Polynomials in one variable

Correct answer: A. \(z^4-2z+1\)

Explanation: In \(z^4-2z+1\), only the variable \(z\) occurs, and its exponents 4, 1, and 0 are non-negative integers. Therefore, it is a polynomial in one variable, \(z\). Options B and C contain a negative power or a variable in the denominator, while option D contains two variables, \(z\) and \(w\). Exam tip: the powers of the variable in a polynomial must be non-negative integers.

Correct answer: A. \(2x^3\)

Explanation: The leading term is the term containing the variable with the highest exponent. In this polynomial, the exponents are 3, 2, 1, and 0, so \(2x^3\) is the leading term. \(-7x^2\) is the next term because its exponent is only 2. Exam tip: when a polynomial is written in descending powers, its first term is the leading term.

Correct answer: A. It is a polynomial

Explanation: The governing criterion is the definition of a polynomial in one variable: it is a finite sum of non-negative integral powers of the variable multiplied by coefficients from the specified number system. Dividing the entire expression by 2 gives (x^2 + 1)/2 = (1/2)x^2 + 1/2. This is a sum of x^2 and a constant, and both coefficients, 1/2 and 1/2, are real numbers. Hence it is a polynomial in x, with degree 2. Option A is correct. A denominator does not automatically prevent an expression from being a polynomial; the problem would arise if x occurred in a denominator or with a negative or fractional exponent. Only x appears, so it does not have two variables. Its degree is also defined because the polynomial is non-zero.

Correct answer: B. \(\frac{3}{x}+1\)

Explanation: In option B, \(\frac{3}{x}=3x^{-1}\). The variable has a negative exponent and occurs in the denominator, so the expression is not a polynomial. The other options contain only non-negative integer powers of the variable and are therefore polynomials. Exam tip: In a polynomial, the powers of the variable must be non-negative integers such as 0, 1, 2, 3, ... .

Correct answer: B. 2

Explanation: Substituting \(x=-1\), \(p(-1)=(-1)^2+2(-1)+3=1-2+3=2\). Therefore, option B is correct. A common mistake is to treat \(2(-1)\) as \(+2\), but it equals \(-2\). In exams, always use brackets when substituting a negative value.

Correct answer: C. 3

Explanation: Combining like terms gives \\(4x^3-2x^3=2x^3\\), so the simplified polynomial is \\(2x^3+x+5\\). The highest power of the variable is 3; therefore, its degree is 3. In an exam, first simplify like terms and then identify the highest exponent of the variable.

Correct answer: C. \(-7\)

Explanation: A constant polynomial does not contain a variable, so its value remains fixed. Since \(-7\) contains no variable \(x\), it is a constant polynomial. The other options contain \(x\), so they are non-constant polynomials. In an exam, first check whether the variable appears in the expression.

Correct answer: B. Quadratic

Explanation: The degree of a polynomial is the greatest exponent of the variable whose coefficient is not zero. In \\(p(x)=ax^2+bx+c\\), the coefficient of \\(x^2\\) is a, and the question states that \\(a\\ne0\\). Therefore the quadratic term is genuinely present, so the degree is 2. A polynomial of degree 2 is called a quadratic polynomial, making option B correct.

The values of b and c may be zero or nonzero, but they cannot remove the nonzero \\(ax^2\\) term. Hence the expression cannot become linear, constant, or cubic. The condition \\(a\\ne0\\) is essential: if a were zero, the degree could fall depending on b and c. Here that condition fixes the highest nonzero power as 2.

Correct answer: B. 1

Explanation: Since a\ne0, the term ax is present, so p(x) is a linear polynomial. Therefore, the highest power of x is 1. Degree 0 would apply only to a constant polynomial, which this is not. Exam tip: The degree of a polynomial is the highest exponent of the variable in a non-zero term.

Correct answer: C. (0)

Explanation: There is no (x)-term, so its coefficient is (0). In exams, treat the coefficient of a missing term as (0).

Correct answer: A. \(\sqrt{2}x^2-3x+1\)

Explanation: In option A, the powers of \(x\) are 2, 1, and 0, all of which are non-negative integers. Since \(\sqrt{2}\) is a real number, it is a polynomial with real coefficients. Option B contains \(x^{1/2}\), while C and D contain negative powers, so they are not polynomials. Exam tip: In a polynomial, the powers of the variable must be 0, 1, 2, 3, and so on.

Correct answer: C. 3

Explanation: The form \(ax^2+bx+c\) contains three coefficients: \(a\), \(b\), and \(c\). In the given polynomial, their values are \(3\), \(4\), and \(5\), respectively, so there are three coefficients. The symbol \(x\) is the variable, not a coefficient. Exam tip: identify the numerical factor multiplying each power of the variable.

Correct answer: A. \(x^4+x^2+1\)

Explanation: The degree of a polynomial is the greatest exponent of the variable having a non-zero coefficient. In \(x^4+x^2+1\), the greatest exponent is \(4\), so its degree is \(4\). The degrees of options B, C and D are \(3\), \(2\) and \(1\), respectively. Exam tip: identify the term with the highest power of the variable and read its exponent.

