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Chapter 6: Problem 59

Factor each trinomial. $$ p^{4}-10 p^{2}+16 $$

Short Answer

Expert verified
\((p^2 - 2)(p^2 - 8)\)

Step by step solution

01

Recognize the structure of the trinomial

First, observe the given trinomial: \( p^{4} - 10 p^{2} + 16 \). Notice that it can be related to a quadratic-like structure if we let \( x = p^2 \). This means we can rewrite the trinomial as \( x^2 - 10x + 16 \).
02

Factor the quadratic expression

Next, factor the quadratic expression \( x^2 - 10x + 16 \). Find two numbers that multiply to 16 and add to -10. These numbers are -2 and -8. Thus, \( x^2 - 10x + 16 \) can be factored as \((x - 2)(x - 8)\).
03

Substitute back the original variable

Now, revert the substitution by replacing \( x \) with \( p^2 \) again. This gives us: \((p^2 - 2)(p^2 - 8)\).
04

Verify the factors

Finally, confirm that \( (p^2 - 2)(p^2 - 8) \) is indeed the correct factorization of the original trinomial by expanding it back and checking if it equals the original expression:

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Key Concepts

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

Quadratic Structure
Many polynomial equations can look intimidating at first, but recognizing a familiar structure can make them easier to solve. In this exercise, we are given the polynomial: \[ p^{4} - 10p^{2} + 16 \]. Upon closer inspection, we can see that the exponents of the variable p are even numbers (4 and 2). This hints that we might be able to treat it similarly to a quadratic equation. To make it simpler, let's use a substitution method. If we let \( x = p^2 \), then our original polynomial transforms to: \[ x^2 - 10x + 16 \]. This new equation will be much easier to handle since it now resembles a standard quadratic equation, allowing us to use familiar methods for factoring it.
Trinomial Factorization
Once we have converted the original polynomial into a quadratic form, our next task is to factorize this new trinomial. The equation we now have is: \[ x^2 - 10x + 16 \]. To factor this, we need to find two numbers that multiply to 16 (the constant term) and add up to -10 (the coefficient of the middle term). After a bit of thinking, we can identify these numbers as -2 and -8. Therefore, we can write the quadratic as the product of two binomials: \[ (x - 2)(x - 8) \]. This step of factorization simplifies our expression and prepares us for reverting back to the original variable.
Substitution Method
After successfully factoring the quadratic, our next step is to return to the original variable. We have factored the expression into \[ (x - 2)(x - 8) \], but recall that we had substituted \( x = p^2 \). Now, substituting back in, we get: \[ (p^2 - 2)(p^2 - 8) \]. Finally, we should verify that this factorization is correct by expanding the product: \[ (p^2 - 2)(p^2 - 8) = p^4 - 8p^2 - 2p^2 + 16 = p^4 - 10p^2 + 16 \]. This confirms that our factorization is indeed correct.

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Most popular questions from this chapter

Concept Check Without actually solving each equation, determine which one of the following has 0 in its solution set. A. \(4 x^{2}-25=0\) B. \(x^{2}+2 x-3=0\) C. \(6 x^{2}+9 x+1=0\) D. \(x^{3}+4 x^{2}=3 x\)

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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. $$ \begin{aligned} 8 x^{3} &+4 x^{2}+27 y^{3}-9 y^{2} \\ &=\left(8 x^{3}+27 y^{3}\right)+\left(4 x^{2}-9 y^{2}\right) \end{aligned} $$ $$ \begin{aligned} &=(2 x+3 y)\left(4 x^{2}-6 x y+9 y^{2}\right)+(2 x+3 y)(2 x-3 y)\\\ &=(2 x+3 y)\left[\left(4 x^{2}-6 x y+9 y^{2}\right)+(2 x-3 y)\right]\\\ &=(2 x+3 y)\left(4 x^{2}-6 x y+9 y^{2}+2 x-3 y\right) \end{aligned} $$ In problems such as this, how we choose to group in the first step is essential to factoring correctly. If we reach a "dead end," then we should group differently and try again. Use the method just described to factor each polynomial. $$ 64 m^{2}-512 m^{3}-81 n^{2}+729 n^{3} $$

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