Problem 28
Question
Write the product in simplest form. $$\left(x^{2}+2 x+1\right) \cdot \frac{x+2}{x^{2}+3 x+2}$$
Step-by-Step Solution
Verified Answer
The simplified form of the given expression is \(\frac{x^{2}+2 x+1}{x + 1}\)
1Step 1: Factoring the polynomial
We can factorize the polynomial found in the denominator, which is \(x^{2} + 3x + 2\). This can be factored into two terms as follows: (x + 1)(x + 2). So the equation becomes: \((x^{2}+2 x+1) \cdot \frac{x+2}{(x + 1)(x + 2)}\)
2Step 2: Cancel the common factors
Observing that \(x + 2\) is common to both the numerator and the denominator, this term cancels out, leading to: \((x^{2}+2 x+1) \cdot \frac{1}{x + 1}\)
3Step 3: Represent in simplest form
Finally, this can be simplified and written as: \(\frac{x^{2}+2 x+1}{x + 1}\)
Key Concepts
Factoring PolynomialsCanceling Common FactorsRational Expressions
Factoring Polynomials
Factoring polynomials is an essential skill in algebra. It helps simplify expressions and solve equations more easily. Let's use the polynomial in the original exercise:
The given polynomial in the denominator is \(x^2 + 3x + 2\). To factor this polynomial, you look for two numbers that multiply to give 2 (the constant term) and add up to 3 (the coefficient of the middle term). After examining your options, you find that 1 and 2 work perfectly:
Factoring breaks down more complicated polynomial expressions into simpler parts, making them easier to work with and reducing the risk of errors in calculations.
The given polynomial in the denominator is \(x^2 + 3x + 2\). To factor this polynomial, you look for two numbers that multiply to give 2 (the constant term) and add up to 3 (the coefficient of the middle term). After examining your options, you find that 1 and 2 work perfectly:
- Multiplying them gives 2: \(1 \times 2 = 2\)
- Adding them gives 3: \(1 + 2 = 3\)
Factoring breaks down more complicated polynomial expressions into simpler parts, making them easier to work with and reducing the risk of errors in calculations.
Canceling Common Factors
In mathematical expressions, particularly involving fractions with polynomials, canceling is a handy step. This process simplifies expressions by removing terms that appear both in the numerator and the denominator.
In the given exercise, the expression is \((x^{2}+2x+1) \, \cdot \, \frac{x+2}{(x+1)(x+2)}\). You can observe that \(x+2\) is present in both the numerator (top) and the denominator (bottom). This means it can be canceled out in order to simplify:
In the given exercise, the expression is \((x^{2}+2x+1) \, \cdot \, \frac{x+2}{(x+1)(x+2)}\). You can observe that \(x+2\) is present in both the numerator (top) and the denominator (bottom). This means it can be canceled out in order to simplify:
- This reduces the expression to \((x^{2}+2x+1) \, \cdot \, \frac{1}{x+1}\).
Rational Expressions
A rational expression is essentially a fraction in which both the numerator and the denominator are polynomials. Working with rational expressions involves similar rules as regular numerical fractions. You simplify them by factoring and canceling common factors.
In the original exercise, the rational expression was initially \(\frac{x+2}{x^{2}+3x+2}\), which we factored to become \(\frac{x+2}{(x+1)(x+2)}\). By canceling the \(x+2\) factor, it was further reduced.
Rational expressions can look complex, but breaking them down into smaller parts through factoring makes them easier to handle.
When simplifying rational expressions, remember:
In the original exercise, the rational expression was initially \(\frac{x+2}{x^{2}+3x+2}\), which we factored to become \(\frac{x+2}{(x+1)(x+2)}\). By canceling the \(x+2\) factor, it was further reduced.
Rational expressions can look complex, but breaking them down into smaller parts through factoring makes them easier to handle.
When simplifying rational expressions, remember:
- Always look for factors common to both numerator and denominator.
- Ensure all terms are factored completely before canceling.
- Check for any restrictions that might arise by setting the denominator equal to zero (since division by zero is undefined).
Other exercises in this chapter
Problem 28
Write the difference in simplest form. $$ \frac{2 x}{5}-\frac{x+1}{4} $$
View solution Problem 28
Solve the equation. Check your solutions. $$ \frac{r+4}{3}=\frac{r}{5} $$
View solution Problem 28
Simplify the expression. If not possible, write already in simplest form. $$ \frac{x^{2}+25}{2 x+10} $$
View solution Problem 29
FACTORING AFTER ADDING OR SUBTRACTING. Simplify the expression. $$ \frac{2 x(x+4)}{(x+1)^{2}}-\frac{3 x-3}{(x+1)^{2}} $$
View solution