Answered By Mahboubeh D

   $\frac{10x^3+3x^2-x}{25x^2-1}=\frac{x\left(10x^2+3x-1\right)}{\left(5x-1\right)\left(5x+1\right)}=\frac{x\left(5x-1\right)\left(2x+1\right)}{\left(5x-1\right)\left(5x+1\right)}=\frac{x\left(2x+1\right)}{5x+1}$10x³+3x²−x25x²−1 =x(10x²+3x−1)(5x−1)(5x+1) =x(5x−1)(2x+1)(5x−1)(5x+1) =x(2x+1)5x+1  

Answered By Emily D

 $\frac{10x^3+3x^2-x}{25x^2-1}$10x³+3x²−x25x²−1  

you can factor an x out of the numerator since it's a factor for every term

 $\frac{x\left(10x^2+3x-1\right)}{25x^2-1}$x(10x²+3x−1)25x²−1  

denominator is a difference of squares! We love differences of squares

  $\frac{x\left(10x^2+3x-1\right)}{\left(5x-1\right)\left(5x+1\right)}$x(10x²+3x−1)(5x−1)(5x+1)   

Now is a good time to check if there are any x values that would make the denominator 0 (we don't want to divide by 0)!

  $\left(5x-1\right)\ne0$(5x−1)≠0  ,  $x\ne\frac{1}{5}$x≠15  

  $\left(5x+1\right)\ne0$(5x+1)≠0  ,   $x\ne-\frac{1}{5}$x≠−15   

factor the polynomial that's left in the numerator: we need two numbers that multiply to -10 (10*-1) and a sum to 3 (5 and -2 should work!)

   $\frac{x\left(10x^2+5x-2x-1\right)}{\left(5x-1\right)\left(5x+1\right)}$x(10x²+5x−2x−1)(5x−1)(5x+1)    

we can factor 5x out of the first two terms in the numerator, we can factor -1 out of the last two terms

    $\frac{x\left(5x\left(2x+1\right)-1\left(2x+1\right)\right)}{\left(5x-1\right)\left(5x+1\right)}$x(5x(2x+1)−1(2x+1))(5x−1)(5x+1)   

you can now factor (2x+1) out of both terms now!

  $\frac{x\left(2x+1\right)\left(5x-1\right)}{\left(5x-1\right)\left(5x+1\right)}$x(2x+1)(5x−1)(5x−1)(5x+1)     

the (5x-1) in the numerator and denominator can cancel eachother out - don't forget, we can't let x be 1/5 or -1/5

 $\frac{x\left(2x+1\right)}{5x+1}$x(2x+1)5x+1