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Chapter 6: Problem 100

In the following exercises, factor completely using trial and error. $$ 5 x^{2}-17 x+6 $$

Short Answer

Expert verified
(5x - 2)(x - 3)

Step by step solution

01

- Identify the equation

The given quadratic equation is \[5x^2 - 17x + 6\].
02

- Find two numbers that multiply to ac

The coefficient of \(x^2\) is 5 (a=5), and the constant term is 6 (c=6). Multiply a and c: \[5 \times 6 = 30\]. Find two numbers that multiply to 30 and add up to the middle term's coefficient, -17.
03

- Identify the pair

The pair of numbers that multiply to 30 and add to -17 is -15 and -2, because \[-15 \times -2 = 30\] and \[-15 + -2 = -17\].
04

- Rewrite the middle term

Rewrite the quadratic expression using -15 and -2: \[5x^2 - 15x - 2x + 6\].
05

- Factor by grouping

Group the terms into two pairs and factor out the common factors: \[5x(x - 3) - 2(x - 3)\].
06

- Factor out the common binomial

Observe that \(x - 3\) is common in both terms: \[(5x - 2)(x - 3)\].

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Key Concepts

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

quadratic equations
A quadratic equation is a polynomial equation of degree 2. This means that the highest exponent of the variable (usually x) is 2. The general form of a quadratic equation is given as: \[ax^2 + bx + c = 0\], where:
  • \(a\), \(b\), and \(c\) are constants
  • \(a eq 0\)
Quadratic equations can appear in various real-world scenarios, such as projectile motion or calculating areas. Solving these equations is crucial for many math-related fields.
factoring by grouping
Factoring by grouping is a method used to factor polynomials with four or more terms. The goal is to rearrange and group terms that have a common factor. For example, given \[5x^2 - 15x - 2x + 6\], you can group the terms as follows: \[(5x^2 - 15x) + (-2x + 6)\].
Next, factor out the greatest common factor (GCF) from each group:
  • In the first group \(5x^2 - 15x\), the GCF is 5x: \[5x(x - 3)\]
  • In the second group \(-2x + 6\), the GCF is -2: \[-2(x - 3)\]
Now, our expression looks like this: \[5x(x - 3) - 2(x - 3)\]. The binomial \(x - 3\) is common in both terms; factor it out to get: \[(5x - 2)(x - 3)\].
trial and error method
The trial and error method for factoring quadratics involves finding two numbers that satisfy specific conditions. In this example, given \[5x^2 - 17x + 6\], we look for two numbers that multiply to \(5 \times 6 = 30\) (product of the coefficient of \(x^2\) and the constant term)
and add to -17 (the coefficient of the middle term). After testing possible pairs, we find that -15 and -2 work because:
  • \(-15 \times -2 = 30\)
  • \(-15 + -2 = -17\)
This trial and error process is essential as it helps set up our equation for the next factoring steps.
polynomial factoring
Polynomial factoring is the process of breaking down a polynomial into simpler components. These factors, when multiplied together, give back the original polynomial.
In our example, factoring \(5x^2 - 17x + 6\) completely resulted in \[(5x - 2)(x - 3)\], which are both simpler polynomials. Factoring polynomials is a foundational skill in algebra and finding zeros of polynomials. It often involves techniques such as:
  • Factoring by grouping
  • Trial and error method
  • Using special products like difference of squares
Mastering these techniques enables solving more complex equations efficiently.

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Most popular questions from this chapter

In the following exercises, factor each expression using any method. $$ 5 r^{2}+25 r+30 $$

In the following exercises, factor completely using trial and error. $$ 4 w^{2}-5 w+1 $$

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

In the following exercises, factor using the 'ac' method. $$ 12 z^{2}-41 z-11 $$

In the following exercises, foctor using substitution. $$ (x-3)^{2}-5(x-3)-36 $$
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