Polynomials in one variable

A zero of a polynomial makes its value equal to zero, so \(p(3)=0\). Thus, \(3k+12=0\), giving \(3k=-12\) and \(k=-4\). Option A has the wrong sign; in such questions, substitute the given zero into the polynomial and equate the result to zero.

The non-zero terms are \(x^4\), \(-6x^2\), and \(9\), so the polynomial has 3 non-zero terms. Option A is incorrect because there are three terms, not two. Exam tip: Count each non-zero term separated by a plus or minus sign, including the constant term.

Substitute \(x=1\) into the polynomial: \(p(1)=1^2+a(1)+a=1+2a\). Since \(p(1)=5\), we get \(1+2a=5\), so \(2a=4\) and \(a=2\). Hence, option B is correct. Exam tip: When \(p(k)\) is given, substitute \(x=k\) directly to form an equation.

A number is a zero of a polynomial only when the value of the polynomial at that number is 0. Here, \(p(3)=3^2-9=0\) and \(p(-3)=(-3)^2-9=0\), so 3 and -3 are zeros. However, \(p(0)=0^2-9=-9\neq 0\); therefore, 0 is not a zero. Exam tip: Use \(x^2-a^2=(x-a)(x+a)\) to find the zeros quickly.

The term with the highest power of the variable is called the leading term. Here, the leading term is \(3x^4\), so its leading coefficient is 3. Although -5 is also a coefficient in the polynomial, it belongs to the lower-degree term \(-5x^2\), not the leading term. Exam tip: identify the highest power first, then select the numerical coefficient of that term.

To find \(p(3)\), substitute 3 for \(x\): \(p(3)=2(3)^2-7(3)+3=18-21+3=0\). Therefore, the correct answer is 0. Exam tip: replace \(x\) with the given value in every term; the value 18 from considering only \(2(3)^2\) is not the value of the complete polynomial.

In the polynomial \(x^3+2x^2+cx+5\), the coefficient of \(x\) is \(c\). Hence, \(c=6\), so \(c+5=6+5=11\). Option 6 is only the value of \(c\), not of \(c+5\). Exam tip: First identify the coefficient, then substitute its value in the required expression.

The governing concept is the definition of a constant polynomial. A constant polynomial has a fixed value and contains no variable term; its degree is 0 for a non-zero constant. The expressions 8, -3, and 0 are all constants, so their values do not change with x. In contrast, 2x + 1 contains the variable x and changes when x changes, so it is a linear polynomial rather than a constant polynomial. Therefore option D is correct. The zero polynomial is also treated as a constant polynomial in this classification, although its degree is often handled separately or described as undefined. The presence of x, not the number of terms, is the decisive distinction here.

Substituting \(x=-2\) gives \(p(-2)=(-2)^2+4(-2)+4=4-8+4=0\). Therefore, \(-2\) is a zero of the polynomial. Option B is incorrect because \(p(0)=4\); similarly, \(p(2)=16\) and \(p(4)=36\). Exam tip: To test whether a number is a zero, substitute it into the polynomial and check whether the result is \(0\).

Substituting x=0 gives p(0)=5(0)^2-3(0)+2=2. Substituting x=1 gives p(1)=5-3+2=4. Therefore, p(0)+p(1)=2+4=6, so option C is correct. In such questions, evaluate each polynomial value separately before adding them.

The governing concept is identifying separate terms in a polynomial. Terms are separated by plus or minus signs, and a sign belongs to the term that follows it. Therefore, x² - 5x + 6 has the three terms x², -5x, and 6, so option A is correct. Option B loses the negative sign, C separates coefficients from variables, and D groups unlike parts together.

The governing concept is substitution of known parameter values into a polynomial and then simplifying. Begin with p(x) = x^3 + ax^2 + bx + 1. Replacing a by 0 gives ax^2 = 0x^2 = 0, so the quadratic term disappears completely. Replacing b by -2 gives bx = -2x. Hence p(x) = x^3 + 0x^2 - 2x + 1, which simplifies to x^3 - 2x + 1. Therefore option A is correct. Option B changes the negative value of b into a positive one. Option C attaches -2 to x^2 instead of x, and option D both mishandles the substitution and omits the constant term 1. Careful matching of each parameter with its term avoids these errors.

In option A, x² + 3x + 1, only x appears as a variable, and its powers are non-negative integers; therefore, it is a polynomial in x and not in y. Options C and D contain both x and y, while option B is a polynomial only in y. Exam tip: identify all letters appearing as variables before deciding whether an expression is in one variable.

To find the zero of the polynomial, set (p(x)=0): (4x-12=0), which gives (4x=12) and hence (x=3). Therefore, 3 is correct. The value 12 is the constant term, not the zero. Exam tip: the zero of a linear polynomial (ax+b) is (-b/a).

A non-zero constant polynomial has the form \(a\), where \(a\ne 0\). Since it contains no variable term, its degree is taken as 0. The degree of the zero polynomial is undefined, but this question refers specifically to a non-zero constant polynomial. In exams, distinguish a non-zero constant polynomial from the zero polynomial.

The degree of a polynomial is the highest power of its variable. Here, the degrees of x^2+1, x^3+x, and 5x-2 are 2, 3, and 1, respectively. Therefore, 5x-2 has the least degree. Exam tip: a polynomial of the form ax+b, where a is nonzero, is linear and has degree 1.

Substituting \(x=-1\) gives \(p(-1)=(-1)^2+(-1)+1=1-1+1=1\). Hence, the correct answer is 1. Note that \((-1)^2=1\), while the minus sign remains in the linear term. In exams, check the sign of each term after substitution.

A term with a zero coefficient can be omitted from a polynomial. Since \(0x^2=0\), the simplified polynomial is \(6x^3-2x+9\). Option C incorrectly changes the coefficient of \(x^2\) to 1, while option D omits the constant term 9. Exam tip: Remove only zero-coefficient terms and retain the signs and all remaining terms unchanged.

\(p(1)=1^2-3(1)+2=1-3+2=0\) and \(p(2)=2^2-3(2)+2=4-6+2=0\). Therefore, both 1 and 2 are zeroes of the polynomial, so option A is correct. In such questions, substitute each given value directly into the polynomial; alternatively, factorising it as \(p(x)=(x-1)(x-2)\) gives the same result.

The leading coefficient of a polynomial is the coefficient of the term with the highest power of the variable. In \(3x^2+x+1\), the highest-degree term is \(3x^2\), so its leading coefficient is 3, not 1. The leading coefficients of the other three polynomials are all 1. Exam tip: Identify the highest-degree term first, then read its coefficient.