Math help please Reducing a rational expression.?

Question

I have an example in my book, but I don't understand how they got the answer.
6x^5(x^2 + 2)^2 - 4x^3(x^2 +2)^3 this is the numerator.
The denominator is x^8.
Then it shows that it is worked out to this:
Numerator is 2x^3(x^2 +2)^2[3x^2 - 2(x^2 + 2)]
denominator is still x^8.
I do not understand how they get the 2x^3 and where the 6x^5 is. Please explain. The final answer to the problem is
numerator is 2(x^2 +2)^2(x-2)(x+2)
denominator is x^5
Please just explain to me this problem.
Thank you so much.

Answer 1

What can you factor out of 6x^5 and 4x^3?

You can factor out a 2 and x^3.

What can you factor out of (x^2+2)^2 and (x^2+2)^3?

You can factor out (x^2+2)^2.

When you factor out 2x^3(x^2+2)^2 out of each of the two terms, what do you have left? Divide both terms by 2x^3(x^2+2)^2 and what you have left is 3x^2-2(x^2+2)

After factoring out an x^3 from the top and having x^8 at the bottom, you know that the three x's on top have to cancel with three x's in the bottom, so at the bottom, you only have five x's left which is x^5.

Answer 2

Well if you remember that x^5=x^2*x^3 (when multiply x^power you add the power numbers)
then taking out the factors 6x^5=2*3*x^3*x^2 and 4x^3=2*2*x^3
You will notice that both have a factor of 2 and a factor of x^3
So you can factorise the equation by taking out a factor 2x^3 from both sides. So the 6x^5 has now become 2x^3(3x^2)

If look at 2x^3 and x^8 you notice they both have a factor of x^3 (x^8=x^5*x^3)
So you can divide both the top and bottom by x^3 which is why the denominator now become x^3 and in the numerator you've got rid of the x^3
So the numerator is now 2[(x^2+2)^2*3x^2-2(x^2+2)^3]
Take out a factor of (x^2+2)^2 from both sides 2[(x^2+2)^2{3x^2-2(x^2+2)}]
Multiply out the {} and refactorise to get the final answer

I'm sure by the time I finished typing this some one may have explained this better

Answer 3

Firstly, if you divide out of each expression in the numerator 2xcubed, express the numerator as 2x^3[3x^2(x^2+2)^2-2(x^2+2)^3}
Second step is to divide through by (x^2+2)^2
Then express the numerator as
2x^3(x^2+2)^2[3x@-2(x^2+2)]
I hope that this helps.

Answer 4

Look at the two terms being subtracted from each other. What's their biggest common factor?

Well, 2 is the biggest common factor of 6 and 4.
x^3 is the biggest common factor of x^3 and x^5.
(x^2+2)^2 is the biggest common factor of the rest.

They just multiplied those three factors.

Answer 5

a million. (((2 / 5y) - (3 / 15y)) / 2) / ((4 / 7) + (7 / 15y)) initiate via fixing the numerator. Multiply (2 / 5y) * (3 / 3), to get (6 / 15y). you on the instantaneous have: (((6 / 15y) - (3 / 15y)) / 2) / ((4 / 7) + (7 / 15y)) ((3 / 15y) / 2) / ((4 / 7) + (7 / 15y)) (3 / 30y) / ((4 / 7) + (7 / 15y)) (a million / 10y) / ((4 / 7) + (7 / 15y)) Now paintings on the denominator. Multiply (4 / 7) via (15y/15y) and (7 / 15y) via (7/7). (a million / 10y) / ((60y / 105y) + (40 9 / 105y)) (a million / 10y) / ((60y + 40 9) / 105y) Multiply the outer fraction via 10y / 10y. a million / ((10y (60y + 40 9)) / 105y) because we've a million interior the numerator, we are in a position to rewrite the outer fraction through utilising truth the reciprocal of the denominator. 105y / (10y (60y + 40 9)) 105y / (600y^2 + 490y) Divide via 5y/5y. 21 / (120y + ninety 8) finally, you're in a position to favor to ingredient out a 2 interior the denominator in case you want. 21 / (2(60y + 40 9)) 2. ((4 / 2z) + (5 / 6z)) / (7 / ((3 / z) - (4 / 6z))) Multiply (4 / 2z) via three/3 and (3 / z) via 6/6. ((12 / 6z) + (5 / 6z)) / (7 / ((18 / 6z) - (4 / 6z))) (17 / 6z) / (7 / (14 / 6z)) Multiply the denominator via (6z / 14) / (6z / 14) (17 / 6z) / (7 (6z / 14)) (17 / 6z) / (42z / 14) (17 / 6z) / 3z 17 / (18z^2)