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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]
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