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Year 8 Maths Algebra

Factorising with Negatives

20 practice questions 2 video lessons Theory + worked examples
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Theory

When the first term is negative, take out a negative HCF so the bracket starts with a positive term. The HCF of \(-4x^2y\) and \(2x\) can be \(2x\) or \(-2x\); with \(-2x\): \(-4x^2y + 2x = -2x(2xy - 1)\). Dividing by a negative changes every sign inside. This Year 8 maths page shows how.

A HCF can be positive or negative. \(-4x^2y = -2x \times 2xy\) and \(2x = -2x \times (-1)\), so \(-2x\) is a common factor, just as \(2x\) is.

Three steps. Find the HCF (number and pronumerals, lowest powers) and make it negative. Divide each term by it. Write it outside the bracket with the answers inside.

Dividing by a negative flips every sign: \(-6x \div (-3) = 2x\) and \(15 \div (-3) = -5\), so \(-6x + 15 = -3(2x - 5)\). A term divided by itself gives \(1\): \(-4k - 4kn = -4k(1 + n)\).

The HCF may be just \(-1\): \(-m + 7 = -(m - 7)\). Check every answer by expanding, using \((-) \times (-) = (+)\).

Dividing by a negative flips the signs The terms -6x and +15 are each divided by -3, giving 2x and -5. So -6x + 15 = -3(2x - 5). −6x+15start 2x−5divide each term by −3: both signs flip
Dividing by \(-3\) flips both signs: \(-6x\) becomes \(2x\) and \(+15\) becomes \(-5\). So \(-6x + 15 = -3(2x - 5)\).

Positive or negative HCF:

ExpressionPositive HCFNegative HCF
\(-12x + 8\)\(4(-3x + 2)\)\(-4(3x - 2)\)
\(-6a - 15\)\(3(-2a - 5)\)\(-3(2a + 5)\)
\(-8y^2 + 20y\)\(4y(-2y + 5)\)\(-4y(2y - 5)\)
Change every sign, not just the first. \(-9x + 6 = -3(3x - 2)\), not \(-3(3x + 2)\).
Check by expanding. \(-3(3x - 2) = -9x + 6\).

How to factorise with a negative HCF

  1. Find the HCF (ignore signs), then put a minus in front.
  2. Divide each term by the negative HCF; every sign flips.
  3. Write HCF(answers) and check by expanding.
Example 1 — A negative HCF with pronumerals
Factorise \(-4x^2y + 2x\) with a negative HCF.
Solution

\(-4x^2y = -2x \times 2xy\) and \(2x = -2x \times (-1)\).

\(-4x^2y + 2x\)\(=\)\(-2x(2xy - 1)\)
−2 x (2 xy −1)

\(-2x(2xy - 1)\)

Example 2 — Reorder, then factorise
Factorise \(3xy^2z - 9xyz\) with a negative HCF.
Solution
\(\)\(=\)\(-3xyz \times (-y) + (-3xyz) \times 3\)
\(\)\(=\)\(-3xyz(-y + 3)\)
\(\)\(=\)\(-3xyz(3 - y)\)
−3 xyz (3− y )

\(-3xyz(3 - y)\)

Example 3 — Negative number as the HCF
Factorise, taking out the negative HCF.
(i) \(-5a + 30\)
(ii) \(-14m - 35\)
Solution
\(-5a + 30\)\(=\)\(-5(a - 6)\)
\(-14m - 35\)\(=\)\(-7(2m + 5)\)
−5( a −6)−7(2 m +5)

(i) \(-5(a - 6)\)
(ii) \(-7(2m + 5)\)

Example 4 — Pronumerals in the HCF
Factorise, taking out the negative HCF.
(i) \(-6y + 10yz\)
(ii) \(-t^2 - t\)
Solution
\(-6y + 10yz\)\(=\)\(-2y(3 - 5z)\)
\(-t^2 - t\)\(=\)\(-t(t + 1)\)
−2 y (3−5 z )− t ( t +1)

(i) \(-2y(3 - 5z)\)
(ii) \(-t(t + 1)\)

Example 5 — Powers and three pronumerals
Factorise, taking out the negative HCF.
(i) \(-20a^2b + 15ab^2\)
(ii) \(-24p^2qr + 8pqr\)
Solution
\(-20a^2b + 15ab^2\)\(=\)\(-5ab(4a - 3b)\)
\(-24p^2qr + 8pqr\)\(=\)\(-8pqr(3p - 1)\)
−5 ab (4 a −3 b )−8 pqr (3 p −1)

(i) \(-5ab(4a - 3b)\)
(ii) \(-8pqr(3p - 1)\)

Example 6 — Checking an answer
Jai factorised \(-18gh + 24g\) as \(-6g(3h + 4)\). Expand to test it, then correct it.
Solution
\(-6g(3h + 4)\)\(=\)\(-18gh - 24g\)
Jai is wrong −6 g (3 h −4)

That is not \(-18gh + 24g\). Since \(24g \div (-6g) = -4\), the answer is \(-6g(3h - 4)\).

Common pitfalls

Changing only the first sign. \(-8x + 4 = -4(2x - 1)\).
Leaving a negative first term in the bracket. Prefer \(-2(x + 3)\) to \(2(-x - 3)\) when a negative HCF is asked for.
Forgetting the 1. \(-k - kw = -k(1 + w)\).
Not checking. Expand to make sure the signs come back right.

Frequently asked questions

When should you take out a negative HCF?

When the first term is negative, or when the question asks for it.

What happens to the signs inside the bracket?

They all change: \(-10a + 5 = -5(2a - 1)\).

Can the HCF be -1?

Yes: \(-p + 3 = -(p - 3)\).

How do you check a negative factorisation?

Expand it: \(-5(2a - 1) = -10a + 5\).

Is a negative HCF still a common factor?

Yes. Both \(2x\) and \(-2x\) divide \(-4x^2y\) and \(2x\) exactly.