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SubjectsMathematics
एक चर वाले बहुपद
In this Class 10 Mathematics topic from the chapter “Polynomials,” students study algebraic expressions involving a single variable, such as x. They learn to identify a polynomial and its degree, understand constant, linear, quadratic, and cubic forms, and find its zeroes or roots. The topic also develops understanding of the relationship between zeroes and coefficients, helping students interpret polynomial equations and solve related problems accurately.
Correct answer: A. (x^3-x^2-6x)
Explanation: The zeroes are (-2,0,3), so the polynomial is (x(x+2)(x-3)=x^3-x^2-6x). Intersections with the (x)-axis give zeroes.
Correct answer: A. \(x^2+2x-8\)
Explanation: Since \(x=1\) is a zero, \((x-1)\) must be a factor of \(p(x)\). Dividing \(x^3+x^2-10x+8\) by \((x-1)\) gives the quotient \(x^2+2x-8\) with remainder \(0\). Verification: \((x-1)(x^2+2x-8)=x^3+x^2-10x+8\). Therefore, option A is correct. Exam tip: if \(a\) is a zero, divide the polynomial by \((x-a)\) to obtain the remaining factor.
Correct answer: A. 0
Explanation: By the Remainder Theorem, the remainder when p(x) is divided by x-a is p(a) . Here the divisor is x-1 , so a=1 . Therefore, p(1)=2(1)^3+3(1)^2-8(1)+3=2+3-8+3=0 . Hence, the correct remainder is 0. Exam tip: For a divisor of the form x-a , evaluate p(a) directly instead of carrying out polynomial division.
Correct answer: A. 0
Explanation: Substituting \(x=3\), \(p(3)=3^3-2(3)^2-5(3)+6=27-18-15+6=0\). Hence, the correct answer is 0. Since \(p(3)=0\), 3 is also a zero of the polynomial. Exam tip: substitute the given value for every occurrence of \(x\), including its powers.
Correct answer: A. Its degree is not defined
Explanation: The zero polynomial has no non-zero term, so its degree is not defined. A non-zero constant polynomial has degree (0).
Correct answer: A. (7)
Explanation: A constant polynomial has no variable term, so (7) is a constant polynomial. A non-zero constant polynomial has degree (0).
Correct answer: A. Cubic polynomial
Explanation: The coefficients (\sqrt{5}) and (-\frac{2}{3}) are real numbers and the highest power is (3). Hence it is a cubic polynomial.
Correct answer: A. \(-\frac{b}{a}\)
Explanation: A zero is the value of \(x\) for which \(p(x)=0\). Setting \(ax+b=0\) gives \(ax=-b\), and hence \(x=-\frac{b}{a}\). Option B misses the negative sign, while options C and D interchange the numerator and denominator. For exams, remember that the zero of \(ax+b\) is \(-\frac{b}{a}\).
Correct answer: A. 3
Explanation: A zero of a polynomial is the value of the variable for which the polynomial equals zero. Setting \(p(x)=0\) gives \(4x-12=0\), so \(4x=12\) and \(x=3\). For the closest distractor, \(p(-3)=-24\), so −3 is not a zero. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-b/a\).
Correct answer: A. No real zeroes
Explanation: For a zero, \(x^2+1=0\), so \(x^2=-1\) must hold. For every real \(x\), \(x^2\geq 0\), hence \(x^2+1\geq 1>0\), and the expression can never be zero. Therefore, there are no real zeroes. Option C is incorrect because \(p(0)=1\), not 0. As an exam tip, when an equation gives \(x^2=-1\), check the domain: its solutions are complex, not real.
Correct answer: A. \(x^2-3x-4\)
Explanation: For zeroes 8-19 and 849, the polynomial is formed as 8x+198x-49. Expanding it gives \(x^2-3x-4\), so option A is correct. Option B has zeroes 1 and -4, so it is not correct. Exam tip: for zeroes \(\alpha\) and \(\beta\), use the monic quadratic form \(x^2-(\alpha+\beta)x+\alpha\beta\).
Correct answer: A. Any non-zero real number
Explanation: For a quadratic polynomial \(x^2+bx+c\), the sum of the zeroes is \(-b\) and their product is \(c\). Here, \(b=-a\) and \(c=a\), so both the sum and product are \(a\). Hence the condition is automatically satisfied for every non-zero real value of \(a\). The values 1, −1, and 2 are only particular examples, not the complete answer. Exam tip: compare \(-b\) and \(c\) directly when using the relationships between zeroes and coefficients.
