help please :)

Question

Reduce these complex fractions as much as possible (answer all please):

 

A) {(3/2x) - (4/5x)} / {(1/3) - (2/x)}

 

B) {(1/x-2) - (3)} / {(5/x+2) + (3)}

 

C) Divide 3x^3 + 2x^2 -4x + 2 by x-1

 

D) Divide 4x^3 -2x by x+2

Answer 1

Is it (3/2)x or 3/(2x)? is it (4/5)x or 4/(5x)?

Problem A: assuming the latter

A) {3/(2x) - 4/(5x)}/ { 1/3 - 2/x } <--- common denominator is 30x;

multiplies everything by 30x

={3 *15 - 4*6}/ { 10x - 2*30 } <--- 30x/2x = 15; 30x/5x=6; 30x/3=10x ; 30x/x = 30

=(45-24)/(10x- 60) =

=21/(10x-60) = 21/ [ 10(x-6)]

assuming the former:

A) {(3/2)x - (4/5)x } / {(1/3) - (2/x)} <-- common denominator again is 30x;

what a co-incidence! multiplies everything by 30x

{ 3*15x*x - 4*6x*x}/ { 10x - 2*30} <-- 30x/2=15x; 30x/5 = 6x; 30x/3 = 10x; 30x/x= 30

{ 45x^2 - 24x^2} / (10x - 60) =

21x^2 / (10x-60) = 21x^2 / [10(x-6)]

notice the answers are almost the same; it is because of the x in the denominator in the

first version of the problem

Problem B:

Again, is it (1/x) -2 or 1/(x-2) ??

is it (5/x) + 2 or 5/(x+2)???

assuming the latter:

B: [1/(x-2) - 3 ] / [ 5/(x+2) + 3] <--- common denominator is (x+2)(x-2)=x^2-4; multiplies EVERYTHING

by (x+2)(x-2)=x^2-4

[x+2 - 3(x^2-4)]/ [ 5(x-2) + 3(x^2-4)] <--- x-2 cancels in the numerator's left fraction;

x+2 cancels in the denominator's left fraction;

[x+2-3x^2+12]/[5x-10 + 3x^2-12]

(-3x^2+x+14)/(3x^2+5x-22)=

-(3x^2-x-14)/(3x^2+5x-22) =

-(3x - 7)(x +2 ) /(3x + 11)(x - 2 ) <--- no common factors :-(

assuming the former:

the problem simplifies:

[(1/x)-5] /[ (5/x) + 5] <-- common denominator is x

(1 -5x)/(5 + 5x) <--- 1/x*x=1; 5/x*x = 5

(1-5x)/(5(1+x)) <--- again, no common factors

C) remainder theorem says, the remainder is, for x=1 and x-1 as a factor, is

3(1)^3 + 2(1)^2 - 4(1) +2 = 3+2-4+2 = 3, so x=1 is NOT a solution

3x^2 + 5x +1

____________________________

x-1 | 3x^3 + 2x^2 - 4x + 2

3x^3 - 3x^2

-------------------------

5x^2 - 4x

5x^2 - 5x

-----------------------

x + 2

x - 1

--------

3

So 3x^2 + 5x + 1 + 3/(x-1) =

[(3x^2+5x+1)(x-1) + 3 ] /(x-1) =

[3x^3 + 5x^2 + x - 3x^2 - 5x - 1+3] / (x-1)

[3x^3 + 2x^2 - 4x + 2]/(x-1) <--- same problem, so it checks

D) remainder theorem says, for x=-2 and x-2 as a factor, is

4(-2)^3 - 2(-2) = 4(-8) + 4 = -32+4 = -28, so x=-2 is NOT a solution

4x^2 -8x+14

_____________________

x+2 | 4x^3 + 0x^2- 2x + 0

4x^3 + 8x^2

-------------------------

-8x^2 - 2x

-8x^2 -16x

-------------------

14x

14x + 28

-----------

-28

(x+2)(4x^2-8x+14)-28 = 4x^3 - 8x^2 + 14x

8x^2 -16x + 28-28

4x^3-2x which is the original dividend, so it checks