
How do you factor\[{x^6} + 10{x^3} + 25\]?
Answer
556.5k+ views
Hint: In this question we need to modify in the form of the sum of cubes identity. First we take the term in the first term for $x$ power. After we rewrite in the form of ${a^2} + 2ab + {b^2}$. Now we rewrite the equation and the expansion of ${a^2} + {b^2}$ is the same here. The $b$ (second term) $5$ takes cube and cube root.
After that we apply the sum of cubes identity.
Complete Step by Step Solution:
First the equation modifies in the form of${a^3} + {b^3}$.
Let, \[{x^6} + 10{x^3} + 25\]
We modify in the form of ${a^3} + {b^3}$
The sum of cubes identity can be written:
${a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})$
Let’s take the given equation,
\[ \Rightarrow {x^6} + 10{x^3} + 25\]
First we take a common term in the first term for $x$ power, hence we get
$ \Rightarrow {({x^3})^2} + 10{x^3} + 25$
In the second term we rewrite$2(5)$, hence we get
$ \Rightarrow {({x^3})^2} + 2(5)({x^3}) + 25$
The third term is square on$5$, so we write the square method, hence we get
$ \Rightarrow {({x^3})^2} + 2(5)({x^3}) + {(5)^2}$
In this equation have in the form of${a^2} + {b^2}$, here $a = ({x^3})$ and $b = 5$
The expansion of ${a^2} + {b^2}$is
${a^2} + 2ab + {b^2}$
Now we rewrite the equation and the expansion of${a^2} + {b^2}$the same here. So, we take${a^2} + {b^2}$, hence we get
\[ \Rightarrow {a^2} + 2ab + {b^2} = {({x^3})^2} + 2(5)({x^3}) + {(5)^2}\]
\[ \Rightarrow {x^6} + 10{x^3} + 25 = {({x^3} + 5)^2}\]
The$b$(second term)$5$takes cube and cube root, hence we get
\[ \Rightarrow {x^6} + 10{x^3} + 25 = {({x^3} + {(\sqrt[3]{5})^3})^2}\]
Now we know that, the sum of cubes identity can be written, hence we get
$ \Rightarrow {a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})$
\[ \Rightarrow {({x^3} + {(\sqrt[3]{5})^3})^2} = {((x + \sqrt[3]{5})({x^2} - x(\sqrt[3]{5}) + {(\sqrt[3]{5})^2}))^2}\]
Now we separate the first and second term in RHS (Right Hand Side), hence we get
\[ \Rightarrow {({x^3} + {(\sqrt[3]{5})^3})^2} = {(x + \sqrt[3]{5})^2}{({x^2} - x(\sqrt[3]{5}) + {(\sqrt[3]{5})^2})^2}\]
Note: Factoring a polynomial involves writing it as a product of two or more polynomials. It reverses the process of polynomial multiplication. We have seen several examples of factoring already. However, for this article, you should be especially familiar with taking common factors using the distributive property.
After that we apply the sum of cubes identity.
Complete Step by Step Solution:
First the equation modifies in the form of${a^3} + {b^3}$.
Let, \[{x^6} + 10{x^3} + 25\]
We modify in the form of ${a^3} + {b^3}$
The sum of cubes identity can be written:
${a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})$
Let’s take the given equation,
\[ \Rightarrow {x^6} + 10{x^3} + 25\]
First we take a common term in the first term for $x$ power, hence we get
$ \Rightarrow {({x^3})^2} + 10{x^3} + 25$
In the second term we rewrite$2(5)$, hence we get
$ \Rightarrow {({x^3})^2} + 2(5)({x^3}) + 25$
The third term is square on$5$, so we write the square method, hence we get
$ \Rightarrow {({x^3})^2} + 2(5)({x^3}) + {(5)^2}$
In this equation have in the form of${a^2} + {b^2}$, here $a = ({x^3})$ and $b = 5$
The expansion of ${a^2} + {b^2}$is
${a^2} + 2ab + {b^2}$
Now we rewrite the equation and the expansion of${a^2} + {b^2}$the same here. So, we take${a^2} + {b^2}$, hence we get
\[ \Rightarrow {a^2} + 2ab + {b^2} = {({x^3})^2} + 2(5)({x^3}) + {(5)^2}\]
\[ \Rightarrow {x^6} + 10{x^3} + 25 = {({x^3} + 5)^2}\]
The$b$(second term)$5$takes cube and cube root, hence we get
\[ \Rightarrow {x^6} + 10{x^3} + 25 = {({x^3} + {(\sqrt[3]{5})^3})^2}\]
Now we know that, the sum of cubes identity can be written, hence we get
$ \Rightarrow {a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})$
\[ \Rightarrow {({x^3} + {(\sqrt[3]{5})^3})^2} = {((x + \sqrt[3]{5})({x^2} - x(\sqrt[3]{5}) + {(\sqrt[3]{5})^2}))^2}\]
Now we separate the first and second term in RHS (Right Hand Side), hence we get
\[ \Rightarrow {({x^3} + {(\sqrt[3]{5})^3})^2} = {(x + \sqrt[3]{5})^2}{({x^2} - x(\sqrt[3]{5}) + {(\sqrt[3]{5})^2})^2}\]
Note: Factoring a polynomial involves writing it as a product of two or more polynomials. It reverses the process of polynomial multiplication. We have seen several examples of factoring already. However, for this article, you should be especially familiar with taking common factors using the distributive property.
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