Week 4

Question 1

-5m(2n - 3m) … Can variables be distributed when simplifying?

Answer 1

Yes: -10mn + 15m^2

Question 2

(6c + 12d) - (10c - 5d) Solution to coefficient of “c” signed or unsigned? Why?

Answer 2

Subtraction is not Commutative thus: (6c) - (10c) = -4c

Question 3

(-12) - (-4) means: (+4) OR (-4) ?

Answer 3

+4 double - turns + (-12) + (-4) is the same as (-4)

Question 4

2n(3n+ 1)(4n+ 2) make into quadratic

Answer 4

Expand the brackets and times by 2n as a whole after = 2n[12n^2+ 6n+ 4n+ 2] = 2n[12n^2+ 10n+ 2] = 24n^3+ 20n^2+ 4n Also can mult by 2n first

Question 5

(2c+ 3)(c−1)(c+ 2) Use foil

Answer 5

Foil first two then foil for remaining two = (2c^2−2c+ 3c−3)(c+ 2) = (2c^2+c−3)(c+ 2) = 2c^3+ 4c^2+c^2+ 2c−3c−6 = 2c^3+ 5c^2−c−6

Question 6

13x−(5x−2)(1−x)

Answer 6

keep parans until fully resolved arrange parans into quadratic = 13x −(5x −5x^2 −2 + 2x) = 13x −(−5x^2 + 7x −2) = 13x + 5x^2 −7x + 2 = 5x^2 + 6x + 2

Question 7

(√7y)(√7y)

Answer 7

Root is negated variable stacks = 7y^2

Question 8

-√7y -√7y

Answer 8

Surds stack -2√7y

Question 9

(4−2a)(4 + 2a) does foil work?

Answer 9

Difference of two squares Foil only gets incomplete answer, must add the index if foil doesnt get that = 4^2−(2a)^2 = 16−4a^2

Question 10

−(x−9)(x+ 9) What does - do?

Answer 10

Minus is applied to the signing of each term. = −(x^2−9^2) = −(x^2−81) = −x^2+81

Question 11

1960 Prime factorise

Answer 11

repeated division of lowest factor 2 | 1960 2 | 980 2 | 490 5 | 245 7 | 49 7 | 7 1 stop at prime

Question 12

HCF? 2 | 1960 2 | 1512 2 | 980 2 | 756 2 | 490 2 | 378 5 | 245 3 | 189 7 | 49 3 | 63 7 | 7 3 | 21 1 7 | 7 1

Answer 12

product of like factors 2^3 * 7 = 56

Question 13

4a^3+12a^2 factorise

Answer 13

4a^2(a+3)

Question 14

8-2n^2 factorise

Answer 14

2(4-n^2) factor out 2(2-n)(2+n) DOTS

Question 15

0.01-x^2 factorise

Answer 15

0.1^2 = 0.01 thus (0.1-x)(0.1+x)

Question 16

4nm^2-256n factorise

Answer 16

4n(m^2-64) factor out 4n(m-8)(m+8) DOTS

Question 17

11-k^2 factorise

Answer 17

root to surd (√11+k)(√11-k)

Question 18

2n^2+14n+20 factorise

Answer 18

Before using crossfire first the coefficient must be factored out 2(n^2+7n+10) then crossfire 2(n+5)(n+2)

Question 19

-x^2-8x-12 factorise

Answer 19

-(x^2-8x-12) factor out crossfire since theres subtraction at head, the answers +6 & +2 can be used with the sub preserved: -(x+6)(x+2)

Question 20

4m^2+2m-6 factorise

Answer 20

2(2m^2+m-3) crossfire 2(2m+3)(m-1)

Question 21

y^4−28y^2+ 75 factorise

Answer 21

To make this easier to factorise, let A=y^2 = A^2−28A+ 75 substitute = (A−3)(A−25) crossfire = (y^2−3)(y^2−25) unsubstitute = (y^2−3)(y+ 5)(y−5) DOTS

