Mathematics Grade 8: Factorisation

Grade 8 Mathematics: Factorisation

The Driving Question If you have a rectangle made of 12 identical square tiles, how many different ways can you arrange them into a rectangle without cutting any tiles—and why does that matter when you’re trying to break down big multiplication problems into smaller, easier ones?

The Core Idea — Built, Not Listed

Imagine you’re packing a lunchbox with 18 identical granola bars. You could stack them all in one long row (1 × 18), or arrange them in two rows of nine (2 × 9), or even three rows of six (3 × 6). Each of these arrangements is a factor pair of 18—two numbers that multiply to give 18. Factorisation is just the process of finding all these pairs, but instead of counting granola bars, you’re breaking down numbers (or algebraic expressions) into their smallest building blocks.

Now, let’s say you have the expression x² + 5x + 6. It’s like a puzzle where you’re trying to find two binomials (like (x + 2) and (x + 3)) that multiply to give you the original expression. The key is to look for two numbers that multiply to the constant term (6) and add up to the coefficient of the middle term (5). Once you find them, you can rewrite the expression as (x + 2)(x + 3), which is much easier to work with—especially when solving equations or simplifying fractions.

Key Vocabulary:
1. Factor - Definition: A number or expression that divides another number or expression evenly (without leaving a remainder). - Example: 4 is a factor of 20 because 20 ÷ 4 = 5. In algebra, (x + 1) is a factor of x² + 3x + 2 because (x + 1)(x + 2) = x² + 3x + 2. - Note for High School: In college algebra, factors can include irrational or complex numbers (e.g., x² + 1 factors into (x + i)(x – i) in the complex plane).

1. Prime Factorisation
2. Definition: Breaking down a number into a product of prime numbers (numbers greater than 1 that have no factors other than 1 and themselves).
3. Example: The prime factorisation of 60 is 2 × 2 × 3 × 5 (or 2² × 3 × 5). For polynomials, x² – 4 factors into (x + 2)(x – 2), both of which are "prime" (irreducible) over the real numbers.

4. Note for High School: In abstract algebra, "prime" takes on a more general meaning, and factorisation can involve rings and fields beyond just integers or real numbers.

5. Greatest Common Factor (GCF)

6. Definition: The largest number or expression that divides two or more numbers or expressions without a remainder.
7. Example: The GCF of 24 and 36 is 12. For the expressions 6x² and 9x³, the GCF is 3x² (the largest term that divides both).

8. Note for High School: In calculus, GCF is used to simplify rational functions before integrating (e.g., factoring out x from x³ + 2x before finding the antiderivative).

9. Factoring by Grouping

10. Definition: A method for factoring polynomials with four or more terms by grouping terms with common factors and then factoring out the GCF from each group.
11. Example: To factor x³ + 3x² + 2x + 6, group as (x³ + 3x²) + (2x + 6), then factor out x² from the first group and 2 from the second: x²(x + 3) + 2(x + 3). Now factor out (x + 3) to get (x + 3)(x² + 2).
12. Note for High School: This technique is foundational for partial fraction decomposition in calculus.

Assessment Translation

How Factorisation Appears on Grade 8 Assessments: - Multiple Choice: Questions often ask you to identify the factored form of an expression (e.g., "Which is the factored form of x² – 9?" with options like (x – 3)(x + 3), (x – 3)², etc.). Distractors might include: - Incorrect signs (e.g., (x + 3)(x + 3) for x² + 6x + 9). - Missing a term (e.g., (x – 3) instead of (x – 3)(x + 3)). - Swapping addition and subtraction (e.g., (x – 3)(x – 3) for x² – 6x + 9). - Short Answer/Constructed Response: You might be asked to factor an expression completely (e.g., "Factor 4x² – 16 completely") or solve an equation by factoring (e.g., "Solve x² + 5x + 6 = 0 by factoring"). - Evidence-Based Writing (Less Common): Some state tests include questions where you must explain why a certain factorisation works (e.g., "Explain how you know that (x + 2) is a factor of x² + 7x + 10").

