Definition of a Polynomial

In the polynomial 3x^4-2x^2+5, there is no x^3 term and no x term. The coefficient of a missing power is taken as 0. Hence, the coefficients of x^3 and x are 0 and 0 respectively. In option A, 3 and -2 are actually the coefficients of x^4 and x^2. Exam tip: Always assign coefficient 0 to any missing power in a polynomial.

In \(0x^5+0x^2+0\), the coefficient of every power of \(x\) is 0, and the constant term is also 0. Hence, it equals 0 for every value of \(x\) and is the zero polynomial. \(0x^5+x=x\) is not a zero polynomial because the coefficient of \(x\) is 1. Exam tip: to identify a zero polynomial, check that every coefficient, including the constant term, is 0.

First, multiplying \(x^2\) by \((x+1)\) gives \(x^3+x^2\). Thus, the expression becomes \(x^3+x^2-x^3+4\). Since \(x^3-x^3=0\), the remaining polynomial is \(x^2+4\). Option A fails to cancel the \(-x^3\) term. Exam tip: after expanding brackets, combine only like terms having the same power of the variable.

In the polynomial \(x^2+4\), the highest power of \(x\) is 2. The constant term \(4\) has degree 0, so it does not increase the degree. Therefore, the degree of the polynomial is 2. Exam tip: Find the degree by identifying the highest exponent of the variable in the simplified polynomial.

(\sqrt{3}) and (-\frac{1}{5}) are real coefficients. The powers of (x) are (0), (2), and (1), so the degree is (2).

Putting (s=2) makes the coefficient of (x^4) equal to (0) and the coefficient of (x^3) equal to (3). So the highest power is (3).

Dividing by (x^2) gives (\frac{x^4}{x^2}=x^2) and (\frac{x^2}{x^2}=1). Therefore the simplified form is (x^2-1).

The coefficient of 0x^8 is 0, so it is a zero term and does not determine the degree of the polynomial. In the remaining polynomial, -5x^4+3x^2-6, the highest power of x is 4. Therefore, its degree is 4. Choosing 8 would be incorrect because a term with coefficient zero is not an effective term of the polynomial. Exam tip: Remove all zero-coefficient terms before finding the degree.

Substitute k=-3 into the polynomial. The coefficient of x^6 becomes k+3=-3+3=0, so the term 0x^6 disappears completely. The expression is then 2x^2+x-7. Its nonzero terms have powers 2, 1, and 0, and the highest of these powers is 2. Therefore the degree of the resulting polynomial is 2, so option A is correct.

It is important not to decide the degree from the original visible exponent 6 before evaluating its coefficient. A term with zero coefficient is absent and cannot determine the degree. The expression is not a constant because the terms 2x^2 and x remain, and it is not degree 6 because the x^6 term vanishes. Thus the substitution and removal of the zero term lead directly to degree 2.

A linear polynomial has degree 1. In the given expression, the degree-3 term is \((m-4)x^3\). Its coefficient must be zero to remove this term: \(m-4=0\). Hence, \(m=4\), and the expression becomes \(7x-2\), which is linear. For \(m=0\) or \(m=-4\), the coefficient of x³ is not zero, so the polynomial remains cubic. Exam tip: To reduce the degree of a polynomial, set the coefficient of its highest-degree term to zero.

(\sqrt{x+4}) is not written using non-negative integer powers of (x). Therefore the whole expression is not a polynomial.

In the polynomial, the term containing x is 0x. The number multiplying x is called its coefficient, so the coefficient of x is 0. Here, 5 is the constant term, while -2 is the coefficient of x². Exam tip: Identify the term with the requested power of x; if that term is absent, its coefficient is 0.

In the original form, (x) is in the denominator, so it is not directly called a polynomial. For simplification, keep the condition (x\neq0) in mind.