Correct answer: A. 64
Explanation: For a quadratic polynomial to have equal zeroes, its discriminant must be zero. Thus, using (b^2-4ac=0), we get (k^2-4\cdot2\cdot8=0), so (k^2=64). Therefore, option A is correct. Exam tip: For equal zeroes of a quadratic, set the discriminant to zero.
Correct answer: A. -(6\sqrt{2}) / (-6\sqrt{2})
Explanation: Let the zeroes be (t) and (2t), then (2t^2=16) gives (t=2\sqrt{2}). The sum is (6\sqrt{2}), so (-k=6\sqrt{2}) and (k=-6\sqrt{2}).
Correct answer: A. 6
Explanation: For a quadratic polynomial \(ax^2+bx+c\), the sum of its zeroes is \(-\frac{b}{a}\). Here, the sum is \(-\frac{-10}{1}=10\). Therefore, if one zero is \(4\), the other zero is \(10-4=6\). The value \(4\) is only the given zero, not the other one. Exam tip: use the sum-of-zeroes formula directly when one zero is known.
Correct answer: A. -6
Explanation: For a quadratic polynomial \(ax^2+bx+c\), the product of its zeroes is \(\frac{c}{a}\). Here, the product is \(\frac{-18}{1}=-18\). Since one zero is \(3\), the other zero is \(\frac{-18}{3}=-6\). Hence, option A is correct. Exam tip: when one zero is given, use the product of zeroes directly rather than the sum unless needed.
Correct answer: A. It cannot have more than 4 real zeroes
Explanation: A non-zero polynomial of degree n has at most n zeroes. Therefore, a degree-4 polynomial can have no more than four real zeroes; it need not have exactly four or even one. Exam tip: distinguish “at most” from “exactly”.
Correct answer: A. (1,4)
Explanation: The zeroes of (p(x)) are (2) and (5), so for (p(x+1)=0), (x+1=2) or (x+1=5). Hence the zeroes of (q(x)) are (1) and (4).
Correct answer: B. (1)
Explanation: For degree (4), the coefficient of (x^6) must be (0) and the coefficient of (x^4) must be non-zero. Both conditions hold for (a=1).
Correct answer: A. (-2)
Explanation: (p(-1)=-k+2+7+5=14-k), so (14-k=16) and (k=-2). Watch signs carefully when substituting a negative value.
Correct answer: C. (4)
Explanation: The direct answer is option C: the degree is \(4\). Add the polynomials term by term: \((6x^5-4x^2+1)+(-6x^5+3x^4+x-9)\). The fifth-degree terms cancel because \(6x^5-6x^5=0\). The sum becomes \(3x^4-4x^2+x-8\). The highest non-zero power is 4, so the degree is 4. Option A, 2, is wrong because the \(x^4\) term remains. Option B, 3, is wrong because there is no non-zero \(x^3\) term and it overlooks the fourth-degree term. Option C is correct. Option D, 5, is wrong because the fifth-degree terms cancel completely. The degree of a sum may be lower than the degrees of the original polynomials when leading terms cancel. Exam cue: always check cancellation of the highest powers.
Correct answer: C. (9)
Explanation: In the product of two non-zero polynomials, degrees add, so (6+3=9). In multiplication, look at the highest-power terms.
Correct answer: A. (-1)
Explanation: ((3x-2)(2x^2+x-5)=6x^3-x^2-17x+10), so the coefficient of (x^2) is (-1). Combine like terms after expansion.
Correct answer: B. 2
Explanation: Substituting \(x=2\), we get \(p(2)=4(2)^3-11(2)^2+6(2)+2=32-44+12+2=2\). Hence, the correct answer is 2. The value 0 may result from mishandling the negative term or the constant term. In an exam, calculate the powers first and carefully preserve the signs while adding the terms.
Correct answer: A. \(3x^2-\sqrt{5}x+7\)
Explanation: In \(3x^2-\sqrt{5}x+7\), the powers of \(x\) are \(2,1,0\), all non-negative integers. \(\sqrt{5}\) is a real coefficient. In B and D, the variable occurs in a denominator, while C has power \(\tfrac12\). Exam tip: check powers of the variable first.