What a Proficient Response Looks Like: - Multiple Choice: The student selects (x – 3)(x + 3) for x² – 9 and can justify it by multiplying the factors to check. - Short Answer: For "Factor 3x² + 6x completely," a proficient response is:

"First, find the GCF of 3x² and 6x, which is 3x. Factor out 3x to get 3x(x + 2). This is completely factored because (x + 2) cannot be factored further." - A developing response might stop at 3(x² + 2x) or forget to factor out the x. - Equation Solving: For "Solve x² + 5x + 6 = 0 by factoring," a proficient response is: "Factor the quadratic as (x + 2)(x + 3) = 0. Set each factor equal to zero: x + 2 = 0-x = –2, and x + 3 = 0-x = –3. The solutions are x = –2 and x = –3."

Model Proficient Response (Short Answer): Prompt: Factor x² – 10x + 25 completely. Response:

"This is a perfect square trinomial because (x – 5)² = x² – 10x + 25. So, the factored form is (x – 5)². I checked by multiplying (x – 5)(x – 5) to confirm it gives the original expression."

Mistake Taxonomy

1. Forgetting to Factor Out the GCF First - Prompt: Factor 6x² + 12x completely. - Common Wrong Response: (6x)(x + 2) or 6(x² + 2x). - Why It Loses Credit: The expression isn’t fully factored because the GCF (6x) wasn’t factored out first. The correct factored form should have the GCF outside the parentheses. - Correct Approach:

"The GCF of 6x² and 12x is 6x. Factor out 6x to get 6x(x + 2). This is completely factored because (x + 2) has no common factors."

2. Mixing Up Signs in Factoring Trinomials - Prompt: Factor x² – 7x + 12. - Common Wrong Response: (x + 3)(x + 4) or (x – 3)(x – 4). - Why It Loses Credit: The signs in the binomials don’t match the original trinomial. The middle term is negative, so both binomials must have subtraction signs (since a negative times a negative is positive). - Correct Approach:

"Find two numbers that multiply to 12 and add to –7. These numbers are –3 and –4. So, the factored form is (x – 3)(x – 4)."

3. Incorrectly Applying the Difference of Squares - Prompt: Factor 4x² – 25. - Common Wrong Response: (2x – 5)(2x – 5) or (4x – 5)(x + 5). - Why It Loses Credit: The first response is a perfect square (not a difference of squares), and the second doesn’t multiply to give the original expression. The difference of squares formula is a² – b² = (a + b)(a – b). - Correct Approach:

"Recognize that 4x² is (2x)² and 25 is 5². Apply the difference of squares formula: (2x + 5)(2x – 5)."

Connection Layer

1. Within Math: Factorisation-Solving Quadratic Equations

2. Once you factor a quadratic like x² + 5x + 6 into (x + 2)(x + 3), you can set each factor equal to zero to find the roots (x = –2 and x = –3). This is the zero product property: if a × b = 0, then a = 0 or b = 0.

3. Across Subjects: Factorisation-Chemistry (Balancing Equations)

4. In chemistry, you balance equations by finding the "factors" (coefficients) that make the number of atoms equal on both sides. For example, H? + O?-H?O becomes 2H? + O?-2H?O—the coefficients are like the factors that scale the molecules.

5. Outside School: Factorisation-Cryptography (Online Security)

6. When you log into a website, your password is protected using RSA encryption, which relies on the fact that it’s easy to multiply two large prime numbers (like 31 × 41 = 1271) but very hard to factor the product back into primes. This "one-way" property keeps your data secure.

The Stretch Question

If you factor a quadratic like x² + bx + c into (x + m)(x + n), you know that m × n = c and m + n = b. But what if the quadratic is x² + bx – c (where c is positive)? How does the relationship between m and n change, and why?

Pointer Toward the Answer: When the constant term is negative, one of the numbers (m or n) must be positive and the other negative. This is because a negative times a positive is negative (so m × n = –c), but their sum (m + n) could be positive or negative depending on which number is larger. For example, x² + x – 6 factors into (x + 3)(x – 2) because 3 × (–2) = –6 and 3 + (–2) = 1. The key is that the product is negative, so the signs must differ.