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Gcse Maths: Factorise Quadratics With Leading Coefficient 1

How to Factorise Quadratics with Leading Coefficient 1

Date : 14/11/2018

Author Information

Nguyen Chuong

Uploaded by : Nguyen Chuong
Uploaded on : 14/11/2018
Subject : Maths

Key Concept:

To factorise a quadratic of the form:

x2 + bx + c


write it as


(x + r1) (x + r2)


where c = r1 r2 and b = r1 + r2

Eight Worked Examples

Question 1

Factorise

x2 + 9x + 14

Solution to Question 1

Comparing x2 + 9x + 14 with x2 + bx + c, you ll see that

c = 14, so you need to find a pair of factors with a product (r1 r2) of 14.

b = 9, so you need to find a pair of factors with a sum (r1 + r2) of 9.

Since c (the product) is positive (14) and b (the sum) is positive (9), you need both factors r1 and r2 to be positive.

Now list the possible factor pairs with a product of 14, and find the pair with a sum of 9.


c = r1 r2 = 14 b = r1 + r2 = 9
1 14 = 14 1 + 14 = 15 2 7 = 14 2 + 7 = 9

The factors 2 and 7 have sum of 9 and product of 14. So use r1 = 2 and r2 = 7. Put these numbers in the brackets:


(x + r1) (x + r2)


(x + 2) (x + 7)


Lets check our answer:


(x + 2) (x + 7)


x2 + 2x + 7x + 14 using distributive rule (FOIL)

x2 + 9x + 14


So x2 + 9x + 14 = (x + 2) (x + 7)

Question 2

Factorise

x2 + 5x + 6

Solution to Question 2

Comparing x2 + 5x + 6 with x2 + bx + c, you ll see that

c = 6, so you need to find a pair of factors with a product (r1 r2) of 6.

b = 5, so you need to find a pair of factors with a sum (r1 + r2) of 5.

Since c (the product) is positive (6) and b (the sum) is positive (5), you need both factors r1 and r2 to be positive.

Now list the possible factor pairs with a product of 6, and find the pair with a sum of 5.

c = r1 r2 = 6 b = r1 + r2 = 5
1 6 = 6 1 + 6 = 7 2 3 = 6 2 + 3 = 5

The factors 2 and 3 have sum of 5 and product of 6. So use r1 = 2 and r2 = 3. Put these numbers in the brackets:

(x + r1) (x + r2)


(x + 2) (x + 3)


Lets check our answer:


(x + 2) (x + 3)


x2 + 2x + 3x + 6 using distributive rule (FOIL)


x2 + 5x + 6

So x2 + 5x + 6 = (x + 2) (x + 3)

Question 3

Factorise

x2 - 10x + 9

Solution to Question 3

Comparing x2 - 10x + 9 with x2 + bx + c, you ll see that

c = 9, so you need to find a pair of factors with a product (r1 r2) of 9.

b = -10, so you need to find a pair of factors with a sum (r1 + r2) of -10.

Since c (the product) is positive (9) and b (the sum) is negative (-10), you need both factors r1 and r2 to be negative.

Now list the possible factor pairs with a product of 9, and find the pair with a sum of -10.

c = r1 r2 = 9 b = r1 + r2 = -10
-1 (-9) = 9 -1 + (-9) = -10
-3 (-3) = 9 -3 + (-3) = -6
The factors -1 and -9 have sum of -10 and product of 9. So use r1 = -1 and r2 = -9. Put these numbers in the brackets:


(x + r1) (x + r2)


(x - 1) (x - 9)


Lets check our answer:


(x - 1) (x - 9)


x2 - x - 9x + 9 using distributive rule (FOIL)

x2 - 10x + 9


So x2 - 10x + 9 = (x - 1) (x - 9)

Question 4

Factorise

x2 - 17x + 30

Solution to Question 4

Comparing x2 - 17x + 30 with x2 + bx + c, you ll see that

c = 30, so you need to find a pair of factors with a product (r1 r2) of 30.

b = -17, so you need to find a pair of factors with a sum (r1 + r2) of -17.

Since c (the product) is positive (30) and b (the sum) is negative (-17), you need both factors r1 and r2 to be negative.

Now list the possible factor pairs with a product of 9, and find the pair with a sum of -10.

c = r1 r2 = 30 b = r1 + r2 = -17
-1 (-30) = 30 -1 + (-30) = -31 -2 (-15) = 30 -2 + (-15) = -17

The factors -2 and -15 have sum of -17 and product of 30. So use r1 = -2 and r2 = -30. Put these numbers in the brackets:


(x + r1) (x + r2)


(x - 2) (x - 15)


Lets check our answer:


(x - 2) (x - 15)


x2 - 2x - 15x + 30 using distributive rule (FOIL)

x2 - 17x + 30


So x2 - 17x + 30 = (x - 2) (x - 15)

Question 5

Factorise

x2 + 16x - 17

Solution to Question 5

Comparing x2 + 16x 17 with x2 + bx + c, you ll see that

c = -17, so you need to find a pair of factors with a product (r1 r2) of -17.

b = 16, so you need to find a pair of factors with a sum (r1 + r2) of 16.

