中二數學題

11. If A(x-8)(x+2) = -x^2+Bx+C, find the constants A,B and C

A. A(x-8)(x+2) = -x^2+Bx+C
A(x^2+2x-8x-16) = -x^2+Bx+C
Ax^2-6Ax-16A = -x^2+Bx+C
By comparing the coefficient of like terms,
A = -1
B = -6(-1)
B = 6
C = -16(-1)
C = 16

Expand the following :
13a) (x-6)(x+6)

A. (x-6)(x+6)
= x^2-6^2
= x^2-36

13b) (3s^2+2s)(-2s+3s^2)

A. (3s^2+2s)(-2s+3s^2)
= (3s^2+2s)(3s^2-2s)
= (3s^2)^2-(2s)^2
= 9s^4-4s^2

13c) 6(3x-4/3)(3x+4/2)

A. not clear

21a) (p+10)^2

A. (p+10)^2
= p^2+2(10)p+10^2
= p^2+20p+100

21c) (3c-10d)^2

A. (3c-10d)^2
= (3c)^2-2(3c)(10d)+(10d)^2
= 9c^2-60cd+100d^2

29a) 6(8x+5)^2

A. 6(8x+5)^2
= 6[(8x)^2+2(8x)5+5^2]
= 6(64x^2+80x+25)
= 384x^2+480x+150

29b) [-4(2+uv)]^2

A. [-4(2+uv)]^2
= (-8-4uv)^2
= (-8)^2-2(-8)(4uv)+(4uv)^2
= 64+64uv+16u^2v^2

Factorize the following :
40a) 6ab-4bc+12abc

A. 6ab-4bc+12abc
= 2b(3a-2c+6ac)

40b) 81hk^2 + 9k^2 h^2 - 27h^3 k^3

A. 81hk^2+9k^2h^2-27h^3k^3
= 9hk^2(9+h-3hk)

40c) 9w(2w+1)+4(2w+1)

A. 9w(2w+1)+4(2w+1)
= (2w+1)(9w+4)

40d) pq^2 - p(q-6)^2

A. pq^2-p(q-6)^2
= pq^2-p(q^2-12q+36)
= pq^2-pq^2+12pq-36p
= 12pq-36p
= 12p(q-3)

49a) ab+bc+2a+2c

A. ab+bc+2a+2c
= b(a+c)+2(a+c)
= (a+c)(b+2)

49b) 5xy-10xz+4y-8z

A. 5xy-10xz+4y-8z
= 5x(y-2z)+4(y-2z)
= (y-2z)(5x+4)

59a) x^2 – 400

A. x^2-400
= x^2-(20)^2
= (x+20)(x-20)

59b) 8x^2 y^2 – 72

A. 8x^2y^2-72
= 8(x^2y^2-3^2)
= 8(xy+3)(xy-3)

59c) (a-b)^2 - (2c+1)^2

A. Already the simplest form:
(a-b)^2-(2c+1)^2
= a^2-2ab+b^2-(4c^2+4c+1)
= a^2-2ab+b^2-4c^2-4c-1----------???

59d) u^2 - 4v^2 + 6(u-2v)

A. u^2-4v^2+6(u-2v)
= (u+2v)(u-2v)+6(u-2v)
= (u-2v)[6(u+2v)]
(= (u-2v)(6u+12v))-----This line is optional

2010-02-16 22:50:05 補充：
68a) p^2 + 14p + 49

A. p^2+14p+49
= (p+7)^2

68b) m^2 - 20m + 100

A. m^2-20m+100
= (m-10)^2

68c) 9x^2 + 24x + 16

A. Cannot be factorize
9x^2+24x+16
= 3^2x^2+2(12)x+4^2

2010-02-16 22:50:45 補充：
Simplify the following expressions :
3a) x-y / 3(x-y)^2

A. (x-y)/3(x-y)^2
= 1/3(x-y)
(= 1/(3x-3y))----------This line is optional

3b) a^2 (b+c) / a(b+c)

