Definition of a Polynomial

In a polynomial, the exponent of the variable \(x\) must be a non-negative integer, while coefficients may be any real numbers. In option A, \(2\), \(-\pi\), and \(\sqrt{13}\) are real coefficients, and the exponents of \(x\) are \(3\), \(2\), and \(0\). Option B has exponent \(-\sqrt{2}\), option C has exponent \(\frac{1}{2}\), and option D has exponent \(-1\); hence they are not polynomials. Exam tip: When \(x\) is under a root or in a denominator, check its exponent first.

The direct answer is A: the expression becomes a constant polynomial when \\(u=5\\). A constant polynomial must contain no variable term with a nonzero coefficient. Here the coefficients of both variable terms \\(x^3\\) and \\(x^2\\) are \\(u-5\\). Set that coefficient equal to zero: \\(u-5=0\\), so \\(u=5\\). Substitution gives \\(0x^3+0x^2+7=7\\), which is a constant polynomial. Option A is correct. Option B gives \\(u=-5\\), so \\(u-5=-10\\); the expression still contains nonzero \\(x^3\\) and \\(x^2\\) terms. Option C gives \\(u=0\\), so the common coefficient is \\(-5\\), and the variable terms remain. Option D is wrong because a suitable value does exist, namely 5. It is enough to make the common coefficient zero because the same coefficient controls both variable terms. The remaining number 7 is allowed as a constant. Exam cue: to make a polynomial constant, set every coefficient of a positive-power variable term equal to zero.

To find the degree after assigning a value to a parameter, substitute that value first and simplify completely. The degree is the greatest exponent of x whose coefficient is not zero. A term with a zero coefficient disappears and must not be counted, even if it originally has a high power.

With \\(r=2\\), the coefficient of \\(x^5\\) is \\(r^2-4=2^2-4=0\\), so that term vanishes. The coefficient of \\(x^4\\) is \\(r-1=2-1=1\\), leaving the polynomial \\(x^4+x\\). Its highest non-zero exponent is 4. Therefore option B is correct. The original power 5 cannot determine the degree after its coefficient becomes zero.

In option D, \(x^0=1\), so the expression is the constant polynomial \(1\), not the zero polynomial. In contrast, option B simplifies as \(x^0-1=1-1=0\), so it is a zero polynomial. Exam tip: simplify terms with exponent \(0\) first.

The degree of a polynomial is the greatest exponent of the variable, and the coefficient of that term is called the leading coefficient. In \(-6x^3+x^2-5\), the greatest exponent is 3 and the coefficient of \(x^3\) is \(-6\), which is negative. Therefore, option B is correct. Option A also has degree 3, but its leading coefficient, \(4\), is positive. Exam tip: identify the highest-power term first, then check the sign of its coefficient.

First apply the distributive property: \(x^2(x^2-3)=x^4-3x^2\). Therefore, the expression becomes \(x^4-3x^2-x^4+2x\). The like terms \(x^4\) and \(-x^4\) add to zero, leaving \(-3x^2+2x\). Option A incorrectly combines the \(x^4\) terms, which have opposite signs. Exam tip: after expanding brackets, combine only like terms with the same power of the variable.

In a polynomial, variable exponents are \(0,1,2,\ldots\). Here, \(x^{-1}=1/x\), so the variable occurs in the denominator and the expression is not a polynomial. Exam tip: check negative exponents first.

Substituting \(h=1\) gives \((1-1)x^6+(1+2)x^3-4x+9=3x^3-4x+9\). The \(x^6\) term disappears because its coefficient is zero. The highest power of \(x\) remaining is \(3\), so the degree is \(3\). Choosing \(6\) would be incorrect because a term with a zero coefficient is not part of the polynomial. Exam tip: after substituting a parameter value, remove all zero-coefficient terms before identifying the highest exponent.

For the polynomial to have degree 2, the coefficient of \(x^4\) must be zero, while the coefficient of \(x^2\), which is 5, remains non-zero. Thus, \(2a+6=0\) gives \(a=-3\). If \(a=-6\), the coefficient of \(x^4\) is \(-6\), so the degree remains 4. Exam tip: In a polynomial with a parameter, first check the coefficient of the highest power.

