Problem 23
Question
For exercises 23-34, use a pattern to factor. Check. Identify any prime polynomials. $$ f^{2}-25 $$
Step-by-Step Solution
Verified Answer
\(f^2 - 25 = (f - 5)(f + 5)\). The polynomial is not prime.
1Step 1: Identify the pattern
The given polynomial is a difference of squares. Recall that the difference of squares formula is \[a^2 - b^2 = (a - b)(a + b)\].
2Step 2: Apply the formula
Compare the given polynomial to the difference of squares formula: \[f^2 - 25\]. Here, \(a = f\) and \(b = 5\). Substitute these values into the formula: \[f^2 - 25 = (f - 5)(f + 5)\].
3Step 3: Check for prime polynomials
The factors \((f - 5)\) and \((f + 5)\) are already in their simplest form since they cannot be factored further.
Key Concepts
Difference of SquaresPrime PolynomialsFactoring Patterns
Difference of Squares
One of the key factoring patterns you need to understand is the difference of squares. This pattern can be recognized when you have a polynomial of the form \(a^2 - b^2\).
The formula for factoring the difference of squares is:
\[ a^2 - b^2 = (a - b)(a + b) \]
This works because when you multiply \(a - b\) by \(a + b\), the middle terms cancel out, leaving you with \(a^2 - b^2\).
Let's look at an example to make this clearer:. If you have the polynomial \( f^2 - 25 \), you can identify this as a difference of squares because
* \(a = f\)
* \(b = 5\)
Now, using our formula, the polynomial can be written as:
\[ f^2 - 25 = (f - 5)(f + 5) \]
Knowing this pattern not only makes factoring quicker but also helps in identifying prime components easily.
The formula for factoring the difference of squares is:
\[ a^2 - b^2 = (a - b)(a + b) \]
This works because when you multiply \(a - b\) by \(a + b\), the middle terms cancel out, leaving you with \(a^2 - b^2\).
Let's look at an example to make this clearer:. If you have the polynomial \( f^2 - 25 \), you can identify this as a difference of squares because
* \(a = f\)
* \(b = 5\)
Now, using our formula, the polynomial can be written as:
\[ f^2 - 25 = (f - 5)(f + 5) \]
Knowing this pattern not only makes factoring quicker but also helps in identifying prime components easily.
Prime Polynomials
Prime polynomials are polynomials that cannot be factored further using integer coefficients. It is similar to prime numbers which are numbers greater than 1 that have no divisors other than 1 and themselves.
When factoring a polynomial, it's essential to determine if it's prime because it means you have reached the simplest form. For instance, in the example of \( (f - 5)(f + 5) \), neither of these factors can be broken down any further using integer coefficients. So, they are considered prime polynomials.
Here are some tips to determine if a polynomial is prime:
If none of these methods work, then the polynomial is likely prime.
When factoring a polynomial, it's essential to determine if it's prime because it means you have reached the simplest form. For instance, in the example of \( (f - 5)(f + 5) \), neither of these factors can be broken down any further using integer coefficients. So, they are considered prime polynomials.
Here are some tips to determine if a polynomial is prime:
- Check for common factors among the terms.
- See if it fits into recognizable factoring patterns like the difference of squares or the sum/difference of cubes.
- If it's a quadratic polynomial, check if you can factor it using the quadratic formula or by finding roots.
If none of these methods work, then the polynomial is likely prime.
Factoring Patterns
Factoring patterns are crucial tools in algebra that help simplify polynomials. By recognizing these patterns, you can factor polynomials quickly and accurately.
Here's a summary of common factoring patterns you should know:
For our given polynomial \( f^2 - 25 \), recognizing the difference of squares pattern allowed us to factor it efficiently. Keep practicing these patterns, and soon you'll recognize them effortlessly.
Here's a summary of common factoring patterns you should know:
- Difference of Squares: \[a^2 - b^2 = (a - b)(a + b)\]
- Perfect Square Trinomials: \[a^2 + 2ab + b^2 = (a + b)^2\] and \[a^2 - 2ab + b^2 = (a - b)^2\]
- Sum/Difference of Cubes: \[a^3 + b^3 = (a + b)(a^2 - ab + b^2)\] and \[a^3 - b^3 = (a - b)(a^2 + ab + b^2)\]
For our given polynomial \( f^2 - 25 \), recognizing the difference of squares pattern allowed us to factor it efficiently. Keep practicing these patterns, and soon you'll recognize them effortlessly.
Other exercises in this chapter
Problem 23
Solve. $$ (x-4)(x-4)=0 $$
View solution Problem 23
Factor completely. Identify any prime polynomials. $$ 90 z^{2}+120 z+40 $$
View solution Problem 23
For exercises 23-50, use the \(a c\) method to factor. Check the factoring. Identify any prime polynomials. $$ 2 c^{2}-c-21 $$
View solution Problem 23
(a) factor out the greatest common factor. Identify any prime polynomials. (b) check. $$ 15 a^{3} b^{6}-90 a b^{7}+35 a^{2} b^{5} $$
View solution