A. a^2(b+c)/a(b+c)
= a

2010-02-16 22:51:01 補充：
3c) 2a-2b / 4ab 除以 a^2-b^2 / b+a

A. [(2a-2b)/4ab]/[(a^2-b^2)/(b+a)]
= [2(a-b)/4ab]/[(a+b)(a-b)/(b+a)]
= [(a-b)/2ab]/(a-b)
= [(a-b)/2ab]*1/(a-b)
= 1/2ab

16a) 4/a+b + 5/a+b

A. 4/(a+b) + 5/(a+b)
= (4+5)/(a+b)
= 9/(a+b)

2010-02-16 22:51:37 補充：
16b) - 2x/2x+y - y/2x+y

A. - 2x/(2x+y) - y/(2x+y)
= (-2x-y)/(2x+y)
= -(2x+y)/(2x+y)
= -1/1
= -1

16d) 2y/3(x-2y) - x/8(2y-x)

A. 2y/3(x-2y) - x/8(2y-x)
= 2y/3(x-2y) + x/8(x-2y)
= 8(2y)/24(x-2y) + 3x/24(x-2y)
= (16y+3x)/24(x-2y)

2010-02-16 22:51:48 補充：
(PS: I do not copy anyone’s answer. There are some difference between my answer and 001’s answer.)

2010-12-03 20:29:12 補充：
Sorry, but I haven't learnt how to solve this kind of question in the first time...

59c) (a-b)^2 - (2c+1)^2
= (a-b+2c+1)(a-b-2c-1)

2010-12-03 20:31:21 補充：
And I know what you mean now...

13c) 6(3x-4/3)(3x+4/2)
= 2(3x-4)(3x+4 /2)
= (3x-4)(3x+4)
= 9x^2-16

11. A(x-8)(x+2) = -x^2+Bx+C
A(x^2-8x+2x+16) = -x^2+Bx+C
Ax^2-6Ax+16A= -x^2+Bx+C
A = -1
B = -6
C = 16+(-1)
= 15

13a) (x-13b) (3s^2+2s)(-2s+3s^2)
=x^2-(6)^2
=x^2-36

13b) (3s^2+2s)(-2s+3s^2)
=(3s^2+2s)(3s^2-2s)
=(3s^2)^2-(2s)^2
=9s^4-4^2



21a) (p+10)^2
=p^2+2(10p)+10^2
=p^2+20p+100

21c) (3c-10d)^2
=(3c)^2-2(30cd)+(-10d)^2
=9c^2-60cd+20d^2

11. A(x-8)(x+2) = -x^2+Bx+C
A(x^2-8x+2x+16) = -x^2+Bx+C
Ax^2-6Ax+16A= -x^2+Bx+C
A = -1
B = -6
C = 16+(-1)
= 15

13a) (x-13b) (3s^2+2s)(-2s+3s^2)
=x^2-(6)^2
=x^2-36

13b) (3s^2+2s)(-2s+3s^2)
=(3s^2+2s)(3s^2-2s)
=(3s^2)^2-(2s)^2
=9s^4-4^2

13c)有冇打錯字???

21a) (p+10)^2
=p^2+2(10p)+10^2
=p^2+20p+100

21c) (3c-10d)^2
=(3c)^2-2(30cd)+(-10d)^2
=9c^2-60cd+20d^2

(遲d再繼續打)

2010-02-16 14:28:19 補充：
29a) 6(8x+5)^2
=6((8x)^2+2(40x)+5^2)
=6(64x^2+80x+25)
=384x^2+480x+150

29b) [-4(2+uv)]^2
=(-4)^2(2+uv)
=16(4+4uv+u^2v^2) (恆等式:(a+b)^2=(a^2+2ab+b^2))
=64+64uv+6u^2v^2

40a) 6ab-4bc+12abc
=2b(3a-2c+6ac)

40b) 81hk^2 + 9k^2 h^2 - 27h^3 k^3
=9hk^2(9+h-3hk)

遲d再繼續打

2010-02-16 14:31:51 補充：
40c) 9w(2w+1)+4(2w+1)
=(2w+1)(9w+4)

40d) pq^2 - p(q-6)^2
=pq^2-p(q^2-12q-36) (skip左一步) (恆等式:(a-b)^2=(a^2-2ab+b^2))
=pq^2-pq^2-12pq+36p (pq^2-pq^2約左)
=-12p(q-3)

49a) ab+bc+2a+2c
=(ab+bc)+(2a+2c)
=b(a+c)+2(a+c)
=(a+c)(b+2)

2010-02-16 14:39:05 補充：
49b) 5xy-10xz+4y-8z
=(5xy-10xz)+(4y-8z)
=5x(y-2z)+4(y-2z)
=(y-2z)(5x+4)

答完咪又係刪....