Polynomials in one variable

(\sqrt{2}) is a real number and the powers of the variable are whole numbers. Real coefficients are allowed in polynomials.

In (x^{\frac{1}{2}}), the power of the variable is not a whole number. In a polynomial, powers must be (0,1,2,\ldots).

There is no (x^3) term, so its coefficient is (0). The coefficient of a missing term is taken as (0).

The governing concept is the degree of a polynomial, defined as the greatest exponent of x having a nonzero coefficient. The zero polynomial has every coefficient equal to zero, so it has no nonzero term and therefore no greatest exponent under the usual school-level definition. Its degree is consequently not defined. This must be distinguished from a nonzero constant polynomial such as 5, whose only nonzero term is 5x⁰ and whose degree is 0. Thus option A is incorrect because it confuses the zero polynomial with a nonzero constant; options B and D assign exponents without a corresponding nonzero term. Option C correctly states the standard result.

Substituting \(x=-1\), we get \(p(-1)=2(-1)^2-3(-1)+4=2+3+4=9\). Hence, 9 is correct. Remember that \((-1)^2=1\) and \(-3(-1)=+3\); treating either sign incorrectly can lead to distractor values. Exam tip: Always place a negative substituted value in brackets.

The polynomial is arranged in descending powers of \(x\). The highest power is 4, so the term \(x^4\) is the leading term. Option C, \(2x^3\), has only degree 3 and therefore cannot be the leading term. In an exam, identify the term containing the highest power of the variable.

The leading term is (3x^2), so the leading coefficient is (3). The coefficient of the highest degree term is the leading coefficient.

A term is counted separately when it is written as an algebraic part of the polynomial. In 4x² + 0x + 9, the middle term 0x is explicitly present, and its coefficient is 0. Therefore exactly one written term has a zero coefficient. The term 4x² has coefficient 4, and the constant term 9 has coefficient 9, so neither qualifies. Although 0x contributes nothing to the numerical value and the expression can be simplified to 4x² + 9, the question asks what is explicitly written before simplification. Thus the correct count is 1. The answer is not 0 because 0x is visible, and it is not 2 or 3 because the other written terms have non-zero coefficients.

If (p(a)=0), then (a) is called a zero of the polynomial. To test a zero, substitute that number for (x).

In \(\frac{5}{x^2}+1=5x^{-2}+1\), the power of \(x\) is \(-2\). In a polynomial, the powers of the variable must be non-negative integers, so this expression is not a polynomial. Option D, \(7x+8\), is a linear polynomial because the power of \(x\) is 1. Exam tip: when the variable appears in the denominator, its power becomes negative, which generally disqualifies the expression from being a polynomial.

To find the zero, set \(p(x)=0\): \(x-6=0\), which gives \(x=6\). Therefore, the correct answer is 6. Choosing -6 results from an incorrect sign; the zero of \(x-a\) is \(a\). Exam tip: Always set the polynomial equal to zero when finding its zero.

There is no (x^2) term, so the coefficient is (0). A missing power can be treated as a term with zero coefficient.

In the polynomial \(9x-3\), the highest power of the variable \(x\) is 1 and its coefficient is non-zero; therefore, its degree is 1. Polynomial A has degree 2, polynomial C has degree 3, and the constant polynomial \(11\) has degree 0. Exam tip: identify the highest power of the variable with a non-zero coefficient.

In the polynomial \(x^2+4x\), the highest power of the variable \(x\) is 2, so its degree is 2. Option B has degree 4, option C has degree 1, and option D is a constant polynomial with degree 0. Exam tip: The degree of a polynomial is the highest exponent of its variable with a non-zero coefficient.

In the given polynomial, the term containing \(x\) to the first power is \(2x\), so its coefficient is 2. The number \(-6\) is the coefficient of \(x^2\), not of \(x\). In exams, first identify the term having the required power of the variable.

(\sqrt{x}=x^{\frac{1}{2}}), and (\frac{1}{2}) is not a whole number. So it is not a polynomial in (x).

Substituting \(x=-2\) gives \(p(-2)=(-2)^2+5=4+5=9\). Therefore, the correct answer is 9. Option 7 results from mishandling the squared term; \((-2)^2\) equals 4, not 2. Exam tip: To evaluate a polynomial, substitute the given value for the variable and simplify the powers before performing the remaining operations.

The governing concept is identification of the coefficient of a specified power. The coefficient of x means the numerical factor multiplying x^1. In x^3 + 4x^2 − 5, the visible terms have powers 3, 2 and 0; there is no x^1 term. A missing term is understood to have coefficient zero, so the polynomial can be written as x^3 + 4x^2 + 0x − 5. Therefore the coefficient of x is 0, making option C correct. Option A is the coefficient of x^3, option B is the coefficient of x^2, and option D is the constant term. The absence of a written x-term does not mean its coefficient is 1; it means the coefficient is zero.

\(p(3)=2(3)+1=7\) and \(p(1)=2(1)+1=3\). Therefore, \(p(3)-p(1)=7-3=4\), so option B is correct. Option C results from adding the two values, whereas the question asks for their difference. Exam tip: calculate \(p(3)\) and \(p(1)\) separately before subtracting.

Substituting \\(x=1\\) gives \\(3(1)^2+2(1)+1=3+2+1=6\\), so the correct value is 6. First evaluate the power; here, \\(1^2=1\\). Option 5 results from forgetting to add the constant term 1.