Welcome! Today we're tackling a clever technique — Proving Composite by Algebraic Factoring.

Here's the challenge:

Suppose someone gives you the expression $7 \times 13 \times 19 + 19$ and asks: is this composite?

The brute-force approach:

Step	What you do
1	Multiply it out → get 1748result
2	Try to factorise 1748

That works, but it's painful.

The cleaner approach:

Use the structure of the expression rather than its value.

No giant multiplications
No tedious trial division
Just algebra

By the end of this section, you'll be able to prove any expression of this form is composite — without ever computing the full product.

1. What it means to prove a number is composite

Before we can prove anything is composite, we need to be precise about what 'composite' means in terms of factors.

The proof strategy comes directly from the definition.

📋 Given Info

Here's what we know:

A prime number has exactly two factors: 1 and itself
A composite number is any integer greater than $1$ that is NOT prime

✍️ Question

Think about this 🤔

To prove that a number is composite, what do you need to exhibit?

Be specific about the conditions on the factors you show.

A composite number is one that can be written as a product of two positive integers, both bigger than $1$ Both factors must exceed 1.

So to PROVEShowing it's NOT prime by breaking it down a number is composite, your job is to find two numbers $A$ and $B$ such that:

$A \times B$ equals the numberFirst condition to check
$A > 1$ Second and third conditions
$B > 1$ Both must exceed 1

That's it. Show these three thingskeyProduct equals number, both factors greater than 1, and you've proven it's composite.

✍️ MCQ

Choose one

If I want to prove that

15

is composite, which of the following would be a valid way to show it?

✓

You don't need to fully factorise the number. You don't need to find all its factors.

You just need to show ONE way to split it into two pieces that are both bigger than 1Both factors must exceed 1.

That's it — just one factorisationkey ruleYou need just one factorisation where both factors are bigger than 1 like $n = a \times b$ where $a > 1$ and $b > 1$ (The moment you find any two factors greater than 1, stop there).

✍️ MCQ

Choose one

To prove that

21

is composite, which of the following is sufficient?

✓

For example, to show 92 is composite: $92 = 4 \times 23$ .

Both 4>1 and 23>1 are greater than 1. Done.

You don't needOne valid factorisation is enough to go further and factorise 4 into $2 \times 2$ .

✍️ MCQ

Choose one

To prove that

51

is composite, which factorisation is sufficient?

✓

The key point: the proof ends the moment you exhibit the product. Everything after that is unnecessaryNo need to factor further or check if factors are prime.

Once you write $N = A \times B$ Writing the number as a product of two factors where both $A > 1$ and $B > 1$ (Both A and B must exceed 1), you've provenThis single step completes the entire proof the number is composite. Full stop.done!The proof ends right here

✍️ MCQ

Choose one

You want to prove that

84

is composite. You write:

84 = 7 \times 12

. What should you do next?

✓

2. Spotting the common factor in a sum expression

You know what you need to show — that an expression is composite.

The question is HOW to find the factorisation when the expression is given as a sum like:

'product of primes plus a number'

— without computing the whole thing first.

📋 Given Info

Consider the expression $7 \times 13 \times 19 + 19$ .

It has two terms:

The product $7 \times 13 \times 19$
The number $19$

Notice that 19 appears in both termscommon factor.

✍️ Question

Your Task 📝

Show that $7 \times 13 \times 19 + 19$ is composite.

⚠️ Important: You must NOT compute the full value. Show the algebraic factoring step clearly.

The expression is $7 \times 13 \times 19 + 19$ . Look at it before computing anything.

The two terms are: $7 \times 13 \times 19$ and $19$ . What do they share? The number 19common19 appears in both terms, spot it and factor it out appears in both — it's one of the factors in the product, and it IS the second term.

✍️ MCQ

Choose one

What can we factor out from the expression

7 \times 13 \times 19 + 19

?

✓

Factor out 19The common factor that makes factoring work:

$7 \times 13 \times 19 + 19 = 19 \times (7 \times 13 + 1)$

Simplify inside the bracket:

$19 \times (7 \times 13 + 1) = 19 \times (91 + 1) =$ $19 \times 92$ Two numbers multiplied together — the form we need

✍️ Yes/No

Yes or No?

We've shown that

7 \times 13 \times 19 + 19 = 19 \times 92

. Is this enough to prove the number is composite?

✓

Now check: is $19 > 1$ ? Yes. Is $92 > 1$ ? Yes.

So the original number equals $19 \times 92$ , where both factors exceed 1. By definition, it's compositeA product of two factors, both greater than 1.

The proof is done. You don't need to simplify $19 \times 92 = 1748$ . You don't need to factorise $92$ further into $4 \times 23$ or $2^{2} \times 23$ .

