Problem 20
Question
For exercises 1-66, simplify. $$ \frac{z^{2}+8 z}{z+8} $$
Step-by-Step Solution
Verified Answer
The simplified form is \( z \).
1Step 1: Factor the numerator
The numerator of the fraction is \( z^2 + 8z \). Notice that \( z^2 + 8z \) can be factored by taking out the greatest common factor, which is \( z \). So, \( z^2 + 8z = z(z + 8) \).
2Step 2: Rewrite the fraction
Substitute the factored form of the numerator into the original fraction: \( \frac{z(z + 8)}{z + 8} \).
3Step 3: Simplify the fraction
Notice that the \( z + 8 \) terms in the numerator and the denominator cancel each other out. Therefore, the fraction simplifies to \( z \).
Key Concepts
Factoring PolynomialsGreatest Common FactorCanceling Terms
Factoring Polynomials
Factoring polynomials is a key skill in simplifying algebraic expressions. When you factor a polynomial, you look for terms that you can pull out as a common factor. This rewriting can help in simplifying rational expressions or solving equations.
In the given exercise, the polynomial in the numerator is \( z^2 + 8z \). Notice that both terms in the polynomial share a common factor, which is \( z \). You can factor this out to get:
\[ z^2 + 8z = z(z + 8) \].
By factoring the polynomial, you have transformed \( z^2 + 8z \) into \( z(z + 8) \), making it easier to simplify the overall expression.
In the given exercise, the polynomial in the numerator is \( z^2 + 8z \). Notice that both terms in the polynomial share a common factor, which is \( z \). You can factor this out to get:
\[ z^2 + 8z = z(z + 8) \].
By factoring the polynomial, you have transformed \( z^2 + 8z \) into \( z(z + 8) \), making it easier to simplify the overall expression.
Greatest Common Factor
The Greatest Common Factor (GCF) is the largest factor that two or more terms share. Finding the GCF is an important step in the factoring process because it allows you to simplify expressions.
In our exercise, the terms \( z^2 \) and \( 8z \) share a GCF of \( z \). This is because:
In our exercise, the terms \( z^2 \) and \( 8z \) share a GCF of \( z \). This is because:
- \( z^2 \) is the product of \( z \times z \)
- \( 8z \) is the product of \(8 \times z \).
- By pulling out the GCF \( z \), you factor the expression into \( z(z + 8) \). Always be sure to look for the GCF first, as it makes factoring easier and leads to simpler expressions.
Canceling Terms
Canceling terms in rational expressions is a method used to simplify the expression by eliminating common factors in the numerator and the denominator.
After factoring the numerator in our example, the expression becomes: \( \frac{z(z + 8)}{z + 8} \). Notice that both the numerator and the denominator have a \( z + 8 \) term. Since \( z + 8 \) is in both the numerator and the denominator, you can cancel them out:
\[ \frac{z(z + 8)}{z + 8} = z \].
Remember that canceling terms only works when the term you are canceling is a factor in both the numerator and the denominator.
Simplification by canceling terms is a powerful technique to make complex expressions more manageable.
After factoring the numerator in our example, the expression becomes: \( \frac{z(z + 8)}{z + 8} \). Notice that both the numerator and the denominator have a \( z + 8 \) term. Since \( z + 8 \) is in both the numerator and the denominator, you can cancel them out:
\[ \frac{z(z + 8)}{z + 8} = z \].
Remember that canceling terms only works when the term you are canceling is a factor in both the numerator and the denominator.
Simplification by canceling terms is a powerful technique to make complex expressions more manageable.
Other exercises in this chapter
Problem 20
For exercises 13-24, rewrite each expression as an equivalent expression with the given denominator. $$ \frac{3}{x^{3}-7 x^{2}-8 x} ; 6 x^{2}(x-8)(x+1) $$
View solution Problem 20
For exercises 7-32, simplify. $$ \frac{k^{2}+12 k+27}{k^{2}+7 k-18} \cdot \frac{k^{2}+5 k-14}{k^{2}+7 k+12} $$
View solution Problem 21
For a fixed length of household copper wire, the relationship of the cross- sectional area, \(x\), and the resistance, \(y\), is an inverse variation. When the
View solution Problem 21
For exercises \(9-24\), evaluate or simplify. $$ \frac{\frac{v^{2}-5 v+4}{v^{2}-6 v+8}}{\frac{v^{2}+2 v-3}{v^{2}+v-6}} $$
View solution