-right and bottom-left regions are pale sage and labelled 3x and 2x; the bottom-right region is pale clay and labelled 6. Dimension labels sit outside the rectangle. The equality (x + 3)(x + 2) = x squared + 5x + 6 appears beneath. x is positive. Not drawn accurately.]

The picture uses positive lengths. The multiplication rule also works for negative values because it comes from distributing over sums, not from measuring lengths.

To expand two binomials, show all four products, then collect like terms. This method is not on the formulae sheet: learn it. A grid is equally valid working.

The next example changes only the sign of the $3$. Two of the four products now change sign.

The first terms need not have coefficient $1$. For example, $(2x + 1)(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3$. Nothing changes about making all four products.

For the next pair of brackets, the negative constant has moved into the second bracket. The four-product rule stays the same.

Finding the two brackets of a quadratic

Factorising a quadratic reverses the four-product expansion: find two numbers whose sum gives the middle coefficient and whose product gives the constant.

From the first example, $(x + 3)(x + 2) = x^2 + (3 + 2)x + 3 \times 2$. The middle coefficient comes from addition, and the final constant comes from multiplication. You must meet both conditions.

List factor pairs of the constant, then test their sum. If the constant is positive, the two numbers have the same sign. If it is negative, their signs differ. For a positive constant and negative middle coefficient, both numbers are negative: $x^2 - 5x + 6 = (x - 2)(x - 3)$, because $-2 - 3 = -5$ and $(-2)(-3) = 6$.

The next example changes only the constant from $+6$ to $-6$. The factors must now have opposite signs.

Not every quadratic has integer factors. No integer pair has product $1$ and sum $1$, so $x^2 + x + 1$ cannot be factorised into brackets with integer constants. If the constant is zero, start with the common factor instead: $x^2 + 5x = x(x + 5)$.

The order of the brackets does not matter. Their two constants must still pass both checks.

The difference of two squares

Some quadratics have no middle term because their cross-products cancel. These are differences of two squares.

Expanding $(x + 4)(x - 4)$ gives $x^2 - 4x + 4x - 16 = x^2 - 16$. The constants are opposites, so their sum is zero. That cancellation is the reason for the pattern.

Identify each square's root before writing the brackets. “Difference” means subtraction; the same pattern cannot turn $x^2 + 16$ into $(x - 4)(x + 4)$, because those brackets expand to $x^2 - 16$.

The next example changes only the first square, from $x^2$ to $(3x)^2$. Take the square root of the whole first term.

There is a fast mental-arithmetic use too. $101^2 - 100^2 = (101 - 100)(101 + 100) = 201$. Consecutive whole-number squares differ by an odd number: the difference is one times their sum. Large squares can hide a very small calculation.

Check the numerical coefficient's square root as carefully as the constant's square root.

Exact square-root factors

A square root can be kept exact inside a bracket. You do not need a decimal approximation to multiply it by itself.

For example, $(\sqrt{5})^2 = 5$ by the definition of a square root. So the difference-of-squares pattern also handles $(x + \sqrt{5})(x - \sqrt{5})$. Treat $\sqrt{5}$ as one constant, just as you treated $4$ in the previous part.

For a positive number $k$, $(\sqrt{k})^2 = k$. This square-root fact and the difference-of-squares identity are not on the formulae sheet: learn them. Here you only need a root multiplied by itself, not any rules for combining different roots.

The next example reverses only the direction: instead of expanding given brackets, find brackets for the given quadratic.

A root outside a bracket is still a factor: $\sqrt{3}(x + \sqrt{3}) = \sqrt{3}\,x + 3$. As before, every term is multiplied. Leaving a little space after the root makes it clear where the next factor starts.

The opening calculation now has a short explanation: $51 \times 49 = (50 + 1)(50 - 1) = 50^2 - 1 = 2499$. Increasing one factor by $1$ and decreasing the other by $1$ costs exactly one from the product.

Quick Recap

• Expanding two binomials produces four products before like terms are collected.
• To factorise $x^2 + bx + c$, find two numbers with sum $b$ and product $c$. Both conditions matter.
• A difference of squares factorises as $A^2 - B^2 = (A - B)(A + B)$; the cross-products cancel.
• Keep square roots exact. $(\sqrt{k})^2 = k$, so the same pattern can give exact square-root factors.

The opening product is $2499$, exactly $1$ less than $50^2$.