The moment you write it as (something $> 1$ ) $\times$ (something else $> 1$ )(This is how you prove composite), the composite proof is completeYou're done once you show this.

This is the golden rule for these problems. One valid factorisationKEYYou don't need all factors = proof done.

✍️ MCQ

Choose one

We showed

7 \times 13 \times 19 + 19 = 19 \times 92

. After writing this, what else do we need to do to complete the proof?

✓

⚠️ The trap most students fall into:Multiplying out loses the structure they compute $7 \times 13 \times 19 + 19 = 1748$ and then spend time trying to factorise 1748.

They see the expression, panic, and immediately reach for the calculator. Then they're stuck with a big number and no clear path forward.

That's the hard way — and in an exam, it's a waste of precious minutes.

The smart approach:Find the shared factor before computing spot the common factorLook for shared terms first before computing. We saw that 19key19 is in both parts of the expression appears in both terms, so we factored it out directly to get $19 \times 92$ This product is clearly composite. No big multiplication needed.

✍️ Yes/No

Yes or No?

If a student computes

7 \times 13 \times 19 + 19 = 1748

first and then tries to factorise 1748, are they doing anything mathematically wrong?

✓

3. The general technique for (product) + k expressions

You've done one example. Now let's extract the general pattern — what to look for in any expression of this form, so you can handle variants without hesitation.

You've seen that in $7 \times 13 \times 19 + 19$ , the key was that 19 appeared both as a factor in the product and as the added term. This let you factor it out.

✍️ Question

Your Turn 🧠

Show that $5 \times 4 \times 3 \times 2 \times 1 + 3$ is composite.

Then state the general rule: for an expression of the form (product of several numbers) $+ k$ , what should you look for?

Let's work through it. The expression is $5 \times 4 \times 3 \times 2 \times 1 + 3$ .

Look at the product: $5 \times 4 \times 3 \times 2 \times 1$ . The added term is $3$ added termThis 3 is what we're adding to the product. Is $3$ one of the factors in the product? Yes — there's a ' $3$ 'factor!The 3 already appears as a factor, so we can factor it out right there in the middle!

✍️ MCQ

Choose one

Since

3

is a common factor, when we factor it out from

5 \times 4 \times 3 \times 2 \times 1 + 3

, what do we get?

✓

Factor out 3We spotted 3 in both parts and pulled it out:

$5 \times 4 \times 3 \times 2 \times 1 + 3 =$ $3 \times (5 \times 4 \times 2 \times 1 + 1)$ factoredThe technique is finding what's shared

$= 3 \times ($ $40 + 1$ $) =$ $3 \times 41$

✍️ Yes/No

Yes or No?

We've written the expression as

3 \times 41

. Is this enough to prove the number is composite?

✓

Both 3 and 41 are greater than 1Each factor must be bigger than 1. Composite. Done.Two factors each bigger than 1 means we're done

✍️ MCQ

Choose one

For an expression of the form (product of several numbers)

+ k

, what should you look for to prove it's composite?

✓

The general rule works like this. Given an expression (A \times B \times C \times \ldots) + k:

Check:This is the essential step to identify Does $k$ keyLook for k among A, B, C appear as one of the factors $A, B, C, \dots$ ?

If yes: Factor $k$ outPull k out from the whole expression. You get $k \times (\text{product of everything else} + 1)$ The bracketed part is always bigger than 1.

The second factor is $(product of everything else + 1)$ . Since the 'everything else' is a product of positive integers (at least 1 each), the second factor is at least 2.
So you have $k > 1$ First factor k and second factor both exceed 1 times something $> 1$ Product of two factors each greater than 1. Composite.Two factors greater than 1 means composite

✍️ MCQ

Choose one

If you're given

(8 \times 7 \times 6 \times 5) + 7

, which number would you factor out to prove it's composite?

✓

The General Rule:Works for any expression of this form, not just this problem

This works for any such expression:

$7 \times 11 \times 23 + 11$ Number appears inside the product, so you can factor it out
$6 \times 5 \times 4 \times 3 \times 2 \times 1 + 4$ Same pattern — the number is inside the product
...or any variation!

What to look for:

For an expression of the form (product of several numbers) $+ k$ :

➡️ Check if $k$ appears as one of the factors in the product.Look for k hiding in the product

If yes, just factor out $k$ Factor it out to get two smaller numbers — the result will always be compositeproven!Two smaller numbers multiplied proves it's composite (a product of two smaller numbers).

✍️ Yes/No

Yes or No?

Consider the expression

8 \times 7 \times 6 \times 5 + 5

. Can this method be used to prove it's composite?

✓