The direct answer is A, \(x^4-1\). The restriction \(x\neq0\) is important because division by zero is not allowed. For nonzero x, divide each term in the numerator by x: \(\frac{x^5}{x}=x^{5-1}=x^4\), and \(\frac{x}{x}=1\). Thus \(\frac{x^5-x}{x}=x^4-1\). Option A is correct. Option B, \(x^5-1\), fails because the first term must lose one power of x after division. Option C, \(x^4-x\), incorrectly leaves an x in the second term; \(x/x=1\), not x. Option D, \(x^3-1\), removes two powers from the first term instead of one. The simplified expression is valid only for the stated domain \(x\neq0\); the original expression itself is undefined at zero. Memory cue: when dividing powers with the same nonzero base, subtract exponents: \(x^m/x=x^{m-1}\).

In the polynomial \(x^4-1\), the highest exponent of \(x\) is \(4\). Therefore, its degree is \(4\). The constant term \(-1\) has degree \(0\), so it does not determine the degree. Exam tip: find the highest exponent of the variable in a polynomial to determine its degree.

The direct answer is D, Never. A constant polynomial must have zero coefficients for every term containing the variable. Here the required equations are v+1=0 and v-2=0. The first gives v=-1, while the second gives v=2. Since one value cannot be both -1 and 2, no single v can remove both variable terms. The remaining constant 3 is acceptable, but it does not solve the incompatible conditions. Option A, v=-1, removes the x² term, but the x coefficient becomes -3, so the expression still contains x. Option B, v=2, removes the x term, but the x² coefficient becomes 3, so it still contains x². Option C asks for both values simultaneously, which is impossible for one parameter. Option D is correct. Memory cue: set every variable coefficient equal to zero, then check whether the resulting values agree.

In \(4x^2-9\), the highest power of \(x\) is \(2\), so its degree is \(2\). There is no linear \(x\)-term, so the coefficient of \(x\) is \(0\). In option B, the coefficient of \(x\) is \(1\), so it is not correct. Exam tip: the coefficient of a missing term is always \(0\).

The degree of a polynomial is the highest power of the variable having a non-zero coefficient. Here, \(0x^9\) is a zero term, so \(x^9\) is not considered. Among the remaining terms, \(6x^5\) has the highest power; therefore, the degree is 5. Option 9 is incorrect because its coefficient is 0. Exam tip: Remove terms with zero coefficients before finding the degree.

The degree of a polynomial is determined only after substituting any given parameter value and removing zero terms. It is the highest exponent of x that still has a non-zero coefficient. A term may appear to have the greatest power initially, but it does not contribute to the degree if its coefficient becomes zero.

For \\(k=5\\), the coefficient of \\(x^7\\) is \\(k-5=5-5=0\\), so the seventh-degree term disappears. The expression becomes \\(4x^3-x+2\\). The non-zero powers present are 3, 1, and 0, and the greatest of these is 3. Thus option C is correct. Options A and B incorrectly retain or misidentify the vanished seventh-degree term.

A linear polynomial has degree 1. In the given expression, the coefficient of \(x^4\) is \(m+6\). For the polynomial to become linear, this coefficient must be zero: \(m+6=0\). Hence, \(m=-6\), and the polynomial becomes \(-2x+9\), which has degree 1. If \(m=0\), the term \(6x^4\) remains, so it is not linear. Exam tip: Before deciding the degree of a polynomial, check whether the coefficient of the highest-power term can become zero.

In a polynomial, the variable may have only non-negative whole-number powers. The term \\(3x^2\\) satisfies this rule because its power of x is 2. However, \\(\\sqrt{2x-1}\\) contains the variable expression 2x minus 1 inside a square root. It cannot be written as an ordinary polynomial term with an allowed whole-number power of x.

A polynomial may have two terms, so having two terms is not the reason for rejection. It may also contain the number 1, and the term \\(3x^2\\) is valid. The radical term is the deciding issue; it introduces a non-polynomial dependence on x. Therefore option C correctly explains why the complete expression is not a polynomial.

In a polynomial, the coefficient is the number multiplying the specified variable term. Here, the term in (x^4) is (0x^4), so its coefficient is 0. The number 5 is the coefficient of (x^5), while -3 is the coefficient of (x), not of (x^4). Exam tip: Find the term with the requested exponent; if it is absent, its coefficient is 0.

