Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
To determine which of the given expressions are polynomials, let's first recall the definition of a polynomial. A polynomial in one variable [tex]\( x \)[/tex] is an expression of the form:
[tex]\[ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0, \][/tex]
where [tex]\( a_n, a_{n-1}, \ldots, a_1, a_0 \)[/tex] are constants (real numbers) and [tex]\( n \)[/tex] is a non-negative integer. A crucial point is that the exponents of [tex]\( x \)[/tex] should be non-negative integers (0, 1, 2, ...).
Now, let's analyze each given expression:
(i) [tex]\( 2x^2 + 5x + 6 \)[/tex]
- This expression consists of terms with [tex]\( x \)[/tex] raised to the powers of 2, 1, and 0, all of which are non-negative integers.
- Therefore, [tex]\( 2x^2 + 5x + 6 \)[/tex] is a polynomial.
(ii) [tex]\( x^2 + 6x \)[/tex]
- This expression consists of terms with [tex]\( x \)[/tex] raised to the powers of 2 and 1, which are non-negative integers.
- Therefore, [tex]\( x^2 + 6x \)[/tex] is a polynomial.
(iii) [tex]\( \sqrt{2} x^2 + \sqrt{3} x \)[/tex]
- This expression has terms with [tex]\( x \)[/tex] raised to the powers of 2 and 1, which are non-negative integers.
- The coefficients are [tex]\( \sqrt{2} \)[/tex] and [tex]\( \sqrt{3} \)[/tex], which are real numbers.
- Therefore, [tex]\( \sqrt{2} x^2 + \sqrt{3} x \)[/tex] is a polynomial.
(iv) [tex]\( 2x + \frac{1}{x^2} \)[/tex]
- The expression [tex]\( 2x \)[/tex] has [tex]\( x \)[/tex] raised to the power of 1, which is a non-negative integer.
- However, [tex]\( \frac{1}{x^2} \)[/tex] can be written as [tex]\( x^{-2} \)[/tex], where [tex]\(-2\)[/tex] is a negative integer.
- Since [tex]\( x^{-2} \)[/tex] is not a valid exponent for polynomials, [tex]\( 2x + \frac{1}{x^2} \)[/tex] is not a polynomial.
(v) [tex]\( x^2 + \sqrt{2x} + 6 \)[/tex]
- The term [tex]\( x^2 \)[/tex] has [tex]\( x \)[/tex] raised to the power of 2, and [tex]\( 6 \)[/tex] is a constant term.
- However, [tex]\( \sqrt{2x} \)[/tex] can be rewritten as [tex]\( (\sqrt{2} \cdot x^{1/2}) \)[/tex], where [tex]\( 1/2 \)[/tex] is not a non-negative integer.
- Since the exponent [tex]\( x^{1/2} \)[/tex] does not meet the criteria for polynomials, [tex]\( x^2 + \sqrt{2x} + 6 \)[/tex] is not a polynomial.
In summary, the expressions that are polynomials are:
(i) [tex]\( 2x^2 + 5x + 6 \)[/tex]
(ii) [tex]\( x^2 + 6x \)[/tex]
(iii) [tex]\( \sqrt{2} x^2 + \sqrt{3} x \)[/tex]
The expressions that are not polynomials are:
(iv) [tex]\( 2x + \frac{1}{x^2} \)[/tex]
(v) [tex]\( x^2 + \sqrt{2x} + 6 \)[/tex]
[tex]\[ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0, \][/tex]
where [tex]\( a_n, a_{n-1}, \ldots, a_1, a_0 \)[/tex] are constants (real numbers) and [tex]\( n \)[/tex] is a non-negative integer. A crucial point is that the exponents of [tex]\( x \)[/tex] should be non-negative integers (0, 1, 2, ...).
Now, let's analyze each given expression:
(i) [tex]\( 2x^2 + 5x + 6 \)[/tex]
- This expression consists of terms with [tex]\( x \)[/tex] raised to the powers of 2, 1, and 0, all of which are non-negative integers.
- Therefore, [tex]\( 2x^2 + 5x + 6 \)[/tex] is a polynomial.
(ii) [tex]\( x^2 + 6x \)[/tex]
- This expression consists of terms with [tex]\( x \)[/tex] raised to the powers of 2 and 1, which are non-negative integers.
- Therefore, [tex]\( x^2 + 6x \)[/tex] is a polynomial.
(iii) [tex]\( \sqrt{2} x^2 + \sqrt{3} x \)[/tex]
- This expression has terms with [tex]\( x \)[/tex] raised to the powers of 2 and 1, which are non-negative integers.
- The coefficients are [tex]\( \sqrt{2} \)[/tex] and [tex]\( \sqrt{3} \)[/tex], which are real numbers.
- Therefore, [tex]\( \sqrt{2} x^2 + \sqrt{3} x \)[/tex] is a polynomial.
(iv) [tex]\( 2x + \frac{1}{x^2} \)[/tex]
- The expression [tex]\( 2x \)[/tex] has [tex]\( x \)[/tex] raised to the power of 1, which is a non-negative integer.
- However, [tex]\( \frac{1}{x^2} \)[/tex] can be written as [tex]\( x^{-2} \)[/tex], where [tex]\(-2\)[/tex] is a negative integer.
- Since [tex]\( x^{-2} \)[/tex] is not a valid exponent for polynomials, [tex]\( 2x + \frac{1}{x^2} \)[/tex] is not a polynomial.
(v) [tex]\( x^2 + \sqrt{2x} + 6 \)[/tex]
- The term [tex]\( x^2 \)[/tex] has [tex]\( x \)[/tex] raised to the power of 2, and [tex]\( 6 \)[/tex] is a constant term.
- However, [tex]\( \sqrt{2x} \)[/tex] can be rewritten as [tex]\( (\sqrt{2} \cdot x^{1/2}) \)[/tex], where [tex]\( 1/2 \)[/tex] is not a non-negative integer.
- Since the exponent [tex]\( x^{1/2} \)[/tex] does not meet the criteria for polynomials, [tex]\( x^2 + \sqrt{2x} + 6 \)[/tex] is not a polynomial.
In summary, the expressions that are polynomials are:
(i) [tex]\( 2x^2 + 5x + 6 \)[/tex]
(ii) [tex]\( x^2 + 6x \)[/tex]
(iii) [tex]\( \sqrt{2} x^2 + \sqrt{3} x \)[/tex]
The expressions that are not polynomials are:
(iv) [tex]\( 2x + \frac{1}{x^2} \)[/tex]
(v) [tex]\( x^2 + \sqrt{2x} + 6 \)[/tex]
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.