Polynomials in one variable

A polynomial in one variable contains only one variable throughout its terms. The variable may have different nonnegative integer powers, and constants are allowed, but another letter cannot appear as an independent variable. The expression must also have the usual polynomial form, with no variable in a denominator or fractional exponent.

Option A, x^2+2x+1, contains only x, so it is a polynomial in one variable x. Option B contains both x and y, option C contains a and b, and option D contains x and y. Those are expressions in two variables, not in one variable. Therefore option A follows directly from counting the independent variables present.

The governing concept is the degree of a non-zero polynomial in one variable. It is the greatest exponent of the variable whose coefficient is non-zero. In 11x⁵ − 3x² + 8, the powers of x are 5, 2, and 0. The coefficient of x⁵ is 11, which is non-zero, so the highest relevant power is 5. Therefore, the degree of the polynomial is 5, making option C correct. The number 3 is only a coefficient, while 8 is the constant term. Also, the polynomial has three terms, but the number of terms does not determine its degree. The correct method is always to identify the largest exponent attached to the variable after checking that its coefficient is not zero.

Substituting \(x=1\) gives \(x^3-1=1^3-1=1-1=0\). Therefore, the correct answer is 0. Option B represents only the value of \(1^3\) and ignores the subtraction of 1. Exam tip: Substitute the given value into the complete polynomial before simplifying.

Substituting \(x=-1\) into the polynomial gives \(p(-1)=(-1)^2+(-1)=1-1=0\). Hence, the correct answer is 0. Option C results from calculating only \((-1)^2=1\) and omitting the \(+x\) term. Exam tip: substitute the given value at every occurrence of \(x\) and carefully handle the sign of a negative number raised to a power.

Substitute (x=2) in the polynomial: (p(2)=2(2)^2-8=2\times4-8=8-8=0). Therefore, the correct answer is 0. The value 8 may result from forgetting to subtract the constant term (8). Exam tip: evaluate powers first, then multiplication, and finally addition or subtraction.

The polynomial is written as x^2-5x+6. Its three terms are x^2, -5x, and 6, so the correct answer is 3. For exams, count the terms separated by plus or minus signs; do not count the exponent or coefficient as a separate term.

A constant polynomial contains no variable term and has a fixed value. Option A contains only the constant number 6, so it is a constant polynomial. Options B, C, and D contain x, so they are not constant polynomials. Exam tip: A non-zero polynomial of degree 0 is a constant polynomial.

Since \(a\ne0\), the term \(ax\) is present, and the highest power of \(x\) is \(1\). Therefore, \(p(x)=ax+b\) is a linear polynomial of degree \(1\). Degree 0 would be possible only if \(a=0\) and \(b\ne0\). Exam tip: The degree of a polynomial is the highest exponent of the variable having a non-zero coefficient.

The governing concept is classification by the degree of a polynomial. In p(x) = ax^2 + bx + c, the condition a ≠ 0 guarantees that the term ax^2 is present with a non-zero coefficient. Thus the greatest exponent of x is 2, so the polynomial has degree 2 and is called a quadratic polynomial. Therefore option B is correct. The values of b and c do not change this conclusion; even if b = 0 or c = 0, the non-zero x^2 term remains. It cannot be linear or constant because the degree-two term is present, and it cannot be cubic because no x^3 term occurs. If a were zero, the classification could change, which explains why the condition a ≠ 0 is essential.

The leading term is the term having the highest power of the variable. Here, the powers are 3, 2, and 1, so the leading term is x^3. The term 2x^2 comes next but has a lower power. Exam tip: Arrange the polynomial in descending powers and identify the first term.

\(p(2)=4\times2-8=0\). Therefore, \(x=2\) is a zero of the polynomial because a zero is a value that makes the polynomial equal to zero. Options B, C and D are incorrect: the degree is 1, the coefficient of \(x\) is 4, and the constant term is \(-8\). In an exam, test a possible zero by substituting it and checking whether the result is 0.

