O-Level Additional Math

[jieheng]

A concerned mom:

Can anyone help to solve the following Sec 2 questions?


1) Factorise the following expression:

a) 4(x + y)^2 – (1 - x)^2

b) 4x^4 – 12x^2 + 9

c) 12x^2 y^3 – 27x^4 y z^2


2) Factorise each of the following:

a) x^2 – 4 + 2x^3 + 4x^2

b) 6cd – 16d - 8c + 12d^2

c) 4c^2 d^2 - 49d^2 - 14cd - 49d

d)16c^2 – 16cd + 4d^2 – 49c^4

Note: ^ denotes \"to the power of\"

Appreciate your help

1)

a)
4(x + y)^2 – (1 - x)^2
=[2(x + y)]^2 - (1 - x)^2
=[2(x + y) + (1 - x)] [2(x + y) - (1 - x)]
=(x + 2y + 1) (3x + 2y - 1)

b)
4x^4 – 12x^2 + 9
=(2x^2 - 3 )^2

c)
12x^2 y^3 – 27x^4 y z^2
=3x^2 y (4y^2 - 9x^2 z^2)
=3x^2 y [(2y)^2 - (3xz)^2]
=3x^2 y (2y + 3xz) (2y - 3xz)

2)

a)
x^2 – 4 + 2x^3 + 4x^2
=2x^3 + 5x^2 - 4
=(x + 2) (2x^2 + x - 2)

b)
6cd – 16d - 8c + 12d^2
=2c(3d - 4) + 4d(3d -4)
=(3d - 4)(2c + 4d)

c)
4c^2 d^2 - 49d^2 - 14cd - 49d
=d^2 [(2c)^2 - (7)^2] - 7d(2c + 7)
=d^2 (2c + 7)(2c - 7) - 7d(2c + 7)
=(2c + 7)[d^2 (2c - 7) - 7d]
=d (2c + 7)(2cd - 7d - 7)

d)
16c^2 – 16cd + 4d^2 – 49c^4
=(4c)^2 - 2(4c)(2d) + (2d)^2 - (7c^2)^2
=(4c - 2d)^2 - (7c^2)^2
=(4c - 2d + 7c^2) (4c - 2d - 7c^2)

[A concerned mom]

Thanks a lot Mr Jie Heng for your solutions!! ^^

[mrsoon]

awesomeguy:

Sec 2 Express Math Qn Proportion


For an adventure camp, food is bought to last for 15 days for a group of 40 students. \tIf 4 students left the camp after 6 days, how many days can the food last for the \trem6aining 36 students?

Please help. Thanks

Inverse Proportion,
Days(x) = 15 - 6 = 9
Students(y) = 40
xy = k
9 x 40 = k
k = 360

xy = 360

when y = 36, x(36) = 360
x = 10 (ans)

[red rose]

jieheng:

red rose:

Would appreciate help for these 2 Qs.


1)

f(x) = 3x^2 -9x +2

f(p) = f(-4q)
3p^2 -9p +2 = 3(-4q)^2 -9(-4q) +2
3p^2 -9p - 48q^2 -36q =0
p^2 -3p - 16q^2 - 12q = 0
p^2 - (4q)^2 - 3(p+4q) = 0
(p+4q)(p-4q) - 3(p+4q) = 0
(p+4q)[(p-4q)-3] = 0
p-4q-3 = 0 as p is not equal to (-4q)
p-4q = 3 (Ans)

2)

Let (Ax+B) be the remainder when f(x) is divided by (x^2+3x+2)

f(x) = (x^2+3x+2) Q(x) + (Ax+B)
= (x+1)(x+2) Q(x) + (Ax+B)

f(-1) = A(-1) + B = 3

f(-2) = A(-2) + B = 5

Solving the two equations ,

A = -2 and B = 1

the remainder = -2x + 1 (Ans)

Thank you so much for your help, Jieheng.:) Would really appreciate your help with another question.

Given that (x+2) is a factor of the expression f(x)=3x^3+hx^2-kx-10, and f(x) leaves a remainder of -4 when divided by (x+1). Show that h=8 and k=1 (I can do this part but need help for parts a and b)

(a) If f(x)=(x+2)(x-2)Q(x)+Ax+B, where Q(x) is a polynomial and A, B are constants, find the remainder when f(x) is divided by (x^2-4).

(b)(i) Find the values of x when f(x)=0
(ii) Hence solve the equation 81x^3+72x^2=3x+10

Thanks in advance!

[jieheng]

red rose:

Thank you so much for your help, Jieheng.:) Would really appreciate your help with another question.

