Polynomials in one variable

Correct answer: A. \(4x^2-2x-4\)

Explanation: When adding polynomials, combine terms with the same power of the variable: \(3x^2+x^2=4x^2\), \(-4x+2x=-2x\), and \(1-5=-4\). Therefore, the sum is \(4x^2-2x-4\). Option B results from incorrectly combining coefficients, while option D incorrectly multiplies powers instead of adding the polynomials. Exam tip: arrange the \(x^2\), \(x\), and constant terms in separate groups before adding.

Correct answer: A. \(3x^2+4x-6\)

Explanation: We need to find \(p(x)-r(x)\): \((5x^2+3x-2)-(2x^2-x+4)\). The minus sign changes the sign of every term in the second polynomial, giving \(5x^2-2x^2=3x^2\), \(3x-(-x)=4x\), and \(-2-4=-6\). Therefore, the result is \(3x^2+4x-6\). Exam tip: when subtracting a polynomial, distribute the minus sign to every term inside the second bracket before combining like terms.

Correct answer: A. \(6x^3-10x^2+2x\)

Explanation: Using the distributive property, multiply \(2x\) by every term in the bracket: \(2x\cdot3x^2=6x^3\), \(2x\cdot(-5x)=-10x^2\), and \(2x\cdot1=2x\). Hence, the simplified form is \(6x^3-10x^2+2x\). In option B, the factor \(x\) has not been multiplied correctly into each term. Exam tip: distribute the outside factor to every term and carefully preserve the signs.

Correct answer: A.
x^2-x-12

Explanation: Using the distributive property,
(x+3)(x-4)=x^2-4x+3x-12=x^2-x-12. Therefore, option A is correct. In option B, the middle terms have been combined incorrectly. Exam tip: multiply every term in one binomial by both terms in the other binomial, then combine like terms.

Correct answer: A. 2x² + 9x − 5

Explanation: Using the distributive property, multiply each term by every term: (2x−1)(x+5) = 2x² + 10x − x − 5 = 2x² + 9x − 5. Therefore, option A is correct. Option B does not combine the −x and −5 terms correctly, while option D has the wrong sign for the constant term. Exam tip: In binomial multiplication, write all four products before combining like terms.

Correct answer: A. 0

Explanation: To find \(p(1)\), substitute \(x=1\) in the polynomial: \(p(1)=1^3-6(1)^2+11(1)-6=1-6+11-6=0\). Hence, 0 is the correct value, and 1 is also a zero of the polynomial. Exam tip: when evaluating a polynomial, replace every occurrence of \(x\) with the given value, including powers.

Correct answer: A. -5

Explanation: Since \(p(2)=0\), substitute \(x=2\) into the polynomial: \(2^2+2k+6=0\), so \(4+2k+6=0\). Hence \(2k=-10\) and \(k=-5\). Therefore, option A is correct. Exam tip: whenever a zero of a polynomial is given, substitute it for \(x\) and set the polynomial equal to zero.

Correct answer: B. 7

Explanation: Substituting \(x=5\) in the given polynomial gives \(p(5)=25-5a+10=0\). Thus, \(35-5a=0\), so \(5a=35\) and \(a=7\). If \(a=10\), then \(p(5)=25-50+10=-15\), so that option is incorrect. Exam tip: substitute the given zero directly into the polynomial.

Correct answer: C. 3

Explanation: If \(x=1\) is a zero of the polynomial, its value at \(x=1\) must be zero. Thus, \(1^2-4(1)+m=0\), which gives \(1-4+m=0\) and hence \(m=3\). Therefore, option C is correct. Exam tip: Substitute the given zero into the polynomial and equate the result to zero.

Correct answer: A. 1 is a zero of the polynomial

Explanation: Substituting x = 1 gives p(1) = 2(1)^2 - 3(1) + 1 = 2 - 3 + 1 = 0. A value of x for which a polynomial becomes zero is called a zero of that polynomial; therefore, 1 is a zero. Option B incorrectly evaluates p(1), while the degree and leading coefficient of the polynomial are both 2, not 1. Exam tip: To test whether a number is a zero, substitute it in the polynomial and check whether the result is 0.

Correct answer: A. 0

Explanation: Substituting (x=-2) gives (p(-2)=(-2)^2+4(-2)+4=4-8+4=0), so option A is correct. Choosing 2 or 4 usually results from mishandling the signs after substitution or addition. In an exam, always use parentheses when squaring a negative value.

