Polynomials in one variable

A polynomial has only non-negative integer powers of (x). In exams check the power of every term.

The degree of a polynomial is the greatest exponent of the variable with a non-zero coefficient. In \(7x^3-2x+9\), the greatest exponent is 3, so the degree is 3. The number 9 is the constant term, not the degree. Exam tip: identify the highest power of the variable.

The term containing \(x\) is \(0x\), so its coefficient is 0. The number 4 is the coefficient of \(x^2\), while 6 is the constant term. In an exam, identify the variable and its power before choosing the coefficient.

(0) is the zero polynomial because every coefficient is zero. Do not apply the usual degree rule to it.

\(5\) is a non-zero constant polynomial. It can be written as \(5x^0\), so the highest power of the variable is \(0\). Therefore, its degree is \(0\). Remember that the degree of the zero polynomial is undefined, but \(5\) is not the zero polynomial.

To find \(p(2)\), substitute 2 for \(x\) in the polynomial: \(p(2)=2\times2+3=4+3=7\). Therefore, the correct answer is 7. Getting 5 would mean ignoring the multiplication. In an exam, substitute the value first, then perform multiplication before addition.

Substitute 3 for \(x\): \(q(3)=3^2-4=9-4=5\). Therefore, the correct value is 5. Option 9 results from calculating only \(3^2\) and forgetting to subtract 4. Exam tip: When evaluating a polynomial, substitute the given value carefully and calculate powers before addition or subtraction.

Substituting \(x=6\) in the polynomial gives \(p(6)=6-6=0\). Therefore, 6 is a zero of the polynomial. Remember that the given value is the input; the polynomial’s value is obtained only after substitution, so 6 is not the answer.

A number is a zero only when the value of the polynomial becomes 0. Since \(p(-4)=-4+4=0\), \(-4\) is the correct answer. Substituting \(4\) gives \(p(4)=8\), not 0. In an exam, verify a possible zero by substituting it into the polynomial and checking whether the result is 0.

The terms of this polynomial are \(3x^2\), \(2x\), and \(-1\), so it has 3 terms. Terms are identified by separating the expression at plus or minus signs. Counting only \(3x^2\) and \(2x\) gives 2, but the constant term \(-1\) is also a separate term. Exam tip: Count negative terms as well as positive terms.

In \(7x^2-3x+1\), the powers of \(x\) are 2, 1 and 0, all non-negative integers, so it is a polynomial in one variable. \(\frac{1}{x}\) has power \(-1\). Exam tip: polynomial exponents must be 0, 1, 2, ... .

The governing concept is classification of a non-zero polynomial by its degree. The degree is the greatest exponent of the variable whose coefficient is not zero. In 9x + 1, the term 9x can be written as 9x^1, so its exponent is 1. The constant term 1 can be viewed as 1x^0, but 0 is smaller than 1. Therefore the greatest exponent is 1, and the polynomial is linear. Option B is correct. It is not constant because its value changes when x changes. It is not quadratic or cubic because there is no non-zero x^2 or x^3 term. The coefficient 9 affects the value and slope, but it does not alter the degree or the polynomial's classification.

To classify a polynomial, identify the greatest power of the variable with a non-zero coefficient. In x² + 5x + 6, the term x² has exponent 2 and coefficient 1, which is non-zero. The remaining terms have degrees 1 and 0. Hence the polynomial has degree 2 and is called a quadratic polynomial. It is not linear because a linear polynomial has highest exponent 1. It is not cubic because there is no x³ term. It is also not the zero polynomial: the expression has several non-zero coefficients and is not identically equal to zero. Although it can be factored as (x + 2)(x + 3), factorisation does not change its degree or its classification.

The constant term is the term that does not contain the variable \(x\). In this polynomial, \(6x^2\) and \(-3x\) contain \(x\), whereas 10 does not; therefore, the constant term is 10. Exam tip: identify the term independent of the variable, not the coefficient of a variable term.

In this polynomial, the term containing \\(x^2\\) is \\(-5x^2\\), so its coefficient is \\(-5\\). The number 8 is the coefficient of \\(x^3\\), while 4 is the constant term. In an exam, match the variable’s power carefully and do not omit the negative sign.

A monomial is a polynomial with exactly one term. \(3x^2\) has only one term, so it is a monomial. \(x+2\) has two terms and is a binomial, while \(x^2+x\) and \(x^2+x+1\) have two and three terms respectively. Exam tip: Count the parts separated by plus or minus signs to identify the number of terms.

A binomial is a polynomial containing exactly two unlike terms. In \(x^2+3\), the terms are \(x^2\) and \(3\), so it is a binomial. Option B has three terms and is a trinomial, while C and D each have only one term. Exam tip: classify a polynomial by counting its non-zero terms.

A polynomial containing three unlike terms is called a trinomial. In \(x^2-2x+1\), the three terms are \(x^2\), \(-2x\), and \(1\), so it is a trinomial. Option C has only two terms and is therefore a binomial. Exam tip: count the terms separated by plus or minus signs, treating the sign as part of the term.

To find \(p(0)\), substitute \(x=0\): \(p(0)=0^2+2\times0+1=1\). Therefore, the correct answer is 1. Option 2 is incorrect because the term \(2x\) becomes 0 when \(x=0\). Exam tip: To evaluate \(p(a)\), replace every occurrence of \(x\) with \(a\).

A zero of a polynomial is the value of \(x\) for which \(p(x)=0\). Thus, \(3x-12=0\), so \(3x=12\) and \(x=4\). Option D is incorrect because the constant term must be divided by the coefficient of \(x\). Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\).

To find the zero, set \(p(x)=0\): \(5x+10=0\), so \(5x=-10\) and \(x=-2\). Therefore, option B is correct. Option A results from missing the negative sign; substituting \(x=2\) gives 20, not zero. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\).

A zero of a polynomial is a value that makes the polynomial equal to zero. Here, \(p(3)=3^2-9=9-9=0\), so 3 is a zero. For example, \(p(2)=4-9=-5\), so 2 is not a zero. In an exam, substitute each option in the polynomial, or solve \(x^2=9\) to obtain \(x=\pm3\).

Substituting \(x=1\) gives \(x^2+1=1^2+1=1+1=2\), so the correct answer is 2. Choosing 1 means ignoring the constant term \(+1\). In the exam, calculate the power first and then perform the addition.

The term with the highest power of \(x\) is \(2x^2\). Its coefficient is 2, so the leading coefficient is 2. The coefficients of the other terms, -7 and 1, are not the leading coefficient.

Exam tip: Arrange the polynomial in descending powers and identify the coefficient of the highest-degree term.

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