Welcome! Today we're tackling Factoring with Surd Coefficients and Difference of Squares — two special cases where the usual splitting method won't work.

Not every quadratic can be factored by the standard splitting method.

When the middle term is missing ( $b = 0$ key), the splitting method has nothing to split.

When coefficients include surds, the arithmetic looks intimidating.

For example:

$\sqrt{2} x^{2} + 5 x + \sqrt{2}$
$\sqrt{3} x^{2} - 4 x + \sqrt{3}$

In this lesson, you will learn two special cases:

Case	When to use
Difference of squares	Middle term is missing ( $b = 0$ b=0)
Surd coefficient technique	$a \times c$ simplifies to a clean integer

1. Recognising and applying the difference-of-squares pattern

When the middle term is absent ( $b = 0$ ), the polynomial might be a difference of squares:

a^2 - b^2 = (a + b)(a - b)

Recognising this pattern is essential because the standard splitting method cannot handle it.

📋 Given Info

You're given the identity:

a^2 - b^2 = (a + b)(a - b)

For polynomials, this means expressions like $x^{2} - 7$ can be rewritten as:

$x^{2} - 7 = x^{2} - (\sqrt{7})^{2}$

The key insight: any positive number can be written as a square of its square root.

✍️ Question

Factor $x^2 - 7$ completely over the reals.

What are the zeros?

When the middle term is missing (b = 0), check if the expression is a difference of two squaresYour first instinct with no middle term.

For $x^2 - 7$ Any number can be written as a square root squared: this is $x^2 - (\sqrt{7})^2$ (Write 7 as square root of 7 squared). It fits the pattern $a^2 - b^2$ This unlocks the factoring pattern with $a = x$ and $b = \sqrt{7}$ .

✍️ MCQ

Choose one

Using the identity

a^2 - b^2 = (a+b)(a-b)

, what is the factored form of

x^2 - 7

?

✓

Using the identity: $a^2 - b^2 = (a + b)(a - b)$ Your go-to tool for subtraction of two squared terms

$x^{2} - (\sqrt{7})^{2} =$ $(x + \sqrt{7})(x - \sqrt{7})$

✍️ MCQ

Choose one

What are the zeros of

(x + \sqrt{7})(x - \sqrt{7}) = 0

?

✓

Zeros: $x = \sqrt{7}$ +√7One of the two zeros we get and $x = -\sqrt{7}$ -√7The opposite zero, always a plus-minus pair

Notice: Even though $7$ isn't a perfect squareDon't think you're stuck without a nice integer root, we can still factor $x^2 - 7$ completely over the realsYou can factor even when the number isn't a perfect square by using $\sqrt{7}$ as our value of $b$ Just use the square root directly.

The factorization (x + √7)(x - √7) tells us exactly where the parabola crosses the x-axis. Let's graph y = x² - 7 to see how the algebraic zeros appear as geometric intercepts.

✍️ MCQ

Choose one

Why are the two x-intercepts (

\sqrt{7}

and

-\sqrt{7}

) equidistant from the origin?

✓

⚠️ Common mistake: saying " $x^2 - 7$ cannot be factored because 7 is not a perfect square.Don't assume it can't be factored"

This is wrongThat's not the deciding factor — it can be factored over the reals using surd factorsSurds handle non-perfect squares.

✍️ Yes/No

Yes or No?

Can

x^2 - 11

be factored over the real numbers?

✓

"Cannot be factored" only applies to sums of squaresThese cannot be factored over reals like $x^{2} + 7$ , where no real factoring exists.

Expression	Can factor over reals?
$x^2 - 7$ (difference)Even with non-perfect square constants	✅ Yes: $(x + \sqrt{7})$ $(x - \sqrt{7})$
$x^2 + 7$ (sum)The sum blocks factoring	❌ No real factorsblockedNo real factors exist

So the rule is simple — it's the plus that stops us, not whether the number is a perfect square!Plus sign is the deciding factor

✍️ T/F

True or False?

x^2 + 7

cannot be factored over the reals because it is a sum of squares, not a difference.

✓

2. Computing a*c with surd coefficients

When a polynomial has surd coefficients, the first step of the splitting method — computing $a \times c$ — looks intimidating.

But here's the good news:

In well-designed problems, the surds in $a$ and $c$ multiply to give a clean integer, making the rest of the process standard.

✍️ Question

Your Turn ✏️

For the polynomial $\sqrt{3}x^2 - 8x + 4\sqrt{3}$ , compute $a \times c$ .

Show the simplification that makes it a clean integer.

When coefficients are surds, the key simplification is:

\sqrt{n} \times \sqrt{n} = (\sqrt{n})^2 = n

(The surd vanishes completely when multiplied by itself)

This is the property that makes surd coefficients manageableSurd coefficients aren't as scary as they look — when you multiply a surd by itself, you get a clean integerMultiplying a surd by itself gives a clean integer.

In our problem: $a \times c = \sqrt{3} \times 4 \sqrt{3} = 4 \times (\sqrt{3})^{2} = 4 \times 3 = 12$

✍️ MCQ

Choose one

If

a = \sqrt{5}

and

c = 3\sqrt{5}

, what is

a \times c

?

