Question

2)
3)
Write the simplest form of am X an.
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Answer 1

Answer:

bunch :)

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Sep 5 '16 at 18:10

Zochonis

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Sep 5 '16 at 18:22

Do you mean 2(x-1)(2x-1) on the second denominator? – Barbosa Sep 5 '16 at 18:16

Your use of parentheses is inconsistent. Is the denominator of the second fraction supposed to be 2(x−1)(2x−1)? – N. F. Taussig Sep 5 '16 at 18:17

No, that's the expanded form – Zochonis Sep 5 '16 at 18:19

Take a look at the second denominator in the first line. Is there a parenthesis missing? – N. F. Taussig Sep 5 '16 at 18:21

Oh, yeah fixed it – Zochonis Sep 5 '16 at 18:22

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Anytime you want to operate fractions like that you need them to have the same denominator. You can get that easily if the denominator is presented as the product of factor, like in your problem.

So

4x−12(x−1)−32(x−1)(2x−1)

is almost ready for the subtraction, all we need is for the first fraction to have a (2x-1) on the denominator, just like the second fraction. But we can get that if we multiple the first fraction by this missing factor like this

(4x−1)(2x−1)2(x−1)(2x−1)−32(x−1)(2x−1)

(multiplying the numerator and denominator of a fraction by the same number doesn't change its value since it is the same as multiplying by one, 2x−12x−1=1).

Now we are ready to operate with the fractions

(4x−1)(2x−1)2(x−1)(2x−1)−32(x−1)(2x−1)=(4x−1)(2x−1)−32(x−1)(2x−1)

Simplifying

(4x−1)(2x−1)−32(x−1)(2x−1)=2(4x2−3x−1)2(x−1)(2x−1)

The solutions for 4x2−3x−1=0 are x=1 and x=−14. Thus,

2(4x2−3x−1)=2(x−1)(x+1/4)=1/2(x−1)(4x+1)

Finally

2(4x2−3x−1)2(x−1)(2x−1)=1/2(x−1)(4x+1)2(x−1)(2x−1)=(4x+1)4(2x−1)