Polynomials in one variable

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.

Option B, 5x³ − 4, contains two distinct terms: 5x³ and −4. Therefore, it is a binomial. Option A has three terms, option C has one term, and option D has four terms. In an exam, count each part separated by a plus or minus sign as one term.

The term containing \(x\) is \(-x\), which can be written as \(-1x\). Therefore, the coefficient of \(x\) is \(-1\). Option C misses the negative sign. Exam tip: when identifying a coefficient, include both the numerical multiplier and its sign.

The degree of a polynomial is the highest power of its variable. In \\(x^5+1\\), the highest power of \\(x\\) is 5, so its degree is 5. The term 1 is a constant and does not increase the degree. In an exam, identify the maximum exponent of the variable rather than adding the exponents.

The direct answer is option A: \(y^2-3y+2\) is a polynomial in \(y\). A polynomial has a variable with only non-negative whole-number powers, such as \(0,1,2,\ldots\), and its coefficients may be real numbers. In option A, the powers of \(y\) are 2, 1, and 0, so it satisfies the rule. Option B contains \(1/y=y^{-1}\), a negative power, so it is not a polynomial. Option C contains \(\sqrt y=y^{1/2}\), a fractional power, so it is not a polynomial. Option D contains \(y^{-2}\), again a negative power, so it is not a polynomial. The important point is that the variable’s powers must be whole numbers; the presence of a variable in a denominator or under a root usually breaks the rule. Exam cue: check powers of the variable first.

\(r\left(\frac{1}{2}\right)=4\left(\frac{1}{2}\right)^2=1\). Use brackets when substituting fractions.

Terms in a polynomial are separated by plus or minus signs. Here, \(7x^2\) and \(5\) are two distinct terms, so the polynomial has 2 terms. In an exam, count each part separated by a plus or minus sign as one term.

Since \(a\neq 0\), the term \(ax^2\) is non-zero, so the highest power of \(x\) is 2. Therefore, the polynomial has degree 2 and is a quadratic polynomial. A linear polynomial has degree 1, while a cubic polynomial has degree 3. Exam tip: Identify the polynomial by its highest power of the variable.

The governing concept is classification by polynomial degree. A polynomial is quadratic exactly when the highest power of its variable with a non-zero coefficient is 2. In p(x) = ax² + bx + c, the coefficient of x² is a. Therefore a must be non-zero, so a ≠ 0 is the required condition and option B is correct. If a = 0, the x² term disappears and the expression becomes bx + c, which may be linear or constant. The values b = 0 or c = 0 are not necessary: for example, 2x² + 0x + 3 is still quadratic, and 2x² + 5x + 0 is also quadratic. Thus the other options do not correctly state the defining condition.

A polynomial is in standard form when its terms are arranged from the greatest exponent of the variable to the least, usually placing the constant term last. In option B, the powers are 3, 1, and 0, which decrease from left to right: 5x³ + 2x - 1. Option A is not descending, C is ascending overall, and D uses increasing powers.

The governing concept is direct identification of coefficients. In a polynomial written in descending powers, the coefficient of x is the number multiplying the first-power term x. Here the expression is p(x) = 2x^2 + 0x + 3, so the x-term is explicitly 0x. Hence its coefficient is 0, and option A is correct. The number 2 is the coefficient of x^2, while 3 is the constant term because it does not contain x. Option D, 5, is not obtained from any coefficient in the given polynomial; adding coefficients is not what the question asks. The zero coefficient also explains why the x-term contributes nothing.

The governing concept is classification by degree. The degree of a non-zero polynomial is the greatest exponent of the variable whose coefficient is non-zero. In p(x) = x² + 1, the coefficient of x² is 1, so the highest exponent is 2. The term 1 is a constant term with degree 0. Consequently, p(x) has degree 2 and is called a quadratic polynomial, making option B correct. It is not linear, because a linear polynomial has degree 1. It is certainly a polynomial because the variable has a non-negative integer exponent and there are finitely many terms. Option D incorrectly treats the constant term or another lower degree as the degree.

Substituting \(x=0\) gives \(p(0)=2(0)^2+3=0+3=3\). Thus, the correct answer is 3. The term containing \(x^2\) becomes zero, leaving the constant term. Exam tip: To find \(p(0)\), replace \(x\) with 0; for a polynomial written in standard form, the result is its constant term.

In option A, the greatest exponent of \(x\) is 4 and its coefficient is non-zero, so its degree is 4. The degrees of options B, C and D are 3, 2 and 1, respectively. Exam tip: The degree of a polynomial is the greatest exponent of the variable with a non-zero coefficient.

The term containing \(x^3\) is \(5x^3\), so the coefficient of \(x^3\) is 5. The numbers 2, -1, and 8 are associated with the \(x^2\) term, the \(x\) term, and the constant term, respectively. Exam tip: identify the numerical multiplier attached to the required power of the variable.

Substituting \(x=-4\), we get \(f(-4)=(-4)^2+4(-4)=16-16=0\). Therefore, the correct answer is 0. The value 16 results from calculating only \((-4)^2\) and ignoring the term \(4x=-16\). Exam tip: the square of a negative number is positive, but the linear term \(4x\) remains negative here.

The constant term is the term that does not contain the variable \(x\). In the given polynomial, \(12x^2\) and \(-9x\) contain \(x\), whereas 4 does not; therefore, the correct answer is 4. Exam tip: Identify the term independent of the variable to find the constant term.

To find the zero, set \(p(x)=0\): \(x+9=0\), which gives \(x=-9\). Therefore, -9 is correct. The distractor 9 has the wrong sign because \(9+9\neq0\). Exam tip: the zero of \(x+a\) is always \(-a\).