✓

For $\sqrt{3}x^2 - 8x + 4\sqrt{3}$ :

$a = \sqrt{3}$ , $c = 4\sqrt{3}$

a \cdot c = \sqrt{3} \times 4\sqrt{3}

$= 4 \times (\sqrt{3} \times \sqrt{3}) =$ $4 \times (\sqrt{3})^2 = 4 \times 3$ The surds cancel out completely $=$ $12$ clean!Surd times itself gives a whole number

✍️ MCQ

Choose one

If

a = \sqrt{5}

and

c = 6\sqrt{5}

, what is

a \cdot c

?

✓

The surds 'cancelled' to give a clean integerThe result is a whole number you can work with. This is by design — well-crafted problemsExam questions are designed this way on purpose always have $a \cdot c$ The product simplifies nicely simplify to an integer, so you can use the standard splitting methodBack to familiar factoring techniques from that point on.

With $a \cdot c = 12$ and $b = - 8$ , you need two numbers with:

Product = 12The two things you're always looking for
Sum = -8Both must be true at the same time

✍️ MCQ

Choose one

Which pair of numbers has a product of

12

and a sum of

-8

?

✓

The pair: -6 and -2Only way to get positive product but negative sum.

(Check: $(-6) \times (-2) = 12$ Always check before moving forward ✓✓Confirming the product and $(-6) + (-2) = -8$ Confirming the sum ✓✓Both conditions satisfied)

3. Completing the surd-coefficient factoring

📋 Given Info

With $a \times c$ as a clean integer, the splitting and grouping proceed as usual — but the extraction step requires recognising surd common factors. This is the final skill for surd-coefficient factoring.

Useful Reference:

For $\sqrt{3} x^{2} - 6 x - 2 x + 4 \sqrt{3}$ , the first pair is $\sqrt{3} x^{2} - 6 x$ .

The GCF involves $\sqrt{3}x$ because:

\sqrt{3}x \times (-2\sqrt{3}) = -2 \times 3 \times x = -6x

The surds multiply out: $\sqrt{3} \times \sqrt{3} = 3$ =3

✍️ Question

Your Turn ✏️

Factor $\sqrt{3}x^2 - 8x + 4\sqrt{3}$ completely.

Show:

The split (find two numbers that multiply to $a \times c$ and add to $b$ )
The grouping with extracted common factors
The zeros of the polynomial

Let's trace through the surd factoring carefully.

We have $\sqrt{3}x^2 - 8x + 4\sqrt{3}$ .

$a \cdot c$ $= \sqrt{3} \times 4 \sqrt{3} =$ $4 \times 3$ Surds simplify when multiplied by themselves $=$ $12$ , and $b = - 8$ .

We need a pair of numbers with product 12 and sum $-8$ Finding the right pair for splitting. That pair is -6 and -2.

Split:The key technique for factoring $\sqrt{3}x^2 - 6x - 2x + 4\sqrt{3}$ (Two terms that add to the same thing but allow grouping)

✍️ MCQ

Choose one

What is the GCF of the first pair

\sqrt{3}x^2 - 6x

?

✓

Group 1: $\sqrt{3} x^{2} - 6 x$

What is the GCFThe GCF can itself contain a surd?

✍️ MCQ

Choose one

For the first group

\sqrt{3}x^2 - 6x

, the GCF involves

\sqrt{3}x

. Why does

\sqrt{3}x

work as a factor of

-6x

?

✓

Notice that $\sqrt{3} x \times (- 2 \sqrt{3})$ = $-2 \times (\sqrt{3})^2 \times x$ Surds multiply themselves to simplify = $- 2 \times 3 \times x$ = $- 6 x$ .

So the GCF is $\sqrt{3}x$ GCFMakes both terms divide out cleanly.

\sqrt{3}x^2 - 6x = \sqrt{3}x(x - 2\sqrt{3})

(Always look at what's left in the brackets)

Watch for matching brackets when factoring the second group

Group 2: $- 2 x + 4 \sqrt{3}$

Now let's find the GCF here. Both terms have a factor of $2$ — and notice the signs. If we factor out $-2$ Factor out negative 2, not positive 2, we get: $- 2 x + 4 \sqrt{3} = - 2 (x - 2 \sqrt{3})$

✍️ MCQ

Choose one

What is the common binomial factor that appears in both groups?

✓

Both groups share $(x - 2\sqrt{3})$ The common factor shared by both groups:

$\sqrt{3} x (x - 2 \sqrt{3}) + (- 2) (x - 2 \sqrt{3})$

Factor out the common binomialFactor out the common binomial:

(x - 2\sqrt{3})(\sqrt{3}x - 2)

(Two binomial factors from surd coefficients)

This is the complete factorisation!

ZerosThe x values that make the expression equal zero:

From $(x - 2 \sqrt{3}) = 0$ :

x = 2\sqrt{3}

(Set each factor to zero and solve)

From $(\sqrt{3} x - 2) = 0$ :

$\sqrt{3} x = 2$

$x = \frac{2}{\sqrt{3}}$

RationalisingNever leave a square root in the denominator:

x = \dfrac{2}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}

(Multiply top and bottom by the same surd)

✍️ MCQ

Choose one

We found that

x = \dfrac{2}{\sqrt{3}}

needs to be rationalised. What do we multiply by to rationalise a denominator of

\sqrt{3}

?

✓