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Factors β Learning Outcomes
Factorise expressions such as: ππ₯, ππ₯π¦, where πββ€ πππ₯π¦+ππ¦, where π, πββ€ π π₯βπ‘π¦+π‘π₯βπ π¦, where π₯,π‘,π₯,π¦ are variable π π₯ 2 +ππ₯, where π,πββ€ π₯ 2 +ππ₯+π, where π,πββ€ π₯ 2 β π 2
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Factorise ππ₯, ππ₯π¦ Recall how to distribute:
Distribution allows operations to break BEMDAS without changing the answer. e.g. 2Γ 1+3 =2Γ1+2Γ3
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Factorise ππ₯, ππ₯π¦ Write down what a factor is.
A factor is a number that divides evenly (βgoes intoβ) another number. Write down the factors of 12 and 20, and find the highest common factor. Write down the factors of 8 and 8π₯, and find the highest common factor. Write down the factors of 6π₯ and 8, then find the highest common factor.
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Factorise ππ₯, ππ₯π¦ Find the highest common factor of 12 and 15π₯.
What is ? What is 15π₯ 3 ? To factorise 12+15π₯, we write 3(4+5π₯) What do you get if you distribute 3(4+5π₯)?
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Factorise ππ₯ππ¦+ππ¦ Find the highest common factor of 16 and 12π₯.
What is ? What is 12π₯ 4 ? Factorise 16+12π₯.
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Factorise ππ₯ππ¦+ππ¦ Find the highest common factor of ππ₯ and ππ₯.
What is ππ₯ π₯ ? What is ππ₯ π₯ ? Factorise ππ₯+ππ₯.
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Factorise ππ₯ππ¦+ππ¦ Factorise each of the following: 14π₯+7 4π₯+20 6π₯+9π¦
5π₯+15π¦ 3π₯β9π¦ 2π₯+12π₯π¦ 4π₯π¦+32π¦π§ 3π₯+6π¦+9π§
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Factorise ππ₯ππ¦+ππ¦ Factorise 2π₯π¦β4π₯π€ Factorise 4ππ+8π Factorise π₯π¦+π€π¦
2003 OL P1 Q5 Factorise 2π₯π¦β4π₯π€ 2005 OL P1 Q5 Factorise 4ππ+8π 2006 OL P1 Q5 Factorise π₯π¦+π€π¦ Factorise 4 π 2 +8π 2015 OL P1 Q9 Factorise fully 7π₯β21π¦
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Factorise π π₯βπ‘π¦+π‘π₯βπ π¦
This is called factorising by grouping. Take ππ‘β4π‘+2πβ8 as an example. There is no common factor for all four terms. Factorise ππ‘β4π‘. Factorise 2πβ8. Write the expression as the sum of the answers to a) and b). π‘ πβ4 +2(πβ4) Note that (πβ4) is now a common factor. Factorising again gives: (πβ4)(π‘+2)
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Factorise π π₯βπ‘π¦+π‘π₯βπ π¦
Write the expression as the sum of a) and b). Note that 3 2+π₯ +π(π₯+2) both have (π₯+2). (π₯+2 and 2+π₯ are the same thing mathematically). Factorising gives (π₯+2)(3+π)
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Factorise π π₯βπ‘π¦+π‘π₯βπ π¦
Factorise each of the following: ππβ3πβππ+3π π₯π+π₯π+π¦π+π¦π 10π₯π¦+14π₯+15π¦+21 10π₯π§+65π₯+8π¦π§β52π¦ 20π₯π¦+9+12π₯+15π¦ π π₯+π¦ βπ₯βπ¦
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Factorise π π₯βπ‘π¦+π‘π₯βπ π¦
2003 OL P1 Q5 Factorise ππβ2ππ+3πβ6π 2004 OL P1 Q5 Factorise 3π₯β3π¦+ππ₯βππ¦ 2005 OL P1 Q5 Factorise ππ+2ππ+5π+10π 2006 OL P1 Q5 Factorise ππ₯βππ¦+ππ₯βππ¦ 2016 OL P1 Q8 Factorise 3ππ₯+ππ¦+3ππ₯+ππ¦
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Factorise π π₯ 2 +ππ₯ e.g. Factorise 2 π₯ 2 +5π₯
What is the highest common factor? 2 π₯ 2 π₯ =2π₯ 5π₯ π₯ =5 Factorising gives: 2 π₯ 2 +5π₯=π₯ 2π₯+5
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Factorise π π₯ 2 +ππ₯ e.g. Factorise 6 π₯ 2 +8π₯
What is the highest common factor? 6 π₯ 2 2π₯ =3π₯ 8π₯ 2π₯ =4 Factorising gives: 6 π₯ 2 +8π₯=2π₯(3π₯+4)
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Factorise π π₯ 2 +ππ₯ Factorise each of the following: 3 π₯ 2 +7π₯
8 π₯ 2 β4π₯ 6 π₯ 2 +24π₯ 6 π₯ 2 +10π₯ 4 π¦ 2 +3π¦ π 4 β3 π 2 6 π₯ 2 +8π₯+12π¦π₯
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Factorise π₯ 2 +ππ₯+π These expressions are called quadratic trinomials. We will cover two methods for factorising quadratic trinomials. Firstly, guide number. e.g. Factorise π₯ 2 +7π₯+12
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Factorise π₯ 2 +ππ₯+π Identify π (the guide number) and set it aside.
factor pairs of π are sets of two numbers which multiply to get π Identify π (the guide number) and set it aside. List out factor pairs of π.
