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Chapter 5: Problem 60

Factor each polynomial. $$ x^{6}-216 $$

Short Answer

Expert verified
\( x^6 - 216 = (x^2 - 6)(x^4 + 6x^2 + 36) \)

Step by step solution

01

- Recognize the difference of cubes

Notice that the polynomial can be factored using the formula for the difference of cubes: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \] Rewrite the given polynomial in the form of a difference of cubes. Recognize that \( x^6 = (x^2)^3 \) and \( 216 = (6)^3 \). So, \( x^6 - 216 \) can be written as \((x^2)^3 - (6)^3\).
02

- Apply the difference of cubes formula

Use the difference of cubes formula with \( a = x^2 \) and \( b = 6 \): \[ (x^2)^3 - 6^3 = (x^2 - 6)((x^2)^2 + x^2 \times 6 + 6^2) \] Simplify the expression inside the second parenthesis.
03

- Simplify the second factor

Expand and simplify the second factor: \[ (x^2)^2 + x^2 \times 6 + 6^2 = x^4 + 6x^2 + 36 \] Write the final factored form.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

difference of cubes
The difference of cubes formula is very useful for factoring certain types of polynomials. It's essential to understand this formula's structure: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
Let's apply this to our example polynomial, \( x^{6} - 216 \). To fit this into the difference of cubes formula, we need to recognize that both terms can be written as cubes: \( x^6 = (x^2)^3 \) and \( 216 = 6^3 \). This representation changes our polynomial from \( x^6 - 216 \) to \( (x^2)^3 - 6^3 \).
It's helpful to rewrite the polynomial in detail:
  • \( x^6 \) is equivalent to \( (x^2)^3 \)
  • \( 216 \) is equivalent to \( 6^3 \)
With this in mind, we recognize that the polynomial fits the difference of cubes pattern where \( a = x^2 \) and \( b = 6 \).
factoring polynomials
Once you understand the difference of cubes, you can factor the polynomial easily.
Let鈥檚 apply the formula with \( a = x^2 \) and \( b = 6 \): \[ (x^2)^3 - 6^3 = (x^2 - 6)((x^2)^2 + x^2 \times 6 + 6^2) \] This gives us two factors:
  • \( x^2 - 6 \)
  • \( (x^2)^2 + x^2 \times 6 + 6^2 \)
Now, we simplify the second factor by expanding and combining like terms:
\[ (x^2)^2 + x^2 \times 6 + 6^2 = x^4 + 6x^2 + 36 \]. Thus, the fully factored form of the polynomial \( x^{6} - 216 \) is: \[ (x^2 - 6)(x^4 + 6x^2 + 36) \].
simplification
Simplification is a crucial final step in polynomial factorization.
After applying the difference of cubes formula, you arrive at: \[ (x^2 - 6)(x^4 + 6x^2 + 36) \]
Ensure each factor is simplified completely and no further factorization is possible. In our case, both factors \( x^2 - 6 \) and \( x^4 + 6x^2 + 36 \) cannot be factored further into simpler polynomial factors. Hence, our final simplified answer remains:
\[ (x^2 - 6)(x^4 + 6x^2 + 36) \].

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Most popular questions from this chapter

Factor each trinomial. \(k^{4}+10 k^{2}+9\)

Skills In some cases, the method of factoring by grouping can be combined with the methods of special factoring discussed in this section. Consider this example. $$ \begin{array}{l} 8 x^{3}+4 x^{2}+27 y^{3}-9 y^{2} \\ \quad=\left(8 x^{3}+27 y^{3}\right)+\left(4 x^{2}-9 y^{2}\right) \\ \quad=(2 x+3 y)\left(4 x^{2}-6 x y+9 y^{2}\right)+(2 x+3 y)(2 x-3 y) \\ \quad=(2 x+3 y)\left[\left(4 x^{2}-6 x y+9 y^{2}\right)+(2 x-3 y)\right] \\ \quad=(2 x+3 y)\left(4 x^{2}-6 x y+9 y^{2}+2 x-3 y\right) \end{array} $$ Use the method just described to factor each polynomial. $$ 125 p^{3}+25 p^{2}+8 q^{3}-4 q^{2} $$

The following exercises are of mixed variety. Factor each polynomial. $$ (p+8 q)^{2}-10(p+8 q)+25 $$

Find two consecutive integers such that the sum of their squares is \(61 .\)

Solve each equation. $$ -x^{2}=9-6 x $$
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