Answer:
3x−1
Explanation:
1 Split the second term in 6{x}^{2}+x-16x
2
+x−1 into two terms.
\frac{6{x}^{2}+3x-2x-1}{4{x}^{2}-1}
4x
2
−1
6x
2
+3x−2x−1
2 Factor out common terms in the first two terms, then in the last two terms.
\frac{3x(2x+1)-(2x+1)}{4{x}^{2}-1}
4x
2
−1
3x(2x+1)−(2x+1)
3 Factor out the common term 2x+12x+1.
\frac{(2x+1)(3x-1)}{4{x}^{2}-1}
4x
2
−1
(2x+1)(3x−1)
4 Rewrite 4{x}^{2}-14x
2
−1 in the form {a}^{2}-{b}^{2}a
2
−b
2
, where a=2xa=2x and b=1b=1.
\frac{(2x+1)(3x-1)}{{(2x)}^{2}-{1}^{2}}
(2x)
2
−1
2
(2x+1)(3x−1)
5 Use Difference of Squares: {a}^{2}-{b}^{2}=(a+b)(a-b)a
2
−b
2
=(a+b)(a−b).
\frac{(2x+1)(3x-1)}{(2x+1)(2x-1)}
(2x+1)(2x−1)
(2x+1)(3x−1)
6 Cancel 2x+12x+1.
\frac{3x-1}{2x-1}
2x−1
3x−1