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Chapter 6: Problem 73

In the following exercises, factor each trinomial of the form \(x^{2}+b x+c .\) $$ 5 p-6+p^{2} $$

Short Answer

Expert verified
(p + 6)(p - 1)

Step by step solution

01

Write the trinomial in standard form

Rewrite the given expression by arranging it in the form of a standard trinomial equation: \[ p^2 + 5p - 6 \].
02

Identify coefficients

Identify the coefficients in the trinomial equation. Here, \[ a = 1, \ b = 5, \ c = -6 \].
03

Find two numbers that multiply to c and add to b

We need to find two numbers that multiply to -6 (the constant term c) and add to 5 (the coefficient of the middle term b). These numbers are 6 and -1 because \[ 6 \times (-1) = -6 \ 6 + (-1) = 5 \].
04

Write the middle term as a sum of the two numbers found

Rewrite the middle term (5p) using the two numbers found: \[ p^2 + 6p - p - 6 \].
05

Factor by grouping

Group the terms in pairs: \[ (p^2 + 6p) + (-p - 6) \]. Factor out the greatest common factor (GCF) from each pair: \[ p(p + 6) - 1(p + 6) \].
06

Factor out the common binomial factor

Factor out the common binomial factor (p + 6): \[ (p + 6)(p - 1) \].
07

Verify the factorization

Multiply the factors to check if the factorization is correct: \[ (p + 6)(p - 1) = p^2 - p + 6p - 6 = p^2 + 5p - 6 \]. The factorization is verified.

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Key Concepts

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

Polynomial Factorization
Polynomial factorization is the process of breaking down a polynomial into simpler polynomials that can be multiplied together to get the original polynomial. This method simplifies many algebraic expressions and equations.
Let's take our polynomial from the exercise: \(p^2 + 5p - 6\).
The polynomial is in standard form, written as \(ax^2 + bx + c\), where:
  • \(a = 1\)
  • \(b = 5\)
  • \(c = -6\)
By identifying these coefficients and finding two numbers that multiply to \(c\) and add up to \(b\), we can break the polynomial into simpler parts. Here, those numbers are \6\ and \-1\, because:
  • \(6 \times -1 = -6\)
  • \(6 + -1 = 5\)
This is the essence of polynomial factorization. Combining these steps, we rewrite in factored form like so: \((p + 6)(p - 1)\).
Trinomial Equations
A trinomial equation is a polynomial equation with three terms. It's often in the form \(ax^2 + bx + c\).

The goal is to factor these equations into the product of two binomials. Our example from the exercise is a perfect case: \(p^2 + 5p - 6\).
We begin by identifying our coefficients \(a, b, \text{and} c\). Next, we find:
  • Two numbers that multiply to \-6\ (coefficient \(c\))
  • And add up to \5\ (coefficient \(b\))
Once found 鈥� here they are \6\ and \-1\ 鈥� we use them to split the middle term: \((5p = 6p - p)\).
Then we group into: \((p^2 + 6p) + (-p - 6)\). Factor by grouping, and we get \((p + 6)(p - 1)\). Finally, verify by multiplying back out to ensure correctness.
Algebraic Expressions
Algebraic expressions represent numbers and operations using variables and coefficients. For example, the expression \(p^2 + 5p - 6\) involves:
  • Variables: \(p\)
  • Coefficients: 1, 5, and -6
To factor such expressions, we perform various algebraic manipulations.
In our given problem, we rewrite the expression using grouping and factoring strategies:
We express the polynomial \(p^2 + 5p - 6\) as: \(p^2 + 6p - p - 6\).
Then, group and factor by grouping: \(p(p + 6) - 1(p + 6) \).
This results in our final factored form: \((p + 6)(p - 1)\). Going through these steps solidifies your understanding of working with algebraic expressions.

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Most popular questions from this chapter

Many trinomials of the form \(x^{2}+b x+c\) factor into the product of two binomials \((x+m)(x+n)\). Explain how you find the values of \(m\) and \(n\).

In the following exercises, factor each expression using any method. $$ 18 m^{2}+15 m+3 $$

In the following exercises, factor completely using trial and error. $$ p^{3}-8 p^{2}-20 p $$

In the following exercises, factor each trinomial of the form \(x^{2}+b x+c .\) $$ n^{2}+19 n+48 $$

In the following exercises, factor using the 'ac' method. $$ 4 k^{2}-16 k+15 $$
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