A4, A18 - Quadratic expansion, factorisation and roots

Expanding two binomials

A rectangle has width $x$ centimetres, length $x+1$ centimetres and area $12\,\mathrm{cm}^2$. How can its area tell you its dimensions?

Its area equation is $x(x+1)=12$. We will return to that problem after learning how products and quadratics fit together. Begin with an easier product you can see as an area.

When both factors contain two terms, each term in one bracket must multiply both terms in the other.

For positive $x$, split a rectangle of sides $x+3$ and $x+2$ into four smaller rectangles. Its area is the product of the outside lengths and also the sum of the four inside areas.

[DIAGRAM: asset_name: Four products in a quadratic expansion; asset_slug: 1ma1_m14_area_expansion; caption: For positive x, the rectangle's four areas add to the product of its two outside lengths.; description: A schematic rectangle has width x plus 3 and height x plus 2, split into four rectangles with areas x squared, 3x, 2x and 6. Blue denotes x squared, clay denotes the two linear areas and sage denotes the constant area.]

The four pieces explain why you must keep all four products before collecting the middle terms.

Change only the sign of $3$. The area picture no longer represents all real inputs, but the distributive law still works.

The same method gives $(x-4)(x+3)=x^2-x-12$ and $(2x+1)(x-3)=2x^2-5x-3$. The coefficient on the squared term comes from multiplying the two letter terms.

Change only the positive constant in Worked example 2.

Factorising quadratics with leading coefficient one

Factorising a quadratic reverses the multiplication of two brackets.

The numbers in the brackets produce the constant by multiplication and the middle coefficient by addition. For instance, $(x+3)(x+4)=x^2+7x+12$, because $3+4=7$ and $3\times4=12$.

You need both conditions. A factor pair that gets only the constant right is not enough.

Change only the sign of the middle coefficient. The product remains positive, so both constants must now be negative.

A negative constant needs opposite signs. For $x^2+x-12$, use $4$ and $-3$, giving $(x+4)(x-3)$. For $x^2-5x-14$, use $-7$ and $2$.

Some expressions have no integer pair that works. In $x^2+x+1$, the integer pairs giving product $1$ have sums $2$ or $-2$, neither $1$. Do not force an integer factorisation by changing a coefficient.

Change only the middle coefficient in Worked example 4.

The difference of two squares

Opposite bracket signs can make the two middle products cancel completely.

Expand $(a-b)(a+b)$: the products $+ab$ and $-ab$ cancel, leaving $a^2-b^2$. This gives a useful factorisation pattern.

The word “difference” matters: it means subtraction between the squares.

Change only the coefficient of $x^2$ from $1$ to $4$. The squared quantity is now $2x$.

A common factor may come first: $3x^2-27=3(x^2-9)=3(x-3)(x+3)$, and $2x^2-32=2(x-4)(x+4)$. Keep the outside factor. The same pattern works in arithmetic: $103^2-97^2=(103-97)(103+97)=6\times200=1200$. Two large squares can have a surprisingly small difference.

Change only the squared-term coefficient in Worked example 6.

Solving equations with the zero-product principle

A product equals zero only when at least one factor equals zero.

If $AB=0$ and $A\ne0$, dividing by $A$ gives $B=0$. If $A=0$, the first factor already explains the zero product. These are the only possibilities.

For example, $x^2-5x+6=0$ becomes $(x-2)(x-3)=0$, so $x=2$ or $x=3$. Each root makes one bracket zero.

Change only the arrangement of the equation: put the constant on the other side. You must restore a zero before applying the principle.

In $x^2-4x=0$, factorising gives $x(x-4)=0$, with roots $0$ and $4$. Dividing by $x$ would lose the zero root. Identical factors, as in $(x-2)^2=0$, give one distinct root $2$, called a repeated root.

Return to the rectangle: $x(x+1)=12$ gives $x^2+x-12=(x+4)(x-3)=0$. The algebraic roots are $-4$ and $3$, but a length must be positive. Its width is $3\,\mathrm{cm}$, its length is $4\,\mathrm{cm}$, and $3\times4=12$ checks the area.

Change only the right-hand constant from Worked example 8.

Quick Recap

• Two binomials produce four products before like terms are collected.
• For a monic quadratic, bracket constants have the required sum and product.
• A difference of squares has opposite-sign factors; a sum of squares does not use that pattern.
• To solve by factorising, make one side zero, factorise and set each factor to zero. Check roots against the context.

The opening rectangle has width $3\,\mathrm{cm}$ and length $4\,\mathrm{cm}$; the negative algebraic root cannot be a length.