Download Class 10 Mathematics Concept Summaries: CBSE Class 10 Maths Polynomials Zeroes and Coefficients Notes and Practice Questions Set 04
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Basic Concepts
A number k is a zero of a polynomial p(x) if p(k) = 0. The graph of a linear polynomial ax + b is a straight line. The graph of a quadratic ax² + bx + c is a parabola that opens upwards (U-shaped) if a > 0 and downwards if a < 0. A polynomial of degree n crosses the x-axis at most n times.
| Polynomial | Sum of zeroes | Product of zeroes |
|---|---|---|
| Quadratic ax² + bx + c, zeroes α, β | α + β = −b/a | αβ = c/a |
| Cubic ax³ + bx² + cx + d, zeroes α, β, γ | α + β + γ = −b/a; αβ + βγ + γα = c/a | αβγ = −d/a |
If α and β are the zeroes of a quadratic, the polynomial is x² − (α + β)x + αβ. For a cubic it is x³ − (α + β + γ)x² + (αβ + βγ + γα)x − αβγ.
Practice Questions with Solutions (1 to 14)
1. Write the zeroes of the polynomial x² − x − 6.
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Answer: x² − x − 6 = (x − 3)(x + 2), so the zeroes are 3 and −2.
Teacher's Note: Use the key term in your answer.
2. Write a quadratic polynomial whose zeroes have sum 3 and product −2.
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Answer: x² − (sum)x + product = x² − 3x − 2.
Teacher's Note: Give an example if the question allows.
3. Is x = −3 a solution of 2x² + 5x + 3 = 0?
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Answer: Put x = −3: 2(9) − 15 + 3 = 6, which is not 0. So x = −3 is not a solution.
Teacher's Note: Write a short, clear answer.
4. If x + a is a factor of 2x² + 2ax + 5x + 10, find a.
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Answer: Put x = −a: 2a² − 2a² − 5a + 10 = 0, so a = 2.
Teacher's Note: Remember this for your exam.
5. Show that x = −2 is a solution of 3x² + 13x + 14 = 0.
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Answer: 3(4) − 26 + 14 = 12 − 26 + 14 = 0. So x = −2 is a solution.
Teacher's Note: Learn this point and write it in your own words.
7. For what value of k is (−4) a zero of x² − x − (2k + 2)?
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Answer: 16 + 4 − (2k + 2) = 0, so 2k = 18 and k = 9.
Teacher's Note: Use the key term in your answer.
8. If 1 is a zero of p(x) = ax² − 3(a − 1)x − 1, find a.
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Answer: p(1) = a − 3(a − 1) − 1 = 0, so −2a + 2 = 0 and a = 1.
Teacher's Note: Give an example if the question allows.
10. (i) If α + β = 6 and αβ = 4, write the polynomial. (iii) If one zero of x² − 4x + 1 is 2 + √3, write the other zero.
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Answer: (i) x² − 6x + 4. (iii) The sum of the zeroes is 4, so the other zero is 4 − (2 + √3) = 2 − √3.
Teacher's Note: Write a short, clear answer.
11. Find the other zeroes of 2x⁴ − 9x³ + 5x² + 3x − 1 if two zeroes are 2 + √3 and 2 − √3.
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Answer: The quadratic factor is (x − 2 − √3)(x − 2 + √3) = x² − 4x + 1. Dividing gives the quotient 2x² − x − 1 = (2x + 1)(x − 1). The other zeroes are −1/2 and 1.
Teacher's Note: Remember this for your exam.
12. If 6x⁴ + 8x³ − 5x² + ax + b is exactly divisible by 2x² − 5, find a and b.
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Answer: Dividing, the quotient is 3x² + 4x + 5 and the remainder is (a + 20)x + (b + 25). The remainder must be zero, so a = −20 and b = −25.
Teacher's Note: Learn this point and write it in your own words.
13. Divide 6x⁴ + 8x³ + 17x² + 21x + 7 by 3x² + 4x + 1.
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Answer: The quotient is 2x² + 5 and the remainder is x + 2.
Teacher's Note: Use the key term in your answer.
14. On dividing 4x⁴ − 5x³ − 39x² − 46x − 2 by g(x), the quotient is x² − 3x − 5 and the remainder is −5x + 8. Find g(x).
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Answer: Dividend = divisor × quotient + remainder, so g(x) = (4x⁴ − 5x³ − 39x² − 46x − 2 + 5x − 8) ÷ (x² − 3x − 5) = 4x² + 7x + 2.
Teacher's Note: Give an example if the question allows.
Practice Questions with Solutions (15 to 30)
15. Show that 1/2 and −3/2 are the zeroes of 4x² + 4x − 3 and verify the relationship.
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Answer: 4x² + 4x − 3 = (2x − 1)(2x + 3), so the zeroes are 1/2 and −3/2. Sum = −1 = −4/4 = −b/a. Product = −3/4 = c/a. The relationship is verified.
Teacher's Note: Write a short, clear answer.
