Chapter 0: Problem 64
Factor each polynomial. $$ 6 z^{2}+5 z+1 $$
Short Answer
Expert verified
(2z + 1)(3z + 1)
Step by step solution
01
Identify terms and structure
Recognize that the given polynomial is in the form of a quadratic equation: \[ 6z^2 + 5z + 1 \]Here, the coefficient of the quadratic term (\(6z^2\)) is 6, the coefficient of the linear term (\(5z\)) is 5, and the constant term is 1.
02
Find the product of \(a\) and \(c\)
In a quadratic equation of the form \(az^2 + bz + c\), multiply the coefficient of the quadratic term (\(a\)) by the constant term (\(c\)). Here, \(a = 6\) and \(c = 1\). Therefore, the product is: \[ 6 \times 1 = 6 \]
03
Find two numbers that multiply to \(ac\) and add to \(b\)
Next, find two numbers that multiply to the product obtained in Step 2 (6) and add up to the coefficient of the linear term (\(b = 5\)). The two numbers are 2 and 3 because: \[ 2 \times 3 = 6 \] and \[ 2 + 3 = 5 \]
04
Split the middle term using the two numbers found
Rewrite the polynomial by splitting the middle term (\(5z\)) into two terms using the numbers found in Step 3 (2 and 3): \[ 6z^2 + 2z + 3z + 1 \]
05
Factor by grouping
Factor the polynomial by grouping the terms in pairs and factoring out the greatest common factor (GCF) from each pair: \[ 6z^2 + 2z + 3z + 1 = 2z(3z + 1) + 1(3z + 1) \]
06
Factor out the common binomial factor
Factor out the common binomial factor (\(3z + 1\)) from the expression obtained in the previous step: \[ 2z(3z + 1) + 1(3z + 1) = (2z + 1)(3z + 1) \]
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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. The general form of a quadratic equation is \[ax^2 + bx + c = 0\], where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. Quadratic equations are very common in algebra and can be solved using various methods, including factoring, completing the square, and the quadratic formula.
In our exercise, we started with the quadratic equation \[6z^2 + 5z + 1\]. Here, \(a = 6\), \(b = 5\), and \(c = 1\). Knowing the standard form makes it easier to apply different techniques to solve or factor the equation.
Understanding the structure and terms of a quadratic equation is the first step towards tackling any problem involving them. Always keep an eye on the coefficients and the constant term, as these values will guide you in choosing the right method to solve the equation.
In our exercise, we started with the quadratic equation \[6z^2 + 5z + 1\]. Here, \(a = 6\), \(b = 5\), and \(c = 1\). Knowing the standard form makes it easier to apply different techniques to solve or factor the equation.
Understanding the structure and terms of a quadratic equation is the first step towards tackling any problem involving them. Always keep an eye on the coefficients and the constant term, as these values will guide you in choosing the right method to solve the equation.
factoring by grouping
Factoring by grouping is a technique used to factor polynomials when you have four terms. It's particularly useful for quadratics with a leading coefficient greater than 1, like our example.
The steps for factoring by grouping start with splitting the middle term into two terms that can be grouped together in pairs. In our polynomial \[6z^2 + 5z + 1\], we split the middle term \(5z\) into \(2z\) and \(3z\): \[6z^2 + 2z + 3z + 1\].
Next, we group the terms: \[(6z^2 + 2z) + (3z + 1)\].
For each pair, factor out the Greatest Common Factor (GCF). This gives us: \[2z(3z + 1) + 1(3z + 1)\]. Notice how \(3z + 1\) is a common factor.
Finally, we factor out the common binomial factor to get: \[(2z + 1)(3z + 1)\]. This method helps simplify and break down complex polynomials into simpler, more manageable factors.
The steps for factoring by grouping start with splitting the middle term into two terms that can be grouped together in pairs. In our polynomial \[6z^2 + 5z + 1\], we split the middle term \(5z\) into \(2z\) and \(3z\): \[6z^2 + 2z + 3z + 1\].
Next, we group the terms: \[(6z^2 + 2z) + (3z + 1)\].
For each pair, factor out the Greatest Common Factor (GCF). This gives us: \[2z(3z + 1) + 1(3z + 1)\]. Notice how \(3z + 1\) is a common factor.
Finally, we factor out the common binomial factor to get: \[(2z + 1)(3z + 1)\]. This method helps simplify and break down complex polynomials into simpler, more manageable factors.
algebraic factoring
Algebraic factoring involves rewriting a polynomial as a product of simpler polynomials. This technique is crucial for solving quadratic equations, simplifying expressions, and finding roots of polynomials.
In our example, we aimed to factor \[6z^2 + 5z + 1\]. We first identified values for \(a\), \(b\), and \(c\) and then looked for two numbers that multiply to \(a \times c\) and add up to \(b\).
Once these numbers were identified (2 and 3 in our case), we split the middle term and grouped the resulting terms.
After factoring by grouping, we obtained the common binomial factor, leading to the final factored form \[(2z + 1)(3z + 1)\].
Algebraic factoring is a key skill in algebra, often required when solving equations, analyzing functions, and simplifying complex expressions. It provides a method for breaking down and understanding polynomial structures.
In our example, we aimed to factor \[6z^2 + 5z + 1\]. We first identified values for \(a\), \(b\), and \(c\) and then looked for two numbers that multiply to \(a \times c\) and add up to \(b\).
Once these numbers were identified (2 and 3 in our case), we split the middle term and grouped the resulting terms.
After factoring by grouping, we obtained the common binomial factor, leading to the final factored form \[(2z + 1)(3z + 1)\].
Algebraic factoring is a key skill in algebra, often required when solving equations, analyzing functions, and simplifying complex expressions. It provides a method for breaking down and understanding polynomial structures.