Given that (x+2) is a factor of the expression f(x)=3x^3+hx^2-kx-10, and f(x) leaves a remainder of -4 when divided by (x+1). Show that h=8 and k=1 (I can do this part but need help for parts a and b)

(a) If f(x)=(x+2)(x-2)Q(x)+Ax+B, where Q(x) is a polynomial and A, B are constants, find the remainder when f(x) is divided by (x^2-4).

(b)(i) Find the values of x when f(x)=0
(ii) Hence solve the equation 81x^3+72x^2=3x+10

Thanks in advance!

a)

f(x)=3x^3+8x^2-x-10

f(-2)=0 as (x+2) is a factor

f(2)=3(2)^3+8(2)^2-(2)-10 = 44

f(x)=(x+2)(x-2)Q(x)+Ax+B

f(-2) = -2A + B = 0

f(2) = 2A + B = 44

Solving these two equations ,

A= 11 and B= 22

the remainder is 11x+22

b)i)

f(x)= 3x^3+8x^2-x-10=0
(x+2)(3x^2+2x-5)=0
(x+2)(3x+5)(x-1)=0
x= -2 , -5/3 or 1

b)ii)

81x^3+72x^2=3x+10
81x^3+72x^2-3x-10=0
3(3x)^3+8(3x)^2-(3x)-10=0
[(3x)+2][3(3x)+5][(3x)-1]=0
x= -2/3 , -5/9 or 1/3

[A concerned mom]

Hi Jie Heng,


Sorry for the trouble and appreciate if you could further enlighten me on the answer
=(x + 2) (2x^2 + x - 2) => how did you manage to get this?


2)a) x^2 – 4 + 2x^3 + 4x^2
=2x^3 + 5x^2 - 4
=(x + 2) (2x^2 + x - 2)

[jieheng]

A concerned mom:

Hi Jie Heng,


Sorry for the trouble and appreciate if you could further enlighten me on the answer
=(x + 2) (2x^2 + x - 2) => how did you manage to get this?


2)a) x^2 – 4 + 2x^3 + 4x^2
=2x^3 + 5x^2 - 4
=(x + 2) (2x^2 + x - 2)

Let f(x) = 2x^3 + 5x^2 - 4 = (ax + b) (Ax^2 + Bx + C)

i)To find a linear factor (ax + b)

The constant term of [f(x) = 2x^3 + 5x^2 - 4] is 4 and the factors of 4 are 1 , 2 and 4

We need to test whether x= 1, -1 , 2 , -2 , 4 and -4 are the solutions of f(x)=0

When x= 1 , f(1) = 2(1)^3 + 5(1)^2 -4 = 2 + 5 - 4 =3

When x= -1 , f(-1) = 2(-1)^3 + 5(-1)^2 -4 = -2 + 5 - 4 =-1

When x= 2 , f(2)= 2(2)^3 + 5(2)^2 -4 = 16 + 20 - 4 =32

When x= -2 , f(-2)= 2(-2)^3 + 5(-2)^2 -4 = -16 + 20 -4 = 0

As f(-2) = 0 , do not need to test f(4) and f(-4)
x= -2
x+2 = 0
(x+2) is a factor of f(x)


ii)To find a quadratic equation (Ax^2 + Bx + C)

f(x) = 2x^3 + 5x^2 - 4 = (x + 2) (Ax^2 + Bx + C)
2x^3 + 5x^2 - 4 = Ax^3 + BX^2 + Cx + 2Ax^2 + 2Bx + 2C
2x^3 + 5x^2 - 4 = Ax^3 + (2A+B)x^2 + (2B+C)x + 2C

Comparing the coefficients of x^3 , A =2

Comparing the coefficients of the constant term , 2C = -4 => C = -2

Comparing the coefficients of x^2 , 2A+B = 5 => 2(2) + B = 5 => B = 1

Ax^2 + Bx + C = 2x^2 + x -2

f(x) = 2x^3 + 5x^2 - 4 = (x + 2) (Ax^2 + Bx + C)
= (x + 2) (2x^2 + x -2)

This method will be taught in A Maths in Sec 3 ; Factor Theorem of Polynomials .

[A concerned mom]

Thank you Jie Heng for your great help!!^^

I think i need to study the syllabus of Sec 3 Add. Maths before analysing the answer.

[Valerie Teoh]

Any tips on doing A Maths Plane Geometry proving questions?

[JadeDry]

I tried solving these questions to no avail, could you please help me?



PQRS is a trapezium where PQ is parallel to SR. Given that the angle PQR = SPR = 90 degrees, PQ = 8.5 and PR = 12.3 calculate

A) PS

B) The area of the trapezium PQRS

And the second question-

ABCD is a trapezium, the angle ABC = the angle BCD = 90 degrees. Given that AB = 8 and, CD = 15, AD = 27 and BD = (3x - 4), calculate the value of X, giving your answer correct to two decimal places.

Thanks in advance!