16. Find b for which (2x + 3) is a factor of 2x³ + 9x² − x − b.
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Answer: Put x = −3/2: 2(−27/8) + 9(9/4) + 3/2 − b = 0, i.e. −27/4 + 81/4 + 6/4 − b = 0, so b = 60/4 = 15.
Teacher's Note: Remember this for your exam.
17. Find the other zeroes of x⁴ − 7x² + 12 if two zeroes are √3 and −√3.
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Answer: Divide by x² − 3. The quotient is x² − 4 = (x − 2)(x + 2). The other zeroes are 2 and −2.
Teacher's Note: Learn this point and write it in your own words.
18. Find all other zeroes of x⁴ − 3x³ − x² + 9x − 6 if two zeroes are √3 and −√3.
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Answer: Divide by x² − 3. The quotient is x² − 3x + 2 = (x − 1)(x − 2). The other zeroes are 1 and 2.
Teacher's Note: Use the key term in your answer.
19. Divide 2x⁴ − 9x³ + 5x² + 3x − 8 by x² − 4x + 1 and verify the division algorithm.
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Answer: The quotient is 2x² − x − 1 and the remainder is −7. Check: (x² − 4x + 1)(2x² − x − 1) − 7 = 2x⁴ − 9x³ + 5x² + 3x − 1 − 7 = 2x⁴ − 9x³ + 5x² + 3x − 8, which is the dividend.
Teacher's Note: Give an example if the question allows.
20. Given that x − √5 is a factor of x³ − 3√5x² − 5x + 15√5, find all the zeroes.
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Answer: Dividing by x − √5 gives x² − 2√5x − 15 = (x − 3√5)(x + √5). So the zeroes are √5, 3√5 and −√5.
Teacher's Note: Write a short, clear answer.
21. If x⁴ − 6x³ + 16x² − 25x + 10 is divided by x² − 2x + k, the remainder is x + a. Find k and a.
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Answer: The remainder is (2k − 9)x + (k² − 8k + 10). Comparing with x + a: 2k − 9 = 1, so k = 5. Then a = 25 − 40 + 10 = −5.
Teacher's Note: Remember this for your exam.
22. What must be subtracted from p(x) = 8x⁴ + 14x³ − 2x² + 8x − 12 so that 4x² + 3x − 2 is a factor?
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Answer: On division the quotient is 2x² + 2x − 1 and the remainder is 15x − 14. So 15x − 14 must be subtracted (or −15x + 14 added).
Teacher's Note: Learn this point and write it in your own words.
23. If 1 and −1 are zeroes of Lx⁴ + Mx³ + Nx² + Rx + P, show that L + N + P = M + R = 0.
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Answer: Putting x = 1: L + M + N + R + P = 0. Putting x = −1: L − M + N − R + P = 0. Adding gives L + N + P = 0 and subtracting gives M + R = 0.
Teacher's Note: Use the key term in your answer.
24. If x + a is a factor of x² + px + q and x² + mx + n, prove that a = (n − q)/(m − p).
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Answer: Putting x = −a in both: a² − pa + q = 0 and a² − ma + n = 0. Subtracting, (m − p)a = n − q, so a = (n − q)/(m − p).
Teacher's Note: Give an example if the question allows.
25. Find a cubic polynomial whose sum of zeroes, sum of products taken two at a time and product of zeroes are 3, −1/2 and 4.
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Answer: x³ − 3x² − (1/2)x − 4, or multiplying by 2: 2x³ − 6x² − x − 8.
Teacher's Note: Write a short, clear answer.
27. α, β, γ are zeroes of kx³ − 5x + 9. If α³ + β³ + γ³ = 27, find k.
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Answer: Here α + β + γ = 0 (no x² term), so α³ + β³ + γ³ = 3αβγ. Also αβγ = −9/k. So 3(−9/k) = 27, giving k = −1.
Teacher's Note: Remember this for your exam.
28. α, β, γ are zeroes of x³ − 12x² + 44x + c and are in AP. Find c.
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Answer: Let the zeroes be a − d, a, a + d. The sum is 3a = 12, so a = 4. As 4 is a zero: 64 − 192 + 176 + c = 0, so c = −48.
Teacher's Note: Learn this point and write it in your own words.
29. Two zeroes of ax³ + 3x² − bx − 6 are −1 and −2. Find the third zero, a and b.
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Answer: Putting x = −1: −a + 3 + b − 6 = 0, so b − a = 3. Putting x = −2: −8a + 12 + 2b − 6 = 0, so 2b − 8a = −6, i.e. b − 4a = −3. Solving, a = 2 and b = 5. The product of zeroes is 6/a = 3, so (−1)(−2)z = 3 and the third zero is 3/2.
Teacher's Note: Use the key term in your answer.
30. α, β, γ are zeroes of x³ − 2x² + qx − r. If α + β = 0, show that 2q = r.
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Answer: The sum of the zeroes is 2, so γ = 2. Also αβ + γ(α + β) = q gives αβ = q, and αβγ = r gives 2q = r.
Teacher's Note: Give an example if the question allows.
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Chapter 02 Polynomials Concepts and Summary for Class 10 Mathematics
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