Problem 12
Question
Simplify each expression. \(\frac{12 p^{2}+6 p-6}{4(p+1)^{2}} \div \frac{6 p-3}{2 p+10}\)
Step-by-Step Solution
Verified Answer
The simplified expression is \(p + 5\).
1Step 1: Simplify the First Fraction
We start by simplifying the numerator \(12p^2 + 6p - 6\) by factoring. Notice that all terms in the numerator have a common factor of 6. Factor this out to get: \[6(2p^2 + p - 1)\]. The denominator is already factored as \(4(p+1)^2\). So, the simplified first fraction is: \[\frac{6(2p^2 + p - 1)}{4(p + 1)^2}\].
2Step 2: Factor the Quadratic
Next, we need to factor the quadratic \(2p^2 + p - 1\) from the first fraction. Looking for two numbers that multiply to \(-2\) and add to \(1\), we find \(2p^2 + 2p - p - 1\), which can be rearranged and factored as:\[2p(p + 1) - 1(p + 1) = (2p - 1)(p + 1)\].
3Step 3: Simplify the Second Fraction
Examine the second fraction \(\frac{6p - 3}{2p + 10}\). Factor the numerator and the denominator: The numerator \(6p - 3\) can be factored as \(3(2p - 1)\), and the denominator \(2p + 10\) can be factored as \(2(p + 5)\). So, the second fraction becomes:\[\frac{3(2p - 1)}{2(p + 5)}\].
4Step 4: Divide the Fractions
Dividing by a fraction is equivalent to multiplying by its reciprocal. So, we rewrite the division as multiplication by flipping the second fraction: \[\frac{6(2p - 1)(p + 1)}{4(p + 1)^2} \times \frac{2(p + 5)}{3(2p - 1)}\].
5Step 5: Cancel Common Factors
Cancel any common factors from the numerators and denominators. We notice that \(2p - 1\) is a common factor, as is \(p + 1\), allowing us to cancel these out:\[\frac{6 \cdot 2 \cdot (p + 5)}{4 \cdot 3} = \frac{12(p + 5)}{12}\].
6Step 6: Simplify the Remaining Expression
The expression now simplifies readily since both the numerator and the denominator have a factor of 12:\[\frac{12(p + 5)}{12} = p + 5\].
Key Concepts
Factoring QuadraticsFraction DivisionCanceling Factors
Factoring Quadratics
Factoring quadratics is an essential skill when working with rational expressions. A quadratic expression is typically in the form of \(ax^2 + bx + c\). The goal of factoring is to express this quadratic as a product of two binomials.
To factor successfully, you need to find two numbers that multiply to \(a \times c\) and add up to \(b\).
In our example, the expression \(2p^2 + p - 1\) needed to be factored. First, we multiply the leading coefficient \(2\) by the constant \(-1\), giving us \(-2\). We seek two numbers that multiply to \(-2\) and add to \(1\). These numbers are \(2\) and \(-1\).
We then rewrite the middle term:
To factor successfully, you need to find two numbers that multiply to \(a \times c\) and add up to \(b\).
In our example, the expression \(2p^2 + p - 1\) needed to be factored. First, we multiply the leading coefficient \(2\) by the constant \(-1\), giving us \(-2\). We seek two numbers that multiply to \(-2\) and add to \(1\). These numbers are \(2\) and \(-1\).
We then rewrite the middle term:
- \(2p^2 + 2p - p - 1\)
- \[2p(p + 1) - 1(p + 1) = (2p - 1)(p + 1)\]
Fraction Division
Dividing fractions is a step in simplifying complex rational expressions. This involves understanding that dividing by a fraction is the same as multiplying by its reciprocal.
The reciprocal of a fraction \(\frac{a}{b}\) is \(\frac{b}{a}\).
In our problem, we had to divide the first fraction by the second. This means:
The reciprocal of a fraction \(\frac{a}{b}\) is \(\frac{b}{a}\).
In our problem, we had to divide the first fraction by the second. This means:
- \[\frac{6(2p^2 + p - 1)}{4(p + 1)^2} \div \frac{3(2p - 1)}{2(p + 5)}\]
- \[ \frac{6(2p^2 + p - 1)}{4(p + 1)^2} \times \frac{2(p + 5)}{3(2p - 1)} \]
Canceling Factors
Canceling common factors is the final critical step in simplifying rational expressions. It requires identifying and removing factors that appear in both the numerator and the denominator of a fraction.
This is possible only if the terms are factored completely.During the simplification process, we need to carefully look for common factors:
This is possible only if the terms are factored completely.During the simplification process, we need to carefully look for common factors:
- In our expression: \[\frac{6 \cdot (2p-1)(p+1)}{4 \cdot (p+1)^2} \times \frac{2(p+5)}{3(2p-1)}\]
- The \((2p-1)\) cancel each other since they appear in both the numerator and the denominator.
- One factor of \((p+1)\) in the numerator cancels with one factor in the denominator.
- \[\frac{12(p + 5)}{12} = p + 5\]
Other exercises in this chapter
Problem 12
Simplify each expression. $$ \frac{x}{x+1}+\frac{3}{x^{2}-4 x-5} $$
View solution Problem 12
ELECTRICITY For Exercises \(9-12\) , use the following information. The current \(I\) in amperes in an electrical circuit with three resistors in series is give
View solution Problem 13
Solve each equation or inequality. Check your solutions. $$ a+1=\frac{6}{a} $$
View solution Problem 13
If \(y\) varies inversely as \(x\) and \(y=16\) when \(x=5,\) find \(y\) when \(x=20\)
View solution