In a polynomial in one variable, the exponent of the variable must be a non-negative integer, such as \(0,1,2,\ldots\). Hence, \(x^0=1\) is acceptable and represents a constant term. \(x^{-1}\) and \(x^{-3}\) have negative exponents, while \(x^{\frac{1}{2}}\) has a fractional exponent; therefore, they cannot be polynomial terms. Exam tip: verify that every variable exponent is a non-negative integer.

Substituting \(x=0\) gives \(3x^2=3(0)^2=0\) and \(2x=2(0)=0\). Therefore, only the constant term \(1\) remains. Exam tip: a term without a variable is called the constant term.

A polynomial of degree 1 is called a linear polynomial, such as \(2x+3\). A constant polynomial has degree 0, a quadratic polynomial has degree 2, and a cubic polynomial has degree 3. In exams, identify the type of polynomial from its degree.

A polynomial whose highest power of the variable is 2, with a non-zero coefficient for that term, is called a quadratic polynomial. For example, \(3x^2+2x-5\) is quadratic. A linear polynomial has degree 1, a cubic polynomial has degree 3, and the degree of the zero polynomial is undefined. Exam tip: identify the polynomial by the highest power of its variable.

A polynomial whose highest power of the variable is 3 is called a cubic polynomial. A polynomial of degree 2 is quadratic, while one of degree 1 is linear, so option A is correct. Exam tip: identify the highest exponent of the variable to determine the polynomial's name.

In the polynomial \\(2x^2+3x+4\\), the term containing \\(x\\) is \\(3x\\), so its coefficient is 3. The term without a variable is 4, so the constant term is 4. Option B incorrectly uses 2, which is the coefficient of \\(x^2\\), not of \\(x\\). In an exam, identify the constant term as the term containing no variable.

Substitute x=-1 into the polynomial: p(-1)=(-1)^2-1=1-1=0. Hence, the correct answer is 0. The distractor -2 results from incorrectly treating (-1)^2 as -1. In an exam, use brackets around a negative value before squaring: (-1)^2=1.

In the standard form of a polynomial, its terms are arranged in descending order of their powers. In option A, the powers are 3, 2, 1 and 0, so it is in standard form. Options B, C and D contain the same terms but are not arranged in descending order. Exam tip: begin with the term having the highest power and then write the remaining terms in decreasing powers.

The polynomial contains \(x^4\), \(3x^2\), \(x\), and the constant term \(2\), but it does not contain an \(x^3\)-term. Therefore, the coefficient of \(x^3\) is considered to be 0. Exam tip: Arrange the terms in descending powers to identify a missing power quickly.

The governing concept is the constant term of a polynomial. A constant term is independent of the variable; equivalently, it is the coefficient of x⁰. The expression p(x) = 7x² has only the term 7x² and contains no separately written term without x. We may therefore write it as 7x² + 0, so the constant term is 0 and option C is correct. The number 7 is the coefficient of x², not the constant term, while 2 is the exponent of x. The number 1 has no role in the expression. A useful check is to put x = 0: p(0) = 7(0)² = 0. For a polynomial, its value at zero equals its constant term, confirming the answer.

Substituting \(x=-2\), we get \(p(-2)=(-2)^2+4(-2)+4=4-8+4=0\). Therefore, the correct answer is 0. The value 4 is only \((-2)^2\), not the value of the complete polynomial. In exams, use brackets carefully when substituting a negative number.

Degree is decided by the highest power (2), not by the number of terms. Avoid this common exam mistake.

There is no (x^2)-term in this polynomial, so its coefficient is (0). The coefficient of a missing term is always taken as (0).

To find \(p(1)\), substitute 1 for every \(x\) in the polynomial: \(p(1)=1^2+3\times1+2=1+3+2=6\). Therefore, the correct answer is 6. The nearby option 5 may result from adding the terms incorrectly or omitting a term. Exam tip: substitute the given value for each occurrence of the variable before simplifying.