Factoring (6x^2 - 5x + 1) /(2x^2-x)

[wikiks]

Hello, regarding this problem I came to conclusion that it must do factoring and pull out common terms. I get stuck on (2x(3x-1) - (3x -1)) /(2x^2 - x). The answer supposed to be (3x-1)/( x). Could come one help me with this? <3

 

[wikiks]

Oh also I got 6x^2-5x+1, I got it by splitting 5x into 2x-3x. Then just put 2x outside so it equals 2x(3x-1) and the rest of the problem 3x - 1 I just put in (..). 2x^2-x I could rewrite as 2x(x-1). And it coul cancel out 2x and could rewrite (3x-1) as 3(x-1) and it would cancel out the rest of the problem, but I have to leave /(x) as the solution suggest

 

[blamocur]

Oh also I got 6x^2-5x+1, I got it by splitting 5x into 2x-3x. Then just put 2x outside so it equals 2x(3x-1) and the rest of the problem 3x - 1 I just put in (..). 2x^2-x I could rewrite as 2x(x-1). And it coul cancel out 2x and could rewrite (3x-1) as 3(x-1) and it would cancel out the rest of the problem, but I have to leave /(x) as the solution suggest

I think you've done the opposite of what might help, i.e. representing the numerator of the fraction as a product of two functions. You've done this for the denominator, so you should be able to solve it after this hint. Good luck!

 

[topsquark]

Oh also I got 6x^2-5x+1, I got it by splitting 5x into 2x-3x. Then just put 2x outside so it equals 2x(3x-1) and the rest of the problem 3x - 1 I just put in (..). 2x^2-x I could rewrite as 2x(x-1). And it coul cancel out 2x and could rewrite (3x-1) as 3(x-1) and it would cancel out the rest of the problem, but I have to leave /(x) as the solution suggest

[imath]2x^2 - x \neq 2x(x-1)[/imath] Multiply it out to check!

You have a good idea (though it would be simpler to just factor the numerator.) [imath]2x^2 - x = x(2x - 1)[/imath] so see if you can find a way to split your numerator into terms with a common 2x - 1.

-Dan

 

[Harry_the_cat]

2x(3x-1) - (3x-1) = 2x(3x-1) - 1(3x-1)

(3x-1) is a common factor.

So, 2x(3x-1) - 1(3x-1) = (3x-1)(2x-1).

Now, factorise the denominator correctly and you will be able to cancel.

 

[pka]

Hello, regarding this problem I came to conclusion that it must do factoring and pull out common terms. I get stuck on (2x(3x-1) - (3x -1)) /(2x^2 - x). The answer supposed to be (3x-1)/( x). Could come one help me with this? <3

[imath]\dfrac{6x^2-5x+1}{2x^2-x}=\dfrac{(3x-1)(2x-1)}{x(2x-1)}[/imath]

 

[Steven G]

Oh also I got 6x^2-5x+1, I got it by splitting 5x into 2x-3x.

two comments here:
1) It is -5x, not 5x.
2) 2x-3x is not -5x. What is it?

 

[Zelda22]

Can you write 6x2+(−2−3)x+ 1 /2x^2-x
then distributy property
6x2−2x−3x+1 /2x^2-x

then group the first two terms, to factor out 2x
(6x2−2x)−3x+1 /2x^2−x

2x(3x−1)−(3x−1) / x(2x−1) (factor out x)
factor out 3x-1
(3x−1)(2x−1) /x(2x−1)
simplify
3x−1 /x

 

[topsquark]

Can you write 6x2+(−2−3)x+ 1 /2x^2-x
then distributy property
6x2−2x−3x+1 /2x^2-x

then group the first two terms, to factor out 2x
(6x2−2x)−3x+1 /2x^2−x

2x(3x−1)−(3x−1) / x(2x−1) (factor out x)
factor out 3x-1
(3x−1)(2x−1) /x(2x−1)
simplify
3x−1 /x

No! You need parenthesis! And to square x you write it as x^2.

Can you write (6x^2+(−2−3)x+ 1) /(2x^2-x)
then distributy property

(6x^2−2x−3x+1)/(2x^2-x)
then group the first two terms, to factor out 2x

( (6x^2−2x)−3x+1 ) /(2x^2−x)
( 2x(3x−1)−(3x−1) ) /( x(2x−1) ) (factor out x)
factor out 3x-1
(3x−1)(2x−1) /( x(2x−1) )
simplify
(3x−1) /x

-Dan

 

[Steven G]

Can you write 6x2+(−2−3)x+ 1 /2x^2-x
then distributy property
6x2−2x−3x+1 /2x^2-x

then group the first two terms, to factor out 2x
(6x2−2x)−3x+1 /2x^2−x

2x(3x−1)−(3x−1) / x(2x−1) (factor out x)
factor out 3x-1
(3x−1)(2x−1) /x(2x−1)
simplify
3x−1 /x

What you wrote is just so awful, that it is painful to look at. You really need to learn to use brackets!

6x2−2x−3x+1 /2x^2-x = \(\displaystyle 6x^2 - 2x - 3x +\dfrac{1}{2}x^2-x\)

(3x−1)(2x−1) /x(2x−1) \(\displaystyle \dfrac{(3x−1)(2x−1)}{x}(2x-1)\)

3x−1 /x = \(\displaystyle 3x - \dfrac{1}{x}\) ...................does NOT follow from line above