Get personalized answers to your unique questions on IDNLearn.com. Discover trustworthy solutions to your questions quickly and accurately with help from our dedicated community of experts.
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 contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.
