Quadratics 1: Expanding, Factorising, and Solving in Standard Form

What is a Quadratic Equation?

Simply put, a quadratic equation is one where $x^2$ is the highest power of $x$. Equations where the highest power is just $x$ (i.e. $x^1$) are linear. Equations where the highest power is $x^3$ are cubic. Equations with even higher powers are all forms of polynomial expressions — a quadratic is a polynomial, but not all polynomials are quadratic.

Standard Form

The standard form of a quadratic equation is:

$$ax^2 + bx + c = 0$$

Where $a \neq 0$ — if $a$ were zero, the $ax^2$ term would vanish and we would just have $bx + c = 0$, which is a linear equation, not a quadratic.

In standard form:

$a$ is the coefficient of $x^2$
$b$ is the coefficient of $x$
$c$ is the constant

For a concrete example, take:

$$x^2 + 5x + 6 = 0$$

Here $a = 1$, $b = 5$, and $c = 6$.

Remember: Just because you can't see a number doesn't mean it isn't there. All three of these are still quadratic equations:

$x^2 - 12 = 0$
$x^2 + 5x = 0$
$x^2 = 0$

Writing in the hidden zero terms to match standard form more closely:

$x^2 + 0x - 12 = 0$
$x^2 + 5x + 0 = 0$
$x^2 + 0x + 0 = 0$

Rearranging into Standard Form

A quadratic equation isn't always handed to you in standard form — you may need to rearrange it first. For example:

$$x^2 + 5x = 6$$

Whatever we do to one side of an equation we must do to the other to keep it balanced. Subtracting 6 from both sides gives:

$$x^2 + 5x - 6 = 0$$

Now it is in standard form and ready to work with.

Expanding

Quadratic expressions can appear in different forms — one common form is two sets of brackets multiplied together, such as:

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

To expand this, every term in the first bracket must be multiplied by every term in the second bracket. A handy mnemonic for two-term brackets is FOIL:

First — the first terms of each bracket: $x \cdot x = x^2$
Outside — the outermost terms of the whole expression: $x \cdot 3 = 3x$
Inside — the innermost terms of the whole expression: $2 \cdot x = 2x$
Last — the last terms of each bracket: $2 \cdot 3 = 6$

Note: FOIL only works for expressions that are two sets of brackets with two terms each. With more brackets the approach is the same in principle — each term in each bracket must be multiplied by each term in every other bracket — but you'll need to be more systematic.

Putting the FOIL results together:

$$x^2 + 3x + 2x + 6$$

The two middle terms both contain $x$ to the same power, so we can collect them:

$$x^2 + 5x + 6$$

Factorising

Factorising is the reverse of expanding. We start with a standard-form expression and want to write it as a product of two brackets. This makes solving the equation much easier (covered in the next section).

The process in six steps:

Multiply $a$ and $c$
Find two numbers that add to $b$ and multiply to $a \times c$
Split the middle term using those two numbers
Group the four terms into pairs
Factorise each pair
Factor out the common bracket

Example 1 — coefficient of $x^2$ is 1

$$x^2 + 13x + 40$$

Here $a = 1$, so $a \times c = 1 \times 40 = 40$. We need two numbers that add to $13$ and multiply to $40$. That's 5 and 8.

Split the middle term:

$$x^2 + 8x + 5x + 40$$

Group into pairs:

$$(x^2 + 8x) + (5x + 40)$$

Factorise each group — the first pair shares $x$, the second shares $5$:

$$x(x + 8) + 5(x + 8)$$

Both terms now contain $(x + 8)$, so factor that out:

$$(x + 5)(x + 8)$$

Example 2 — coefficient of $x^2$ is not 1

$$3x^2 + 11x + 6$$

Here $a = 3$, so $a \times c = 3 \times 6 = 18$. We need two numbers that add to $11$ and multiply to $18$. That's 9 and 2.

Split the middle term:

$$3x^2 + 9x + 2x + 6$$

Group into pairs:

$$(3x^2 + 9x) + (2x + 6)$$

Factorise each group — the first shares $3x$, the second shares $2$:

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

Remember: If the brackets don't match at this point, check your signs and your common factors. Sometimes you need to take out a negative to make them match.

Both terms contain $(x + 3)$, so factor that out:

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

Solving

Recall that the standard form is $ax^2 + bx + c = 0$, so our factorised equation is:

$$(3x + 2)(x + 3) = 0$$

For this product to equal zero, at least one of the two brackets must be zero. We can therefore solve each bracket independently:

$$3x + 2 = 0 \implies 3x = -2 \implies x = -\tfrac{2}{3}$$ $$x + 3 = 0 \implies x = -3$$

The solution is $x = -\dfrac{2}{3}$ or $x = -3$.