
How do you factor \[128{{x}^{3}}-1024\]?
Answer
547.8k+ views
Hint: This question is from the topic of algebra. We will find the factor of \[128{{x}^{3}}-1024\]. For solving this question, we will first find out the prime factorization of 128. After that, we will find out the prime factorization of 1024. After solving further, we will take out a common term. After that, we will use the formula of \[{{a}^{3}}-{{b}^{3}}\] which can be written as \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]. After that, we will get our answer.
Complete step by step answer:
Let us solve this question.
In this question, we have asked to find out the factor of given term which is given in the question.
The term which we have to solve is
\[128{{x}^{3}}-1024\]
Let us first find the prime factorization of 128.
\[\begin{align}
& 2\left| \!{\underline {\,
128 \,}} \right. \\
& 2\left| \!{\underline {\,
64 \,}} \right. \\
& 2\left| \!{\underline {\,
32 \,}} \right. \\
& 2\left| \!{\underline {\,
16 \,}} \right. \\
& 2\left| \!{\underline {\,
8 \,}} \right. \\
& 2\left| \!{\underline {\,
4 \,}} \right. \\
& 2\left| \!{\underline {\,
2 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align}\]
So, we can say that the prime factorization of 128 will be \[2\times 2\times 2\times 2\times 2\times 2\times 2\times 1\].
Now, let us find out the prime factorization of 1024.
\[\begin{align}
& 2\left| \!{\underline {\,
1024 \,}} \right. \\
& 2\left| \!{\underline {\,
512 \,}} \right. \\
& 2\left| \!{\underline {\,
256 \,}} \right. \\
& 2\left| \!{\underline {\,
128 \,}} \right. \\
& 2\left| \!{\underline {\,
64 \,}} \right. \\
& 2\left| \!{\underline {\,
32 \,}} \right. \\
& 2\left| \!{\underline {\,
16 \,}} \right. \\
& 2\left| \!{\underline {\,
8 \,}} \right. \\
& 2\left| \!{\underline {\,
4 \,}} \right. \\
& 2\left| \!{\underline {\,
2 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align}\]
We can say that the prime factorization of 1024 will be \[2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 1\]
So, the term \[128{{x}^{3}}-1024\] can also be written as
\[128{{x}^{3}}-1024=2\times 2\times 2\times 2\times 2\times 2\times 2\times 1\times {{x}^{3}}-2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 1\]
Taking \[2\times 2\times 2\times 2\times 2\times 2\times 2\times 1\] as common in the above equation, we can write
\[\Rightarrow 128{{x}^{3}}-1024=\left( 2\times 2\times 2\times 2\times 2\times 2\times 2\times 1 \right)\times \left( {{x}^{3}}-2\times 2\times 2 \right)\]
The above equation can also be written as
\[\Rightarrow 128{{x}^{3}}-1024=128\left( {{x}^{3}}-2\times 2\times 2 \right)\]
As we know that \[2\times 2\times 2={{2}^{3}}\], so we can write the above equation as
\[\Rightarrow 128{{x}^{3}}-1024=128\left( {{x}^{3}}-{{2}^{3}} \right)\]
Now, using the formula \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\], we can write the above equation as
\[\Rightarrow 128{{x}^{3}}-1024=128\left( x-2 \right)\left( {{x}^{2}}+2\times x+{{2}^{2}} \right)\]
The above equation can also be written as
\[\Rightarrow 128{{x}^{3}}-1024=128\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)\]
Now, we have found the factor of the term \[128{{x}^{3}}-1024\]. The factor of the term \[128{{x}^{3}}-1024\] is \[128\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)\].
Note: As we can see that this question is from the topic of algebra, so we should have a better knowledge in the topic of algebra. For solving this type of question, we should know how to find the prime factorization of any number. We should remember the following formula:
\[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]
Remember that the term \[x\times x\times x........n\text{ times}\] can also be written as \[{{x}^{n}}\].
Complete step by step answer:
Let us solve this question.
In this question, we have asked to find out the factor of given term which is given in the question.
The term which we have to solve is
\[128{{x}^{3}}-1024\]
Let us first find the prime factorization of 128.
\[\begin{align}
& 2\left| \!{\underline {\,
128 \,}} \right. \\
& 2\left| \!{\underline {\,
64 \,}} \right. \\
& 2\left| \!{\underline {\,
32 \,}} \right. \\
& 2\left| \!{\underline {\,
16 \,}} \right. \\
& 2\left| \!{\underline {\,
8 \,}} \right. \\
& 2\left| \!{\underline {\,
4 \,}} \right. \\
& 2\left| \!{\underline {\,
2 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align}\]
So, we can say that the prime factorization of 128 will be \[2\times 2\times 2\times 2\times 2\times 2\times 2\times 1\].
Now, let us find out the prime factorization of 1024.
\[\begin{align}
& 2\left| \!{\underline {\,
1024 \,}} \right. \\
& 2\left| \!{\underline {\,
512 \,}} \right. \\
& 2\left| \!{\underline {\,
256 \,}} \right. \\
& 2\left| \!{\underline {\,
128 \,}} \right. \\
& 2\left| \!{\underline {\,
64 \,}} \right. \\
& 2\left| \!{\underline {\,
32 \,}} \right. \\
& 2\left| \!{\underline {\,
16 \,}} \right. \\
& 2\left| \!{\underline {\,
8 \,}} \right. \\
& 2\left| \!{\underline {\,
4 \,}} \right. \\
& 2\left| \!{\underline {\,
2 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align}\]
We can say that the prime factorization of 1024 will be \[2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 1\]
So, the term \[128{{x}^{3}}-1024\] can also be written as
\[128{{x}^{3}}-1024=2\times 2\times 2\times 2\times 2\times 2\times 2\times 1\times {{x}^{3}}-2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 1\]
Taking \[2\times 2\times 2\times 2\times 2\times 2\times 2\times 1\] as common in the above equation, we can write
\[\Rightarrow 128{{x}^{3}}-1024=\left( 2\times 2\times 2\times 2\times 2\times 2\times 2\times 1 \right)\times \left( {{x}^{3}}-2\times 2\times 2 \right)\]
The above equation can also be written as
\[\Rightarrow 128{{x}^{3}}-1024=128\left( {{x}^{3}}-2\times 2\times 2 \right)\]
As we know that \[2\times 2\times 2={{2}^{3}}\], so we can write the above equation as
\[\Rightarrow 128{{x}^{3}}-1024=128\left( {{x}^{3}}-{{2}^{3}} \right)\]
Now, using the formula \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\], we can write the above equation as
\[\Rightarrow 128{{x}^{3}}-1024=128\left( x-2 \right)\left( {{x}^{2}}+2\times x+{{2}^{2}} \right)\]
The above equation can also be written as
\[\Rightarrow 128{{x}^{3}}-1024=128\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)\]
Now, we have found the factor of the term \[128{{x}^{3}}-1024\]. The factor of the term \[128{{x}^{3}}-1024\] is \[128\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)\].
Note: As we can see that this question is from the topic of algebra, so we should have a better knowledge in the topic of algebra. For solving this type of question, we should know how to find the prime factorization of any number. We should remember the following formula:
\[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]
Remember that the term \[x\times x\times x........n\text{ times}\] can also be written as \[{{x}^{n}}\].
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