The direct answer is A: \\(x^3+2\\). Since \\(x\\neq0\\), division by x is permitted. Split the numerator term by term: \\(\\frac{x^4+2x}{x}=\\frac{x^4}{x}+\\frac{2x}{x}\\). Using \\(x^m/x=x^{m-1}\\) for nonzero x, \\(x^4/x=x^3\\). Also, \\(2x/x=2\\). Therefore the result is \\(x^3+2\\). Option A is correct. Option B, \\(x^4+2\\), fails to reduce the first power and incorrectly leaves the second term as 2 after removing x inconsistently. Option C, \\(x^3+2x\\), simplifies the first term but fails to cancel x in the second term. Option D, \\(x^2+2\\), subtracts 2 from the exponent instead of 1 in the first term, so it is incorrect. The condition \\(x\\neq0\\) matters because division by zero is not defined. One may also factor the numerator: \\(x^4+2x=x(x^3+2)\\), and then cancel x to get \\(x^3+2\\). Memory cue: divide each term by x, reducing the exponent by one when the term contains x.

The degree of a polynomial is the greatest exponent of the variable in any of its terms. Here, the terms are x^3 and 2, and the greatest power of x is 3. Hence, the degree is 3. Option 0 is the degree only of a non-zero constant polynomial. Exam tip: First write the polynomial in simplified form, then identify the highest exponent of the variable.

(|x|) is not written as a term with non-negative integer powers of (x). Therefore it is not a polynomial.

The direct answer is D, Never. A constant polynomial must have no variable term. Here the coefficients of both variable terms must therefore be zero: first, r-1=0 gives r=1; second, r+2=0 gives r=-2. One number r cannot be both 1 and -2 at the same time. The constant term 5 is already acceptable, but both variable terms must disappear. Option A, r=1, removes the x² term, but the x coefficient becomes 3, so the expression is still variable. Option B, r=-2, removes the x term, but the x² coefficient becomes -3, so it is still not constant. Option C states both values together, but one parameter cannot have two different values simultaneously; hence it is not a valid single value. Option D is correct because no single value satisfies both equations. Memory cue: for a constant polynomial, set every variable coefficient to zero and check whether all conditions have one common value.

The polynomial contains only the terms \(4x^6\), \(-3x^4\), and \(2\). There is no \(x^5\) term or \(x\) term, and the coefficient of a missing term is taken as \(0\). Hence, both coefficients are \(0\). The numbers \(6\) and \(1\) are exponents, not coefficients. Exam tip: if a term of a given power is absent, its coefficient is always \(0\).

First multiply \(x^3\) by \((x-2)\): \(x^3(x-2)=x^4-2x^3\). Thus, the expression becomes \(x^4-2x^3-x^4+6x\). Since \(x^4-x^4=0\), the remaining polynomial is \(-2x^3+6x\). Option A incorrectly leaves the \(-x^4\) term uncancelled. Exam tip: after expanding brackets, combine only like terms with the same power of the variable.

The degree of a polynomial is the greatest exponent of the variable in any term with a non-zero coefficient. In
(-2x^3+6x), the exponents of x are 3 and 1. The greatest exponent is 3, so its degree is 3. The number 6 is a coefficient, not an exponent. Exam tip: To find a degree, look at the powers of the variable, not the coefficients.

(\pi) and (-\frac{2}{7}) are real coefficients. The powers of (x) are (0), (3), and (1), so the degree is (3).

The degree of a polynomial is determined by the highest power of its variable. In \(x^2+x-1\), the highest power of \(x\) is \(2\), so it is a quadratic polynomial. A linear polynomial has highest power \(1\), whereas a cubic polynomial has highest power \(3\). Exam tip: first write the polynomial in simplified form, then identify its highest exponent.

Since \(x^0=1\), the expression becomes \(x^5+x^3+1\). The degree of a polynomial is the highest power of its variable, which is \(5\) here. \(3\) is only the power of the term \(x^3\), not the degree of the whole polynomial. Exam tip: simplify terms such as \(x^0\) first, then identify the greatest exponent of the variable.

The terms are \(6x^4\), \(-2x^3\), \(0x^2\), \(5x\), and \(-9\). Since the coefficient of \(0x^2\) is 0, it is not a non-zero term. The other four terms are non-zero, so the answer is 4. Choosing 5 is a common mistake because it counts the zero term as well. Exam tip: Count only terms whose coefficients are not zero.

With respect to (x), (3y^2-5y+1) has no (x) term. So it can be treated as a constant polynomial in (x).