- A quadratic expression is one of the most important algebraic patterns you will meet in Standard Mathematics because it appears in models of area, motion, profit, and many other contexts.
- To work confidently with quadratics, you need to be fluent in two opposite processes:
- Expanding (multiplying out brackets to remove them)
- Factorizing (rewriting an expression as a product of brackets)
Quadratic Expressions Are Defined By Their Highest Power
Quadratic expression
An algebraic expression in one variable whose highest exponent is 2 (and with a non-zero $x^2$ term). It can be written in the form $ax^2+bx+c$ where $a\neq 0$.
- A useful way to recognize a quadratic is to look for the highest power of the variable.
- $x^2-5x-14$ is quadratic (highest power is 2).
- $3x^2+2x$ is quadratic.
- $2y-5y^2+7$ is quadratic (it still has a $y^2$ term, even if the order is mixed).
- Not quadratic:
- $x^3-5x-14$ is cubic (highest power is 3).
- $8t+12$ is linear (highest power is 1).
- $x^2+3y-25$ is not a quadratic in one variable because it has two different variables.
- A common mistake is to think an expression is quadratic just because it contains an $x^2$ term.
- If there is also a higher power (like $x^3$), then it is not quadratic.
Expanding Brackets Creates An Equivalent Expression
- When you expand, you do not change the value of the expression, you only change its form.
- In algebra we often show this with the symbol $\equiv$, meaning identically equal.
$$3(x-2) \equiv 3x-6$$
This is always true, for any value of $x$.
- An equation is only true for particular values (solutions).
- For instance, $3x+5=17$ is true only for specific $x$, so it is not an identity.
Expanding Two Binomials (FOIL / Grid Method)
- A very common quadratic form comes from multiplying two binomials (two-term brackets): $$(x+p)(x+q)$$
- Expand by multiplying every term in the first bracket by every term in the second: $$(x+p)(x+q)=x\cdot x+x\cdot q+p\cdot x+p\cdot q$$
- Combine like terms: $$(x+p)(x+q)=x^2+(p+q)x+pq$$
When you expand $(x+p)(x+q)$, the coefficient of $x$ is the sum $p+q$, and the constant term is the product $pq$.
Expand $(x-7)(x+2)$.
Solution
Multiply out:
$$x\cdot x=x^2$$
$$x\cdot 2=2x$$
$$-7\cdot x=-7x$$
$$-7\cdot 2=-14$$
$$ \implies (x-7)(x+2)\equiv x^2-7x+2x-14\equiv x^2-5x-14$$
Standard Form And Factor Form Are Connected By Patterns
- Quadratics are often written in standard form: $$ax^2+bx+c$$
- and sometimes in factor form: $$(x+p)(x+q)$$
- When the factor form is exactly $(x+p)(x+q)$, the leading coefficient is $a=1$.
- The patterns are:
- $a=1$
- $b=p+q$
- $c=pq$
- Think of $p$ and $q$ as two "hidden numbers."
- When you expand, they leave two fingerprints: their sum becomes the $x$ coefficient, and their product becomes the constant.
Factorizing Means Rewriting A Quadratic As A Product
- Factorizing is the reverse of expanding.
- You are trying to "undo" the multiplication.
Factorizing When $a = 1$
- To factorize $x^2+bx+c$, you look for two numbers $p$ and $q$ such that:
- $p+q=b$
- $pq=c$
- Then $$x^2+bx+c=(x+p)(x+q)$$
Factorize $x^2+8x+15$.
Solution
We need $p$ and $q$ with:
$$p+q=8$$
$$pq=15$$
$3$ and $5$ work (sum $8$, product $15$), so
$$x^2+8x+15=(x+3)(x+5)$$
Factorize $x^2-x-12$.
Solution
- We need $p+q=-1$ and $pq=-12$.
- Try factor pairs of $12$: $(1,12)$, $(2,6)$, $(3,4)$.
- Because the product is negative, one is positive and one is negative.
- $3$ and $-4$ work since $3+(-4)=-1$ and $3\cdot(-4)=-12$.
$$x^2-x-12=(x+3)(x-4)$$
- Sign errors are the most common factorizing mistake.
- If $c$ is negative, the two numbers must have opposite signs.
Special Pattern: Difference Of Two Squares
A very important quadratic pattern is the difference of two squares: $$a^2-b^2=(a-b)(a+b)$$
$$x^2-4 = x^2-2^2=(x-2)(x+2)$$
$$9x^2-25=(3x)^2-5^2=(3x-5)(3x+5)$$
- Always check first whether a quadratic is a difference of two squares.
- It can save time and avoids trial-and-error.
Factorizing When $a\neq 1$
- When the coefficient of $x^2$ is not 1, you often factorize by finding two numbers whose product is $ac$ and whose sum is $b$.
- For $ax^2+bx+c$, steps are:
- Compute $ac$.
- Find two numbers $m$ and $n$ such that $mn=ac$ and $m+n=b$.
- Rewrite $bx$ as $mx+nx$.
- Factorize by grouping.
Factorize $2x^2+7x+3$.
Solution
$$ac=2\cdot 3=6$$
- Numbers with product $6$ and sum $7$ are $6$ and $1$.
- Split the middle term: $$2x^2+7x+3=2x^2+6x+x+3$$
- Group: $$=2x(x+3)+1(x+3)=(2x+1)(x+3)$$
Common Factor First: A Key Preparation Skill
- Before factorizing a quadratic, always check for a common factor and factor that out first.
- It makes the remaining quadratic simpler.
Factorize $6x^2+3x+12$.
Solution
All terms share a factor of $3$: $$6x^2+3x+12=3(2x^2+x+4)$$
The bracket may or may not factorize further over integers, but taking out the common factor is always correct and often required.
Factorize $4x(3x-5)+7(3x-5)$.
Solution
The common factor is $(3x-5)$: $$4x(3x-5)+7(3x-5)=(3x-5)(4x+7)$$
Checking Your Factorization By Expanding
After factorizing, you should quickly expand to verify.
- A fast check for $(x+p)(x+q)$ is: $x^2+(p+q)x+pq$.
- If your middle and constant terms match, the factorization is correct.
- Is $5x-14$ quadratic? Explain using the "highest power" idea.
- Expand $(x+2)(x-5)$ and identify $a$, $b$, and $c$.
- Factorize $x^2-10x+16$.
- Factorize $3x^2-12$ using the difference of two squares (after taking out any common factor).