Correct answer: B. \(4x^3+x-9\)

Explanation: Option B, \(4x^3+x-9\), has only non-negative integer powers of the variable \(x\), so it is a polynomial. Its highest power is 3, making it a cubic polynomial rather than a quadratic polynomial. Option A has degree 2 and is therefore quadratic. Option C contains \(x^{-1}\), and option D contains \(x^{1/2}\); hence neither is a polynomial. Exam tip: check that all variable exponents are non-negative integers, then identify the highest exponent as the degree.

Correct answer: B. 2

Explanation: The zero of a polynomial is the value of \(x\) for which \(p(x)=0\). Thus, \(7x-14=0\), so \(7x=14\) and \(x=2\). Therefore, option B is correct. Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\); here, \(-\frac{-14}{7}=2\).

Correct answer: B. -4

Explanation: To find the zero, set p(x) equal to 0: 3x + 12 = 0. Thus, 3x = -12 and x = -4. Therefore, -4 is correct. Option 4 results from missing the negative sign. Exam tip: for ax + b, the zero is -b/a.

Correct answer: B. \(-\frac{b}{a}\)

Explanation: The zero of a polynomial is the value of \(x\) for which \(p(x)=0\). Thus, \(ax+b=0\) gives \(ax=-b\), so \(x=-\frac{b}{a}\). Option A has the wrong sign. In an exam, remember that moving the constant term to the other side changes its sign.

Correct answer: B. \(x^3\)

Explanation: The polynomial contains the terms \(x^4\), \(2x^2\), and the constant term \(1=x^0\), but it has no \(x^3\) term. Therefore, among the given options, B is correct. The coefficient of a missing term is taken as 0; in an exam, first list the powers that are explicitly present.

Correct answer: C. Non-negative integers

Explanation: In a polynomial, powers of the variable are non-negative integers like (0,1,2,\ldots). This is the key identification rule.

Correct answer: C. \(x^2+y+1\)

Explanation: A polynomial in one variable contains only one variable, raised only to non-negative integer powers. Option C, \(x^2+y+1\), contains two different variables, \(x\) and \(y\), so it is not a polynomial in one variable. Options A, B and D contain only one variable each—\(x\), \(t\) and \(z\), respectively—and are therefore polynomials in one variable. Exam tip: count the distinct variables, regardless of their names.

Correct answer: D. 6

Explanation: Substituting \(x=0\), we get \(p(0)=0^2-5(0)+6=6\). The terms containing \(x\) become zero, leaving only the constant term, 6. Therefore, option D is correct. Exam tip: For any polynomial, \(p(0)\) equals its constant term.

Correct answer: A. (0)

Explanation: \(p\left(\frac{3}{2}\right)=4\cdot\frac{9}{4}-18+9=0\). While substituting a fraction, square it carefully first.

Correct answer: C. 0

Explanation: The polynomial has no x-term, so the missing term is understood as 0x. Therefore, the coefficient of x is 0. Here, 2 is the coefficient of x^3, 3 is the coefficient of x^2, and -5 is the constant term. Exam tip: The coefficient of an absent power is taken as 0.

Correct answer: A. (2)

Explanation: The coefficient of (x) is (-2) in the first polynomial and (4) in the second, so the sum is (2). Add coefficients of like powers only.

Correct answer: B. (2)

Explanation: For the degree to be (2), the coefficient of (x^3) must be (0), so (m-2=0) and (m=2). In exams, check the coefficient of the highest power first.

Correct answer: A. (0)

Explanation: (p(0)=-4) and (p(1)=2-5+7-4=0), so (p(0)+p(1)=-4). Always substitute each value separately before adding.

Correct answer: B. (1)

Explanation: The direct answer is option B, 1. Recognize the identity \(x^3-3x^2+3x-1=(x-1)^3\). A zero of a polynomial is an input that makes its value 0. Therefore \((x-1)^3=0\) gives \(x-1=0\), so \(x=1\). Direct substitution confirms it: \(p(1)=1^3-3(1)^2+3(1)-1=1-3+3-1=0\). Option A, 0, gives \(-1\), not 0. Option B is correct. Option C, 2, gives \(8-12+6-1=1\), so it is not a zero. Option D, 3, gives \(27-27+9-1=8\), also not zero. The polynomial is a perfect cube, so recognizing the binomial pattern is quickest. Memory cue: \(a^3-3a^2+3a-1=(a-1)^3\).