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Factorise π₯ 2 +ππ₯+π Add up the numbers in each factor pair.
Note which pair add up to π.
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Factorise π₯ 2 +ππ₯+π Replace π with the two factors you picked.
Factorise by grouping.
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Factorise π₯ 2 +ππ₯+π Secondly, big X. e.g. Factorise π₯ 2 +7π₯+12
Draw a big X.
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Factorise π₯ 2 +ππ₯+π On each of the left ends, put an π₯.
factor pairs of π are sets of two numbers which multiply to get π On each of the left ends, put an π₯. On the right ends, put a factor pair of π. π₯ 1 e.g. Factorise π₯ 2 +7π₯+12 π₯ 12
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Factorise π₯ 2 +ππ₯+π Multiply down each line of the X. Add the results.
1 1π₯ +12π₯ 13π₯ e.g. Factorise π₯ 2 +7π₯+12 π₯ 12
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Factorise π₯ 2 +ππ₯+π If the sum matches π, youβve found your factors.
If not, try other factors until you get π. π₯ 1 π₯ 2 (π₯ + 3) e.g. Factorise π₯ 2 +7π₯+12 π₯ 12 π₯ 6 (π₯ + 4) 1π₯ +12π₯ 13π₯ 2π₯ +6π₯ 8π₯ 3π₯ +4π₯ 7π₯ So π₯ 2 +7π₯+12=(π₯+3)(π₯+4)
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Factorise π₯ 2 +ππ₯+π Factorise each of the following: π₯ 2 +10π₯+24
π 2 +12π+27 π¦ 2 +8π¦+15 π 2 +6π+8 π 2 +9π+14 π 2 +13π+36 π§ 2 +5π§+6 π§ 2 +7π§+6 10π+25+ π 2
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Factorise π₯ 2 +ππ₯+π Be careful if some signs are negative:
e.g. Factorise π₯ 2 βπ₯β12 = π₯ 2 +3π₯β4π₯β12 =π₯ π₯+3 β4 π₯+3 =(π₯+3)(π₯β4) Aside βππ 1+β12 β11 2+β6 β4 3+β4 β1 4+β3 1 6+β2 4 12+β1 11 Guide number / grouping (π₯ + 3) 3π₯ β4π₯ βπ₯ Big X (π₯ β 4)
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Factorise π₯ 2 +ππ₯+π Factorise each of the following: π₯ 2 β2π₯β15
π 2 +2πβ3 π¦ 2 β6π¦+8 π 2 β9π+20 π 2 +πβ6 π§ 2 +3π§β18 π§ 2 β3π§β18 π§ 2 β9π§+18 42β13π+ π 2
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Factorise π₯ 2 +ππ₯+π Factorise π₯ 2 +2π₯β8 Factorise π₯ 2 +2π₯β15
2003 OL P1 Q5 Factorise π₯ 2 +2π₯β8 2005 OL P1 Q5 Factorise π₯ 2 +2π₯β15 2007 OL P1 Q5 Factorise π₯ 2 βπ₯β90 2008 OL P1 Q5 Factorise π₯ 2 β2π₯β24 2017 OL P1 Q11 Factorise π₯ 2 +4π₯β5. One of the factors is (π₯+5).
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Factorise π₯ 2 β π 2 When an expression can be written as a square minus a square, (i.e. π₯ 2 β π¦ 2 ) it is called a difference of two squares. The factorisation can be written simply (π₯βπ¦)(π₯+π¦). e.g. π₯ 2 β π 2 = π₯βπ π₯+π e.g. π 2 β π 2 =(πβπ)(π+π) e.g β 5 2 = 7β = =24 If the two terms are not squares, they must be turned into squares first: e.g. π₯ 2 β16= π₯ 2 β 4 2 =(π₯β4)(π₯+4)
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Factorise π₯ 2 β π 2 Factorise each of the following: π₯ 2 β4 π¦ 2 β9
π§ 2 β25 π 2 β121 64β π 2 10 2 β 3 2 4.2 2 β 3.8 2
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Factorise π₯ 2 β π 2 Factorise 36β π¦ 2 Factorise π₯ 2 β25
2003 OL P1 Q5 Factorise 36β π¦ 2 2004 OL P1 Q5 Factorise π₯ 2 β25 2005 OL P1 Q5 Factorise π₯ 2 β π¦ 2 2006 OL P1 Q5 Factorise π 2 β36 2015 OL P1 Q9 Factorise π₯ 2 β25
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