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Question

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

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Solution

## Step I - Let's find the factors of the numerator $4{x}^{2}-4x+1$.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,$4{x}^{2}-2x-2x+1$Step III - Now let's take the common monomial from the first two terms and then the last two terms. We get,$2x\left(2x-1\right)-1\left(2x-1\right)$Here, $\left(2x-1\right)$is common monomial in each term.So this can be written as $4{x}^{2}-2x-2x+1$ = $\left(2x-1\right)$$\left(2x-1\right)$Step IV - Put the factors of the numerator and divide them by their denominator.$\frac{4{x}^{2}-4x+1}{2x-1}$ = $\frac{\left(2x-1\right)\left(2x-1\right)}{\left(2x-1\right)}$Now, cancelling out $\left(2x-1\right)to\left(2x-1\right)$. We get $\left(2x-1\right)$Therefore, the factors of $\frac{4{x}^{2}-4x+1}{2x-1}$ = $\left(2x-1\right)$.

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