Correct answer: C. 5

Explanation: In the polynomial p(x)=2x^2+kx+3, the term containing x is kx, so its coefficient is k. Since this coefficient is given as 5, k=5. Exam tip: identify the coefficient as the numerical factor multiplied by the variable.

Correct answer: C. \(4\)

Explanation: A constant term is a term that does not contain the variable \(x\). In this polynomial, both \(x^2\) and \(ax\) contain \(x\), whereas \(4\) does not; therefore, the constant term is \(4\). Exam tip: the constant term can be viewed as the coefficient of \(x^0\).

Correct answer: A. Zero polynomial

Explanation: A polynomial whose value is zero for every \(x\), with all its coefficients equal to zero, is called the zero polynomial. Its degree is undefined, so it is not classified as linear, quadratic, or cubic. Exam tip: Interpret \(p(x)=0\) as the zero polynomial when it is true identically for every \(x\).

Correct answer: A. 0

Explanation: To evaluate the polynomial, substitute \(x=1\): \(p(1)=1^3-1=1-1=0\). Therefore, option A is correct. Option B results from forgetting to subtract 1. Exam tip: To find \(p(a)\), substitute \(a\) directly for \(x\) in the polynomial.

Correct answer: B. \(x^2+\frac{1}{3}x+2\)

Explanation: In a polynomial, the powers of the variable must be non-negative integers. In option B, the powers of \(x\) are 2, 1, and 0, and \(\frac{1}{3}\) is a valid real coefficient; therefore, it is a polynomial. Options A and D contain fractional powers, while option C has \(x\) in the denominator, giving it a negative power. Exam tip: an expression is not a polynomial if the variable has a fractional or negative exponent.

Correct answer: D. 0

Explanation: The polynomial has no \(x^2\) term, so its coefficient is 0. The value \(-5\) is the coefficient of \(x^3\), while 1 is the coefficient of \(x\). Exam tip: if a power is missing from a polynomial, its coefficient is taken as 0.

Correct answer: B. 5

Explanation: Substitute \(x=3\) into the polynomial: \(p(3)=3^2-4=9-4=5\). Therefore, option B is correct. Option C is only the value of \(3^2\), without subtracting 4. Exam tip: evaluate the power first, then apply the constant term.

Correct answer: A. 12

Explanation: The governing definition states that every non-zero constant polynomial has degree 0, because it can be written as a constant times x^0 and contains no higher power of x. Option A, the polynomial 12, is non-zero and constant, so its degree is 0. The polynomial x + 12 has highest exponent 1 and therefore degree 1. The polynomial x^2 + 12 has highest exponent 2 and therefore degree 2. The zero polynomial in option D is special: its degree is generally left undefined in school-level algebra because it has no highest non-zero power. Thus option D is not a second answer, and option A is the only unambiguous correct choice. The non-zero condition is essential when identifying a constant polynomial of degree zero.

Correct answer: B. 2

Explanation: The polynomial \(7x^2-1\) has two terms: \(7x^2\) and \(-1\). The minus sign belongs to the second term; it is not counted as a separate term. Therefore, the correct answer is 2, and the polynomial is a binomial. Exam tip: Count the distinct monomial parts separated by plus or minus signs.

Correct answer: B. (x^3+2x+5)

Explanation: A polynomial in standard form is written in descending order of powers of the variable. The term with the highest exponent comes first, followed by terms with successively smaller exponents, and the constant term comes last. In option B, the terms are arranged as \(x^3+2x+5\), with powers 3, 1, and 0. Therefore option B is the expression in standard polynomial form.

The missing \(x^2\) term is understood to have coefficient zero, so it does not need to be written. The other options contain the same terms but place the constant or lower-power terms before the cubic term. They are mathematically equivalent expressions, but they are not arranged in the requested standard descending-power order.

Correct answer: A. \(0\)

Explanation: Substitute \(x=-2\) into the polynomial: \(p(-2)=2(-2)^2+3(-2)-2=2(4)-6-2=0\). Therefore, option A is correct. Option D usually results from a calculation error, such as mishandling \(3(-2)\) or omitting the constant term \(-2\). Exam tip: the square of a negative number is positive, so \((-2)^2=4\).

Correct answer: A. It is a polynomial

Explanation: A polynomial in x may have real-number coefficients, including irrational numbers such as \\(\\sqrt3\\). What matters is that the powers of x are nonnegative whole numbers and that x is not placed in a denominator or under a variable-dependent radical. In \\(x^2+\\sqrt3x+1\\), the powers are 2, 1, and 0, so it is a polynomial. Thus option A is correct.

The coefficient of x is \\(\\sqrt3\\), which is a fixed real number; it is not a variable expression. The terms have the required forms \\(x^2\\), \\(\\sqrt3x\\), and 1. Its highest nonzero power is 2, so its degree is also defined and equals 2. The presence of an irrational coefficient does not disqualify a polynomial.