Factor polynomial $$ 8x^{9}+x^{6}-8x^{3}-1 $$
Answer
Explanation
Step 1 :
To factor $ 8x^{9}+x^{6}-8x^{3}-1 $ we can use factoring by grouping:
Group $ \color{blue}{ 8x^{9} }$ with $ \color{blue}{ x^{6} }$ and $ \color{red}{ -8x^{3} }$ with $ \color{red}{ -1 }$ then factor each group.
$$ \begin{aligned} 8x^{9}+x^{6}-8x^{3}-1 = ( \color{blue}{ 8x^{9}+x^{6} } ) + ( \color{red}{ -8x^{3}-1 }) &= \\ &= \color{blue}{ x^{6}( 8x^{3}+1 )} + \color{red}{ -1( 8x^{3}+1 ) } = \\ &= (x^{6}-1)(8x^{3}+1) \end{aligned} $$Step 2 :
Rewrite $ x^{6}-1 $ as:
$$ x^{6}-1 = (x^{3})^2 - (1)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = x^{3} $ and $ II = 1 $ , we have:
$$ x^{6}-1 = (x^{3})^2 - (1)^2 = ( x^{3}-1 ) ( x^{3}+1 ) $$Step 3 :
To factor $ x^{3}-1 $ we can use difference of cubes formula:
$$ I^3 - II^3 = (I - II)(I^2 + I \cdot II + II^2) $$After putting $ I = x $ and $ II = 1 $ , we have:
$$ x^{3}-1 = ( x-1 ) ( x^{2}+x+1 ) $$Step 4 :
Step 4: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:
$$ \color{blue}{ b = 1 } ~ \text{ and } ~ \color{red}{ c = 1 }$$Now we must discover two numbers that sum up to $ \color{blue}{ 1 } $ and multiply to $ \color{red}{ 1 } $.
Step 5: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.
| PRODUCT = 1 | |
| 1 1 | -1 -1 |
Step 6: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.
Step 5 :
To factor $ x^{3}+1 $ we can use sum of cubes formula:
$$ I^3 - II^3 = (I + II)(I^2 - I \cdot II + II^2) $$After putting $ I = x $ and $ II = 1 $ , we have:
$$ x^{3}+1 = ( x+1 ) ( x^{2}-x+1 ) $$Step 6 :
Step 6: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:
$$ \color{blue}{ b = 1 } ~ \text{ and } ~ \color{red}{ c = 1 }$$Now we must discover two numbers that sum up to $ \color{blue}{ 1 } $ and multiply to $ \color{red}{ 1 } $.
Step 7: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.
| PRODUCT = 1 | |
| 1 1 | -1 -1 |
Step 8: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.
Step 7 :
Step 7: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:
$$ \color{blue}{ b = -1 } ~ \text{ and } ~ \color{red}{ c = 1 }$$Now we must discover two numbers that sum up to $ \color{blue}{ -1 } $ and multiply to $ \color{red}{ 1 } $.
Step 8: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.
| PRODUCT = 1 | |
| 1 1 | -1 -1 |
Step 9: Because none of these pairs will give us a sum of $ \color{blue}{ -1 }$, we conclude the polynomial cannot be factored.
Step 8 :
To factor $ 8x^{3}+1 $ we can use sum of cubes formula:
$$ I^3 - II^3 = (I + II)(I^2 - I \cdot II + II^2) $$After putting $ I = 2x $ and $ II = 1 $ , we have:
$$ 8x^{3}+1 = ( 2x+1 ) ( 4x^{2}-2x+1 ) $$Step 9 :
Step 9: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:
$$ \color{blue}{ b = 1 } ~ \text{ and } ~ \color{red}{ c = 1 }$$Now we must discover two numbers that sum up to $ \color{blue}{ 1 } $ and multiply to $ \color{red}{ 1 } $.
Step 10: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.
| PRODUCT = 1 | |
| 1 1 | -1 -1 |
Step 11: Because none of these pairs will give us a sum of $ \color{blue}{ 1 }$, we conclude the polynomial cannot be factored.
Step 10 :
Step 10: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:
$$ \color{blue}{ b = -1 } ~ \text{ and } ~ \color{red}{ c = 1 }$$Now we must discover two numbers that sum up to $ \color{blue}{ -1 } $ and multiply to $ \color{red}{ 1 } $.
Step 11: Find out pairs of numbers with a product of $\color{red}{ c = 1 }$.
| PRODUCT = 1 | |
| 1 1 | -1 -1 |
Step 12: Because none of these pairs will give us a sum of $ \color{blue}{ -1 }$, we conclude the polynomial cannot be factored.
Step 11 :
Step 11: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:
Step 12: Multiply the leading coefficient $\color{blue}{ a = 4 }$ by the constant term $\color{blue}{c = 1} $.
$$ a \cdot c = 4 $$Step 13: Find out two numbers that multiply to $ a \cdot c = 4 $ and add to $ b = -2 $.
Step 14: All pairs of numbers with a product of $ 4 $ are:
| PRODUCT = 4 | |
| 1 4 | -1 -4 |
| 2 2 | -2 -2 |
Step 15: Find out which factor pair sums up to $\color{blue}{ b = -2 }$
Step 16: Because none of these pairs will give us a sum of $ \color{blue}{ -2 }$, we conclude the polynomial cannot be factored.