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Chapter 6: Problem 42

Factor each trinomial. $$ 4 p^{3}+24 p^{2}-64 p $$

Short Answer

Expert verified
The trinomial factors to \(4p(p + 8)(p - 2)\).

Step by step solution

01

Identify Common Factors

Start by identifying the greatest common factor (GCF) of all the terms in the trinomial. The terms are: \(4p^3\), \(24p^2\), and \(-64p\). The GCF is \(4p\).
02

Factor Out the GCF

Factor the GCF (\(4p\)) from each term in the trinomial. This changes the expression from \(4p^3 + 24p^2 - 64p\) to: \[4p(p^2 + 6p - 16)\].
03

Factor the Quadratic Trinomial

Now focus on the quadratic trinomial inside the parenthesis: \(p^2 + 6p - 16\). Find two numbers that multiply to \(-16\) and add to \(6\). These numbers are \(8\) and \(-2\).
04

Write in Factored Form

Write the quadratic trinomial as a product of two binomials using the numbers identified: \((p + 8)(p - 2)\). Thus, the factored form of the original trinomial is: \[4p(p + 8)(p - 2)\].

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Key Concepts

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

Greatest Common Factor
When factoring trinomials, the first step is to identify the Greatest Common Factor (GCF) of all the terms.
The GCF is the largest factor that divides each term in the expression.
For the trinomial given, the terms are: \(4p^3\), \(24p^2\), and \(-64p\).
To find the GCF:
  • Identify the smallest power of each common variable: here it is \(p\).

  • Note the largest integer that can divide all the coefficients: here it is 4.


So, the GCF of \(4p^3\), \(24p^2\), and \(-64p\) is \(4p\).
Factoring out the GCF from the trinomial: \[ 4p^3 + 24p^2 - 64p = 4p(p^2 + 6p - 16) \]
Quadratic Trinomials
Quadratic trinomials are polynomials of the form \(ax^2 + bx + c\).
In our case, after factoring out the GCF, we have the quadratic trinomial \(p^2 + 6p - 16\).
To factor quadratic trinomials, we need to find two numbers that multiply to give the constant term (\(-16\)) and add to give the linear coefficient (\(6\)).

  • Here, the two numbers are \(8\) and \(-2\) because \(8 \times -2 = -16\) and \(8 + (-2) = 6\).

We can then write the trinomial as two binomials: \((p + 8)(p - 2)\).
So, the factored form of the quadratic trinomial \(p^2 + 6p - 16\) is \((p + 8)(p - 2)\).
Binomials
A binomial is a polynomial with exactly two terms.
It can often be factored or used in factoring trinomials.
After factoring the quadratic trinomial \(p^2 + 6p - 16\) from the previous step, you get the binomials \((p + 8)\) and \((p - 2)\).

These binomials represent the simpler factors that our original trinomial can be broken down into.
For our initial problem, the complete factored form would be: \[ 4p(p^2 + 6p - 16) = 4p(p + 8)(p - 2) \]
This process simplifies solving equations or finding roots.

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Most popular questions from this chapter

Factor each trinomial. $$ 27 z^{2}+42 z-5 $$

Factor each trinomial. $$ 4(x-y)^{2}-23(x-y)-6 $$

Find each product. $$ (p+3 q)^{2} $$

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. $$ 125 p^{3}+25 p^{2}+8 q^{3}-4 q^{2} $$

Factor each trinomial. $$ 15 p^{2}+24 p q+8 q^{2} $$
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