Week 4 Flashcards

1
Q

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

A

Yes: -10mn + 15m^2

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2
Q

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

A

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

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3
Q

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

A

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

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4
Q

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

A

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

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5
Q

(2c+ 3)(c−1)(c+ 2) Use foil

A

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

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6
Q

13x−(5x−2)(1−x)

A

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

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7
Q

(√7y)(√7y)

A

Root is negated variable stacks = 7y^2

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8
Q

-√7y -√7y

A

Surds stack -2√7y

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9
Q

(4−2a)(4 + 2a) does foil work?

A

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

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10
Q

−(x−9)(x+ 9) What does - do?

A

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

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11
Q

1960 Prime factorise

A

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

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12
Q

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

A

product of like factors 2^3 * 7 = 56

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13
Q

4a^3+12a^2 factorise

A

4a^2(a+3)

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14
Q

8-2n^2 factorise

A

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

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15
Q

0.01-x^2 factorise

A

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

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16
Q

4nm^2-256n factorise

A

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

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17
Q

11-k^2 factorise

A

root to surd (√11+k)(√11-k)

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18
Q

2n^2+14n+20 factorise

A

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

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19
Q

-x^2-8x-12 factorise

A

-(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)

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20
Q

4m^2+2m-6 factorise

A

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

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21
Q

y^4−28y^2+ 75 factorise

A

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

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22
Q

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

A

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

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23
Q

y-1/5 + y+3/2

A

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

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24
Q

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

A

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

