Question

Factorise $\frac{4x^2-4x+1}{2x-1}$.

Answer 1

Step I - Let's find the factors of the numerator $4x^2-4x+1$.

Find the two factors of 4 for which 4 should come in adding and 4 in multiplication.

So the two factors are 2 and 2.

Step II - Now, split the middle term. We get,

$4x^2-2x-2x+1$

Step III - Now let's take the common monomial from the first two terms and then the last two terms. We get,

$2x(2x-1)-1(2x-1)$

Here, $(2x-1)$is common monomial in each term.

So this can be written as $4x^2-2x-2x+1$ = $(2x-1)$$(2x-1)$

Step IV - Put the factors of the numerator and divide them by their denominator.

$\frac{4x^2-4x+1}{2x-1}$ = $\frac{(2x-1)(2x-1)}{(2x-1)}$

Now, cancelling out $(2x-1)to(2x-1)$. We get $(2x-1)$

Therefore, the factors of $\frac{4x^2-4x+1}{2x-1}$ = $(2x-1)$.