The direct answer is A, because x² appears in the denominator in the original form. By definition, a polynomial is written using non-negative integer powers of the variable, not a variable in a denominator. Indeed, for x≠0 the expression can be simplified: (x⁵+3x²)/x²=x³+3. The simplified expression is a polynomial, but the original rational expression has the restriction x≠0 and is not identical as a function at x=0, where the original expression is undefined. Option A states the relevant original-form reason. Option B is wrong because x⁵ is a perfectly allowed non-negative integer power. Option C is wrong because 3x² is also an ordinary polynomial term. Option D is wrong because addition is allowed in polynomials. Keep the distinction clear: simplification may produce a polynomial expression, but the original domain restriction remains.

The direct answer is A, x³+3. Because x≠0, cancellation by x² is allowed. Work term by term: x⁵/x²=x^(5−2)=x³, and 3x²/x²=3x^(2−2)=3. Therefore the whole expression becomes x³+3. Option A matches this calculation. Option B incorrectly keeps the original fifth power and also loses the effect of dividing both terms by x². Option C incorrectly leaves 3x² unchanged, although x² cancels completely from that term. Option D incorrectly reduces x⁵/x² to x² instead of x³ and therefore gives the wrong result. The condition x≠0 matters because division by x² and cancellation are not valid at x=0. The result is a polynomial in the simplified form, with degree 3 and constant term 3. Memory trick: when dividing powers with the same non-zero base, subtract exponents.

The degree of a polynomial is the greatest exponent of the variable in any of its terms. In \(x^3+3\), the term \(x^3\) has exponent 3, while the constant term 3 has degree 0. Therefore, the degree of the polynomial is 3. Option 0 would apply only to a non-zero constant polynomial. Exam tip: simplify first, then identify the highest power of the variable.

The correct answer is C: 4x^2-5|x|+8 is not a polynomial. In a polynomial, every variable term must use a non-negative integer power of x. The terms 4x^2 and 8 are acceptable, but |x| is not a power such as x^0, x^1, x^2, and so on. It is a modulus expression, whose rule changes according to whether x is positive or negative, so it is not a polynomial term in the Grade 9 definition. Option A is wrong because having x^2 does not repair another invalid term. Option B is wrong because a constant may occur in a non-polynomial expression too. Option C is correct. Option D is wrong because the number of terms does not decide whether an expression is a polynomial. Memory cue: check every term, not just one attractive term.

Direct answer: Option D, never, is correct. A constant polynomial must have no x-term or higher-power term. In the expression, the coefficient of x^2 is r + 4, so we need r + 4 = 0, giving r = −4. The coefficient of x is r − 1, so we also need r − 1 = 0, giving r = 1. One number r cannot be both −4 and 1. Therefore no value makes both variable terms disappear, and the expression cannot become constant. Option A removes the x^2 term, but leaves the x coefficient equal to −5, so the expression is still variable. Option B removes the x term, but leaves the x^2 coefficient equal to 5. Option C states both conditions together, but they are contradictory, not simultaneously possible. Option D correctly expresses the result. Exam cue: for a constant polynomial, set every coefficient of x, x^2, and higher powers to zero, then check whether one value satisfies all equations.

With respect to \(x\), \(y\) is treated as a coefficient. In option A, the powers of \(x\) are \(3,1\), and \(0\), all of which are non-negative integers. Therefore, \(y^2x^3-2yx+5\) is a polynomial in \(x\). In option B, \(x\) is in the denominator, so its exponent is \(-1\); in C, the exponent of \(x\) is \(\frac12\); and in D, it is \(-2\). Hence, these are not polynomials in \(x\). Exam tip: check exponents only of the variable in question; they must be \(0,1,2,\ldots\).

The polynomial \(9x^7-2x^5+4\) contains only an \(x^7\) term, an \(x^5\) term, and the constant term \(4\). It has no \(x^6\), \(x^4\), or \(x\) term, so their coefficients are \(0, 0, 0\), respectively. The number \(-2\) is the coefficient of \(x^5\), not of \(x^4\). Exam tip: if a power of the variable is missing from a polynomial, its coefficient is \(0\).