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25
n+1/n+2 - 2n+1/3n-2 factorise
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)
26
x/25 \* 10/y
Q: x/25 \* 10/y = x/5 \* 2/y = 2x/5y
27
20/21a \* 7c/10
Q: 20/21a \* 7c/10 = 2/3a \* c/1 = 2c/3a
28
3nm/5 / 9n/10
note: variable n cancels Q: 3nm/5 / 9n/10 = 3nm/5 \* 10/9n = m/1 \* 2/3 = 2m/3
29
Q: x-7/1-x^2 / 2x-14/x+1
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)
30
−3(a+ 6)
−3(a+ 6) = −3a−18
31
2(2h+ 4k)
2(2h+ 4k) = 4h+ 8k
32
−5m(2n−3m)
−5m(2n−3m) =−10mn+ 15m2
33
4(x+ 2y) + 2(2x−y)
4(x+ 2y) + 2(2x−y) = 4x+ 8y+ 4x−2y = 8x+ 6y
34
6(c+ 2d)−5(2c+d)
6(c+ 2d)−5(2c+d) = 6c+ 12d−10c−5d =−4c+ 7d
35
2a+b−3(a−2b)
2a+b−3(a−2b) = 2a+b−3a+ 6b = −a+ 7b
36
−2(n−4)−(3n+ 1)
37
x(2x−3) + 4(x2+ 2)
x(2x−3) + 4(x2+ 2) = 2x2−3x+ 4x2+ 8 = 6x2−3x+ 8
38
2a(5a−1) + 3(4a+ 2)
2a(5a−1) + 3(4a+ 2) = 10a2−2a+ 12a+ 6 = 10a2+ 10a+ 6
39
2g(1−2g)−5(g−2)
2g(1−2g)−5(g−2) = 2g−4g2−5g+ 10 =−4g2−3g+ 10
40
4(x−3)−2(x−2)
4(x−3)−2(x−2) = 4x−12−2x+ 4 = 2x−8
41
12(n−3) +n(−2−2n)
12(n−3) +n(−2−2n) = 12n−36−2n−2n2 =−2n2+ 10n−36
42
(a+ 4)2
(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
43
(2x−3)2
(2x−3)2 = (2x−3)(2x−3) = 4x2−6x−6x+ 9 = 4x2−12x+ 9
44
(y−√7)2
(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.
45
(n+ 6)(n−6)
(n+ 6)(n−6) =n2−6n+ 6n−36 =n2−36
46
(4−2a)(4 + 2a)
(4−2a)(4 + 2a) = 16 + 8a−8a−4a2 = 16−4a2 Alternatively, this can also be written as −4a2+ 16
47
−(x−9)(x+ 9)
−(x−9)(x+ 9) =−(x2+ 9x−9x−81) =−(x2−81) =−x2+ 81
48
(x−4)(x−5)
(x−4)(x−5) = x2−5x−4x+ 20 = x2−9x+ 20
49
(2y+ 7)(4y−1)
(2y+ 7)(4y−1) = 8y2−2y+ 28y−7 = 8y2+ 26y−7
50
(3b+ 8)(10b−3)
(3b+ 8)(10b−3) = 30b2−9b+ 80b−24 = 30b2+ 71b−24
51
(7h−2)(3h−5)
(7h−2)(3h−5) = 21h2−35h−6h+ 10 = 21h2−41h+ 10
52
(a−2)(a+ 5)
(a−2)(a+ 5) =a2+ 5a−2a−10 =a2+ 3a−10
53
(2x−1)(3x+ 4)
(2x−1)(3x+ 4) = 6x2+ 8x−3x−4 = 6x2+ 5x−4
54
2n(3n+ 1) (4n+ 2)
[2n(3n+ 1)] (4n+ 2) = [6n2+ 2n](4n+ 2) = 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
55
−5y(3−y)(y−2)
−5y(3−y)(y−2) =−5y(3y−6−y2+ 2y) =−5y(−y2+ 5y−6) = 5y3−25y2+ 30y
56
(2c+ 3)(c−1)(c+ 2)
(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
57
13x−(5x−2)(1−x)
13x−(5x−2)(1−x) = 13x−(5x−5x2−2 + 2x) = 13x−(−5x2+ 7x−2) = 13x+ 5x2−7x+ 2 = 5x2+ 6x+ 2
58
(8 + 9h)(2h−3)
(8 + 9h)(2h−3) = 16h−24 + 18h2−27h = 18h2−11h−24
59
−2a(a+ 3)(a−3)
−2a(a+ 3)(a−3) = −2a(a2−3a+ 3a−9) = −2a(a2−9) = −2a3+ 18a
60
x2−8x
x2−8x = x(x−8)
61
4a3+ 12a2
4a3+ 12a2 = 4a2(a+ 3)
62
3a2b−15ab2
3a2b−15ab2 = 3ab(a−5b)
63
y2−9
y2−9 = (y+ 3)(y−3) | (Difference of two squares)
64
8−2n2
8−2n2 = 2(4−n2) = 2(2 +n)(2−n)
65
1−x2
1−x2 = (1 +x)(1−x)
66
4nm2−256n
4nm2−256n = 4n(m2−64) = 4n(m+ 8)(m−8)
67
5x2−80
5x2−80 = 5(x2−16) = 5(x+ 4)(x−4)
68
x2+ 5x+ 6
x2+ 5x+ 6 = (x+ 2)(x+ 3) cross method
69
a2−11a−12
a2−11a−12 = (a−12)(a+ 1) cross
70
y2−9y+ 20
y2−9y+ 20 = (y−4)(y−5) cross
71
2n2+ 14n+ 20
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.
72
3k2+ 33k+ 30
3k2+ 33k+ 30 = 3(k2+ 11k+ 10) = 3(k+ 1)(k+ 10) cross
73
−x2−8x−12
−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.
74
5c2+ 105c+ 100
5c2+ 105c+ 100 = 5(c2+ 21c+ 20) = 5(c+ 1)(c+ 20)
75
4m2+ 2m−6
4m2+ 2m−6 = 2(2m2+m−3) = 2(2m+ 3)(m−1)
76
y4−28y2+ 75
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)
77
(a+ 3)2+ 8(a+ 3) + 12
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)
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