an introduction to factorisation

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An Introduction to Factorisation Slideshow 14 Mathematics Mr Sasaki Room 307

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An Introduction to Factorisation. Slideshow 14 Mathematics Mr Sasaki Room 307. Objectives. Review the expansion of brackets Be able to factorise by removing factors from terms (the simple case). Review. It’s time to test your expansion skills. Good luck!. Answers. - PowerPoint PPT Presentation

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An Introduction to Factorisation

Slideshow 14  Mathematics

Mr Sasaki    Room 307

Objectives

•Review the expansion of brackets•Be able to factorise by removing factors from terms (the simple case)

Factorisation (Simple)Do you remember what factorisation is?

Factorisation is the opposite of expansion! We remove factors from an expression and place them outside of it.ExampleFactorise .

3 𝑥+12=¿3 (𝑥+4)Both terms here can be divided by 3.We did this in Grade 8!

Factorisation (Simple)If terms share common unknowns, they can also be placed outside the expression.

ExampleFactorise .

8 𝑥2+14 𝑥𝑦−6 𝑥=¿2 𝑥()4 𝑥+7 𝑦−3Try to remove whatever you can from each term. Good luck!

Answers – Part 1, Easy

3 (3 𝑥−1)3 (6 𝑦−5) 14 𝑥 (𝑦−2)𝑦 (𝑦−𝑥) 5(4 𝑥−7 𝑦 ) 21 𝑥(3 𝑦−1)3 𝑥 (2𝑥 𝑦2−3 𝑦+1)𝑥2(15𝑥2−11)−3 (𝑥+2 𝑦 )2 𝑥(− 𝑥𝑦+3 𝑦+2)2𝑏(7 𝑎𝑏+2𝑏−1) 𝑥 (−4 𝑥2+3𝑥+2)35 𝑥 (𝑥−𝑦 ) 8 𝑥2(2𝑥−𝑦+3)−4 𝑥 (𝑥+7 𝑦 )−27 𝑥 (𝑥2+3 𝑦 2)36 𝑥 (2 𝑥+3 𝑦 ) 14 𝑎𝑏(3𝑎+7𝑏)16(4 𝑎+7𝑏+5𝑐) 27 𝑥2(−5 𝑥2 𝑦2+4 𝑥𝑦+3)

Answers – Part 1, Hard

3 (𝑥+2)2(3𝑥−1)5(𝑥+3)5(2𝑥+3)7 (𝑥+2)

2(2𝑥−1)6 (𝑥+5)5(𝑥−1)4(𝑥−7)4(2𝑥−7)

2 𝑥(𝑥+1)2 𝑥(𝑥+8)

7 (𝑥2−𝑦 )12𝑥 (𝑥−7 )

Expanding QuadraticsLet’s look back at expanding quadratics again.

We know we multiply each combination of terms out…but what would a middle step look like?

(𝑥+𝑦 ) (𝑎+𝑏)=¿¿ 𝑥𝑎+𝑥𝑏+𝑦𝑎+𝑦𝑏

How many lots of are there?There is of them…plus of them.

𝑥 (𝑎+𝑏 )+𝑦 (𝑎+𝑏)

So… .

Other FactorisationsSo basically, we can combine separate pairs of brackets in this way.

ExampleFactorise .

3 (𝑥+2 )− 𝑥 (𝑥+2 )=¿(3−𝑥 )(𝑥+2)The brackets must contain the same polynomial so we may need to adjust them.

ExampleFactorise .

𝑥 (𝑦+1 )−2(3 𝑦+3)=¿𝑥 (𝑦+1 )−6(𝑦+1)¿ (𝑥−6)(𝑦+1)

Answers – Part 2

3 (𝑥+3)(5𝑥−2)2(𝑥−2)(𝑥+1)(𝑥−2 )2(𝑥+2)(𝑥+9)2(𝑥+6)(𝑥+𝑦 )−(𝑥+2)(𝑥+𝑦)(𝑎−3)(𝑎+𝑏)(𝑥+2)(𝑥+𝑦)3 𝑥 (𝑥2 𝑦−2 𝑥+𝑦2−2 𝑦 )2 𝑥(5 𝑥+3 𝑦 )2(𝑥2+3 𝑥+𝑦2−3 𝑦)5(2𝑥+𝑦 2−2 𝑦)

6 (𝑥−2)(𝑥+3) or