Chapter 6: Problem 50
Use a pattern to factor. Check. Identify any prime polynomials. $$ h^{2}+4 h+4 $$
Short Answer
Expert verified
\((h^2 + 4h + 4)\) factors to \((h + 2)^2\), and it is not a prime polynomial.
Step by step solution
01
- Identify the quadratic form
Notice that the polynomial is a quadratic trinomial in the form of \(ax^2 + bx + c\). Here, \(a = 1\), \(b = 4\), and \(c = 4\).
02
- Check for a perfect square trinomial
A perfect square trinomial takes the form \((x + k)^2 = x^2 + 2kx + k^2\). Compare this form to \(h^{2} + 4h + 4\). Here, \(x = h\) and we need to check if the middle term \(4h\) can be written as \(2 \times h \times k\) and the last term \(4\) as \(k^2\).
03
- Find the value of \(k\)
We solve \(2hk = 4h\) to find \(k\). By setting \(2h \times k = 4h\), we get \(k = 2\). Also, \(k^2 = 4\) is true for \(k = 2\). Hence, \(h^2 + 4h + 4\) can be factored as \((h + 2)^2\).
04
- Verify the factorization
Expand \((h + 2)^2\) to confirm the factorization: \((h + 2)(h + 2) = h^2 + 2h + 2h + 4 = h^2 + 4h + 4\). This matches the original polynomial, so \((h + 2)^2\) is correct.
05
- Check if it's a prime polynomial
A prime polynomial cannot be factored further. Since \(h^2 + 4h + 4\) has been factored as \((h + 2)^2\), it is not a prime polynomial.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
perfect square trinomial
A perfect square trinomial simplifies factoring. It follows a specific pattern. ewline ewline A perfect square trinomial has the form ewline ewline \((x + k)^2 = x^2 + 2kx + k^2\) ewline ewline Recognizing this pattern is crucial. When factoring \(h^2 + 4h + 4\), compare it to \((x + k)^2\). ewline Here, \(x = h\) and you compare terms to find \(k\). ewline Notice, 4h fits as \(2hk\). Therefore, \(k\) is determined as 2 because \(2h \times 2 = 4h\). ewline The last term matches \(k^2\). So \(k = 2\) because \(2^2 = 4\). This formation confirms a perfect square trinomial.
quadratic form
Quadratic form is represented by the equation \(ax^2 + bx + c\). ewline In this problem, \(h^2 + 4h + 4\), each coefficient is clear: ewline \(a = 1\), \(b = 4\), and \(c = 4\). ewline Recognizing the quadratic nature helps in identifying a pattern for factoring. Examining how these coefficients relate is essential. ewline When \(a = 1\), your trinomial might be a perfect square. This makes it easier to use simplification methods.
factoring polynomials
Factoring polynomials means breaking them into products of simpler polynomials. ewline For instance, factoring \(h^2 + 4h + 4\) results in \((h + 2)^2\). ewline To factor, notice the quadratic form first. Then use patterns or methods like grouping or the quadratic formula. ewline ewline Here's a common approach:
- Identify coefficients
- Check for special forms like perfect squares
- Apply patterns accordingly
prime polynomials
Prime polynomials cannot be factored further using real numbers. ewline In this exercise, checking whether \(h^2 + 4h + 4\) is prime involves factoring initially. ewline After factoring to \((h + 2)^2\), recognize it's not prime as it can be simplified. ewline If you can't simplify more, then the polynomial is prime. Here, the polynomial \(h^2 + 4h + 4\) breaks down into simpler factors, proving it's not prime. Confirming it's not prime ensures comprehension and correct completion of factoring tasks.