Question 22

(a+ 3)^2+ 8(a+ 3) + 12

Answer 22

= B^2+ 8B+ 12 = (B+ 2)(B+ 6) = ((a+ 3) + 2)((a+ 3) + 6) = (a+ 5)(a+ 9)

Question 23

y-1/5 + y+3/2

Answer 23

= 2y-2/10 + 5y+15/10 = 7y+13/10

Question 24

2c/(c+5) + 4/(c-1)

Answer 24

Q. 2c/(c+5) + 4/(c-1) = 2c(c-1)/(c+5)(c-1) + 4(c+5)/(c+5)(c-1) = 2c(c-1)+4(c+5)/(c+5)(c-1) = 2c^2-2c+4c+20/(c-1)(c+5) = 2c^2+2c+20/(c-1)(c+5) This is complete, quad wont factorise further

Question 25

n+1/n+2 - 2n+1/3n-2 factorise

Answer 25

Q. n+1/n+2 - 2n+1/3n-2 = (n+1)(3n-2)/(n+2)(3n-2) - (2n+1)(n+2)/(n+2)93n-2) = (n+1)(3n-2) - (2n+1)(n+2)/(n+2)(3n-2) = 3n^2-2n+3n-2-(2n^2+4n+n+2)/(n+2)(3n-2) = 3n^2+n-2-2n^2-5n-2/(n+2)(3n-2) = n^2-4n-4/(n+2)(3n-2)

Question 26

x/25 * 10/y

Answer 26

Q: x/25 * 10/y = x/5 * 2/y = 2x/5y

Question 27

20/21a * 7c/10

Answer 27

Q: 20/21a * 7c/10 = 2/3a * c/1 = 2c/3a

Question 28

3nm/5 / 9n/10

Answer 28

note: variable n cancels Q: 3nm/5 / 9n/10 = 3nm/5 * 10/9n = m/1 * 2/3 = 2m/3

Question 29

Q: x-7/1-x^2 / 2x-14/x+1

Answer 29

Q: x-7/1-x^2 / 2x-14/x+1 = x-7/1-x^2 * x+1/2x-14 = x-7/(1+x)(1-x) * x+1/2(x-7) = 1/1-x * 1/2 = 1/2(1-x)

Question 30

−3(a+ 6)

Answer 30

−3(a+ 6)

= −3a−18

Question 31

2(2h+ 4k)

Answer 31

2(2h+ 4k)

= 4h+ 8k

Question 32

−5m(2n−3m)

Answer 32

−5m(2n−3m)

=−10mn+ 15m²

Question 33

4(x+ 2y) + 2(2x−y)

Answer 33

4(x+ 2y) + 2(2x−y)

= 4x+ 8y+ 4x−2y

= 8x+ 6y

Question 34

6(c+ 2d)−5(2c+d)

Answer 34

6(c+ 2d)−5(2c+d)

= 6c+ 12d−10c−5d

=−4c+ 7d

Question 35

2a+b−3(a−2b)

Answer 35

2a+b−3(a−2b)

= 2a+b−3a+ 6b

= −a+ 7b

Question 36

−2(n−4)−(3n+ 1)

Question 37

x(2x−3) + 4(x²+ 2)

Answer 37

x(2x−3) + 4(x²+ 2)

= 2x²−3x+ 4x²+ 8

= 6x²−3x+ 8

Question 38

2a(5a−1) + 3(4a+ 2)

Answer 38

2a(5a−1) + 3(4a+ 2)

= 10a²−2a+ 12a+ 6

= 10a²+ 10a+ 6

Question 39

2g(1−2g)−5(g−2)

Answer 39

2g(1−2g)−5(g−2)

= 2g−4g²−5g+ 10

=−4g²−3g+ 10

Question 40

4(x−3)−2(x−2)

Answer 40

4(x−3)−2(x−2)

= 4x−12−2x+ 4

= 2x−8

Question 41

12(n−3) +n(−2−2n)