Since c (the product) is negative (-17) and b (the sum) is positive (16), you need one factor to be negative and the other to be positive.

Now list the possible factor pairs with a product of -17, and find the pair with a sum of 16.

c = r1 r2 = -17 b = r1 + r2 = 16
1 (-17) = -17 1 + (-17) = -16 -1 17 = -17 -1 + 17 = 16

The factors -1 and 17 have sum of 16 and product of -17. So use r1 = -1 and r2 = 17. Put these numbers in the brackets:


(x + r1) (x + r2)


(x - 1) (x + 17)


Lets check our answer:


(x - 1) (x + 17)


x2 - x + 17x - 17 using distributive rule (FOIL)

x2 +16x - 17


Therefore x2 +16x 17 = (x - 1) (x + 17)

Question 6

Factorise

x2 + 7x - 78

Solution to Question 6

Comparing x2 + 7x 78 with x2 + bx + c, you ll see that

c = -78, so you need to find a pair of factors with a product (r1 r2) of -78.

b = 7, so you need to find a pair of factors with a sum (r1 + r2) of 7.

Since c (the product) is negative (-78) and b (the sum) is positive (7), you need one factor to be negative and the other to be positive.

Now list the possible factor pairs with a product of -78, and find the pair with a sum of 7.

c = r1 r2 = -78 b = r1 + r2 = 7
1 (-78) = -78 1 + (-78) = -77 2 (-39) = -78 2 + (-39) = -37 3 (-26) = -78 3 + (-26) = -23 6 (-13) = -78 6 + (-13) = -7 -6 (13) = -78 -6 + 13 = 7

The factors -6 and 13 have sum of 7 and product of -78. So use r1 = -6 and r2 = 13. Put these numbers in the brackets:


(x + r1) (x + r2)


(x - 6) (x + 13)


Lets check our answer:


(x - 6) (x + 13)


x2 - 6x + 13x - 78 using distributive rule (FOIL)

x2 + 7x - 78


Therefore x2 + 7x 78 = (x - 6) (x + 13)

Question 7

Factorise

x2 - 2x - 24


Solution to Question 7

Comparing x2 - 2x 24 with x2 + bx + c, you ll see that

c = -24, so you need to find a pair of factors with a product (r1 r2) of -24.

b = -2, so you need to find a pair of factors with a sum (r1 + r2) of -2.

Since c (the product) is negative (-24) and b (the sum) is negative (-2), you need one factor to be negative and the other to be positive.

Now list the possible factor pairs with a product of -24, and find the pair with a sum of -2.

c = r1 r2 = -24 b = r1 + r2 = -2
1 (-24) = -24 1 + (-24) = -23 2 (-12) = -24 2 + (-12) = -10 3 (-8) = -24 3 + (-8) = -5 4 (-6) = -24 4 + (-6) = -2

The factors 4 and -6 have sum of -2 and product of -24. So use r1 = 4 and r2 = -6. Put these numbers in the brackets:


(x + r1) (x + r2)


(x + 4) (x - 6)


Lets check our answer:


(x + 4) (x - 6)


x2 + 4x 6x - 24 using distributive rule (FOIL)

x2 - 2x - 24


So x2 - 2x 24 = (x + 4) (x - 6)

Question 8

Factorise

x2 - 6x 27

Solution to Question 8

Comparing x2 - 6x 27 with x2 + bx + c, you ll see that

c = -27, so you need to find a pair of factors with a product (r1 r2) of -27.

b = -6, so you need to find a pair of factors with a sum (r1 + r2) of -6.

Since c (the product) is negative (-27) and b (the sum) is negative (-6), you need one factor to be negative and the other to be positive.

Now list the possible factor pairs with a product of -27, and find the pair with a sum of -6.

c = r1 r2 = -27 b = r1 + r2 = -6
1 (-27) = -27 1 + (-27) = -26 3 (-9) = -27 3 + (-9) = -6

The factors 3 and -9 have sum of -6 and product of -27. So use r1 = 3 and r2 = -9. Put these numbers in the brackets:


(x + r1) (x + r2)


(x + 3) (x - 9)


Lets check our answer:


(x + 3) (x - 9)


x2 + 3x 9x - 27 using distributive rule (FOIL)

x2 - 6x 27

So x2 - 6x 27 = (x + 3) (x - 9)

This resource was uploaded by: Nguyen Chuong