Question

How do you factor \[64{{x}^{2}}-16x+1\]?

Answer 1

Hint: This type of problem is based on the concept of factoring a polynomial. First, we have to consider the polynomial with degree 2. We have to split the middle term of the polynomial in such a way that we get common terms from the first and last term. Here, the middle term is -16x. We can split the middle as an addition of -8x twice. Then, we need to take 8x common from the first two terms and -1 common from the last two terms. We find that (8x-1) is common in the two terms. On taking the common terms, we get the factors of the polynomial.

Complete step by step solution:
According to the question, we are asked to find the factors of \[64{{x}^{2}}-16x+1\].
We have been given the polynomial is \[64{{x}^{2}}-16x+1\]. ---------(1)
The given polynomial is of degree 2 and variable x.
To find the factors of the polynomial, we have to consider the middle term and split the middle term in such a way that the addition of these two terms will be -16 and multiplication will be 64.
We know that -8-8=-16 and \[-8\times -8=64\].
Therefore, we get
\[64{{x}^{2}}-16x+1=64{{x}^{2}}+\left( -8-8 \right)x+1\]
Using the distributive property in the middle term, that is \[a\left( b+c \right)=ab+ac\].
Here, a=x, b=-8 and c=-8.
\[\Rightarrow 64{{x}^{2}}-16x+1=64{{x}^{2}}+\left( -8 \right)x+\left( -8 \right)x+1\]
\[\Rightarrow 64{{x}^{2}}-16x+1=64{{x}^{2}}-8x-8x+1\]
We can express the polynomial as
\[64{{x}^{2}}-16x+1={{8}^{2}}{{x}^{2}}-8x-8x+1\]
We find that 8x are common in the first two terms of the simplified polynomial and -1 are common in the last two terms of the simplified polynomial.
Let us take 8x and -1 common out of the bracket respectively.
\[\Rightarrow 64{{x}^{2}}-16x+1=8x\left( 8x-1 \right)-1\left( 8x-1 \right)\]
We find that 8x-1 is common in both the terms of the equation. On taking (8x-1) common, we get
\[64{{x}^{2}}-16x+1=\left( 8x-1 \right)\left( 8x-1 \right)\]
Here, we find that the polynomial is converted as a product of two identical linear polynomials which are the factors of the given polynomial.
Therefore, the factors of \[64{{x}^{2}}-16x+1\] are 8x-1 and 8x-1.

Note: We can also find the factors in other methods.
We can express the polynomial as \[{{8}^{2}}{{x}^{2}}-2\times 8x+1\].
\[\Rightarrow 64{{x}^{2}}-16x+1={{8}^{2}}{{x}^{2}}-2\times 8x+1\]
On further simplification, we get
\[64{{x}^{2}}-16x+1={{\left( 8x \right)}^{2}}-2\times 8x\times 1+{{\left( -1 \right)}^{2}}\]
This is of the form \[{{a}^{2}}-2ab+{{b}^{2}}\] where a=8x, b=1.
We know that \[{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}\].
Therefore, we get
\[64{{x}^{2}}-16x+1={{\left( 8x-1 \right)}^{2}}\]
Hence, the factors are 8x-1 and 8x-1.