# maths problems

Plz help me to solve the following problems and EXPLAIN,thx in advance!

1. find the highest common factor of 4x^3 - 11x^2 +25x +7 and 2x^3 - 5x^2 +11x +7

2. find the lowest common multiple of a^4 + a^2b^2 and a^3 - b^3 and a^3 + b^3

3. the net taxable income is calculated by subtracting the basic allowance,which is \$108000 this year,from the total annual income.It has been announced that next year,the basic allowance would be reduced to \$100000.Eric and Erica said that they had to pay \$960 and \$810 more respectively.

-Use an inequality to represent the possible range of net taxable income of Eric this year.Hence,state Eric's monthly income this year.

-Calculate Erica’s monthly income this year.

-Erica needs to pay rates as she owns a flat.The tax rate on ratable value is 5% this year.If she needs to pay \$10000 in total for 2 types of tax this year,find the ratable value of her flat.

Below is the table for salaries tax,

Net taxable incomes: On the first \$35000 – Rate 2%

On the next \$35000 – Rate 7%

On the next \$35000 – Rate 12%

On the remaining – Rate 17%

Update:

4. The line segment joining A(a,a) and B (b,b) is divided into n equal parts.The coordinates of the rth point of division from A is?(plz tell me what is this question talking about as well ORZ)

Update 2:

5. The line segment joining A(8,1) and B(0,3/5) meets the x-axis at P (p,0).Find the ratio AP:PB.

xin2122 what r u talking about

Update 3:

im so sorry!i got Q.5 wrong.B should be (0,-3.5).i follow ur method but i cannot calculate the correct ans.would u mind to work on it again?

Update 4:

DOES ANYONE KNOW HOW TO DO Q.3?

Rating

Q1. Let the HCF be H

so 4x^3 - 11x^2 + 25x + 7 = HQ(x) where Q(x) is a function of x.

2x^3 - 5x^2 + 11x + 7 = Hf(x) where f(x) is also a function of x.

so (4x^3 - 11x^2 + 25x + 7) - (2x^3 - 5x^2 + 11x + 7) = HQ(x) - Hf(x)

= H[Q(x) - f(x)] = 2x^3 - 6x^2 + 14x = 2x(x^2 -3x + 7).

since both 2 and x are not common factors of the two polynomials, so their HCF is x^2 - 3x + 7.

Q2. LCM is the product of the 3 items since they don't have common factors.

Q4. Let the rth division point be P(x, y).

So number of division of AP = r, number of division of PB = n - r.

Based on the formula of point of division, we get

so x = [rb + a(n -r)]/[r + (n -r)] = (rb + an - ar)/n = y since A is (a,a) and B is (b,b).

Q5.

Equation of line joining AB is

(y - 1)/(3/5 - 1) = (x - 8)/(0 - 8)

(y -1)/(-2/3) = (x -8)/-8

8(y -1) = 2/3(x - 8).

when y = 0, x = p

so -8 = 2/3(p - 8)

-24 = 2p - 16

-8 = 2p, p = - 4.

By similar triangle AP : PB = 8 + 4 : 4 = 12 : 4 = 3 : 1.

2009-03-19 19:47:57 補充：

Correction: Q5 Line 2 (3/5 - 1) should be = -2/5, not -2/3, so the equation should be 8(y-1) = 2/5(x -8), so p should be -12, so AP : PB = 12 + 8 : 12 = 20 : 12 = 5 : 3.

2009-03-19 21:16:03 補充：

Q5 can be solved by a much simpler way, again by using similar triangle but in the y direction, since A is (8,1), B is (0,3/5) and P is (p, 0), so AP : PB = 1 : 3/5

= 5 : 3.