Answer 41

12(n−3) +n(−2−2n)

= 12n−36−2n−2n²

=−2n²+ 10n−36

Question 42

(a+ 4)²

Answer 42

(a+ 4)²

= (a+ 4)(a+ 4)

=a²+ 4a+ 4a+ 16

=a²+ 8a+ 16

Alternatively, this could be expanded by

i. Squaring the first term: a²
ii. Twice the product of the terms: 2×a×4 = 8a
iii. Squaring the second term: 4²= 16

Question 43

(2x−3)²

Answer 43

(2x−3)²

= (2x−3)(2x−3)

= 4x²−6x−6x+ 9

= 4x²−12x+ 9

Question 44

(y−√7)²

Answer 44

(y−√7)²

= (y−√7)(y−√7)

=y²−y√7−y√7 + 7

=y²−2y√7 + 7

Alternatively, this can also be written as y²−2√7y+ 7.

Note that in the second term,the square root sign is only over the 7 and not they. When writing this by hand,make sure that it is clear what is under the square root sign.

Question 45

(n+ 6)(n−6)

Answer 45

(n+ 6)(n−6)

=n²−6n+ 6n−36

=n²−36

Question 46

(4−2a)(4 + 2a)

Answer 46

(4−2a)(4 + 2a)

= 16 + 8a−8a−4a²

= 16−4a²

Alternatively, this can also be written as −4a²+ 16

Question 47

−(x−9)(x+ 9)

Answer 47

−(x−9)(x+ 9)

=−(x²+ 9x−9x−81)

=−(x²−81)

=−x²+ 81

Question 48

(x−4)(x−5)

Answer 48

(x−4)(x−5)

= x²−5x−4x+ 20

= x²−9x+ 20

Question 49

(2y+ 7)(4y−1)

Answer 49

(2y+ 7)(4y−1)

= 8y²−2y+ 28y−7

= 8y²+ 26y−7

Question 50

(3b+ 8)(10b−3)

Answer 50

(3b+ 8)(10b−3)

= 30b²−9b+ 80b−24

= 30b²+ 71b−24

Question 51

(7h−2)(3h−5)

Answer 51

(7h−2)(3h−5)

= 21h²−35h−6h+ 10

= 21h²−41h+ 10

Question 52

(a−2)(a+ 5)

Answer 52

(a−2)(a+ 5)

=a²+ 5a−2a−10

=a²+ 3a−10

Question 53

(2x−1)(3x+ 4)

Answer 53

(2x−1)(3x+ 4)

= 6x²+ 8x−3x−4

= 6x²+ 5x−4

Question 54

2n(3n+ 1) (4n+ 2)

Answer 54

[2n(3n+ 1)] (4n+ 2)

= 6n²+ 2n

= 24n³+ 12n²+ 8n²+ 4n

= 24n³+ 20n²+ 4n

or

2n[(3n+ 1)(4n+ 2)]

= 2n[12n²+ 6n+ 4n+ 2]

= 2n[12n²+ 10n+ 2]

= 24n³+ 20n²+ 4n

Question 55

−5y(3−y)(y−2)

Answer 55

−5y(3−y)(y−2)

=−5y(3y−6−y²+ 2y)

=−5y(−y²+ 5y−6)

= 5y³−25y²+ 30y

Question 56

(2c+ 3)(c−1)(c+ 2)

Answer 56

(2c+ 3)(c−1)(c+ 2)

= (2c²−2c+ 3c−3)(c+ 2)

= (2c²+c−3)(c+ 2)

= 2c³+ 4c²+c²+ 2c−3c−6

= 2c³+ 5c²−c−6

Question 57

13x−(5x−2)(1−x)

Answer 57

13x−(5x−2)(1−x)

= 13x−(5x−5x²−2 + 2x)

= 13x−(−5x²+ 7x−2)

= 13x+ 5x²−7x+ 2

= 5x²+ 6x+ 2

Question 58

(8 + 9h)(2h−3)

Answer 58

(8 + 9h)(2h−3)

= 16h−24 + 18h²−27h

= 18h²−11h−24

Question 59

−2a(a+ 3)(a−3)

Answer 59

−2a(a+ 3)(a−3)

= −2a(a²−3a+ 3a−9)

= −2a(a²−9)

= −2a³+ 18a

Question 60

x²−8x

Answer 60

x²−8x

= x(x−8)

Question 61

4a³+ 12a²

Answer 61

4a³+ 12a²

= 4a²(a+ 3)

Question 62

3a²b−15ab²

Answer 62

3a²b−15ab²

= 3ab(a−5b)

Question 63

y²−9

Answer 63

y²−9 = (y+ 3)(y−3)

(Difference of two squares)

Question 64

8−2n²

Answer 64

8−2n²

= 2(4−n²)

= 2(2 +n)(2−n)

Question 65

1−x²

Answer 65

1−x²

= (1 +x)(1−x)

Question 66

4nm²−256n

Answer 66

4nm²−256n

= 4n(m²−64)

= 4n(m+ 8)(m−8)

Question 67

5x²−80

Answer 67

5x²−80

= 5(x²−16)

= 5(x+ 4)(x−4)

Question 68

x²+ 5x+ 6

Answer 68

x²+ 5x+ 6

= (x+ 2)(x+ 3)

cross method

Question 69

a²−11a−12

Answer 69

a²−11a−12

= (a−12)(a+ 1)

cross

Question 70

y²−9y+ 20

Answer 70

y²−9y+ 20

= (y−4)(y−5)

cross

Question 71

2n²+ 14n+ 20

Answer 71

2n²+ 14n+ 20

= 2(n²+ 7n+ 10)

= 2(n+ 5)(n+ 2)

Note that while the cross method has been used, the first step was to factorise usinga common factor. This made the cross method much easier.

Question 72

3k²+ 33k+ 30

Answer 72

3k²+ 33k+ 30

= 3(k²+ 11k+ 10)

= 3(k+ 1)(k+ 10)

cross

Question 73

−x²−8x−12

Answer 73

−x²−8x−12

=−(x²+ 8x+ 12)

=−(x+ 2)(x+ 6)

As all terms were negative, a common factor of −1 was first introduced. This allowed for the cross method to be used where all terms were positive.

Question 74

5c²+ 105c+ 100

Answer 74

5c²+ 105c+ 100

= 5(c²+ 21c+ 20)

= 5(c+ 1)(c+ 20)

Question 75

4m²+ 2m−6

Answer 75

4m2+ 2m−6

= 2(2m²+m−3)

= 2(2m+ 3)(m−1)

Question 76

y⁴−28y²+ 75

Answer 76

We have been asked to factorise y⁴−28y²+ 75. This question could be done several different ways. To make it easier to factorise,

let A=y²

y⁴−28y²+ 75

= A²−28A+ 75

= (A−3)(A−25)

= (y²−3)(y²−25)

= (y²−3)(y+ 5)(y−5)

Question 77

(a+ 3)²+ 8(a+ 3) + 12

Answer 77

We have been asked to factorise (a+ 3)²+ 8(a+ 3) + 12. To make it easier to factorise,let B= (a+ 3)

(a+ 3)²+ 8(a+ 3) + 12

=B²+ 8B+ 12

= (B+ 2)(B+ 6)

= ((a+ 3) + 2)((a+ 3) + 6)

= (a+ 5)(a+ 9)

Alternatively, you could expand, simplify, and then factorise.

(a+ 3)²+ 8(a+ 3) + 12

=a²+ 6a+ 9 + 8a+ 24 + 12

=a²+ 14a+ 45

= (a+ 5)(a+ 9)

Question 78

Answer 78

Question 79

Answer 79

Question 80

Answer 80

Question 81

Answer 81

Question 82

Answer 82

Question 83

Answer 83

Question 84

Answer 84

Question 85

Answer 85

Question 86

Answer 86

Question 87

Answer 87