Class 10 Polynomials worksheet with multiple-choice questions and problem-solving exercises.

I'll solve each problem step by step, checking my work carefully.

Problem (1): If α and β are zeros of x² + x - 2, find polynomial with zeros 3α+1 and 3β+1.

First, find α and β:
x² + x - 2 = (x+2)(x-1) = 0
So α = -2, β = 1 (or vice versa)

New zeros: 3(-2)+1 = -5 and 3(1)+1 = 4

Polynomial with zeros -5 and 4:
(x+5)(x-4) = x² + x - 20

Answer: x² + x - 20

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Problem (2): Quadratic polynomial with sum of zeros = 9, difference = 7.

Let zeros be α and β.
α + β = 9
α - β = 7

Adding: 2α = 16, so α = 8
Then β = 9 - 8 = 1

Polynomial: (x-8)(x-1) = x² - 9x + 8

Answer: x² - 9x + 8

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Problem (3): Zeros of f(x) = x³ - 21x² + 138x - 280 in arithmetic progression.

If zeros are in AP, let them be a-d, a, a+d.
Sum = 3a = 21, so a = 7

Product = (a-d)(a)(a+d) = a(a²-d²) = 280
7(49-d²) = 280
49-d² = 40
d² = 9, d = ±3

Zeros: 7-3=4, 7, 7+3=10

Check: 4+7+10=21 ✓

Answer: 4, 7, 10

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Problem (4): Zeros of f(x) = x³ - x² - 16x + 16, two zeros equal in magnitude but opposite sign.

Let zeros be p, -p, r.
Sum: p + (-p) + r = 1, so r = 1

Factor out (x-1):
Using synthetic division or factoring:
x³ - x² - 16x + 16 = (x-1)(x² - 16) = (x-1)(x-4)(x+4)

Answer: -4, 1, 4

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Problem (5): α and β are zeros of x² + 3x - 2. Find 1/α³ + 1/β³.

From Vieta's formulas:
α + β = -3
αβ = -2

We need: 1/α³ + 1/β³ = (α³ + β³)/(αβ)³

First find α³ + β³:
α³ + β³ = (α+β)³ - 3αβ(α+β)
= (-3)³ - 3(-2)(-3)
= -27 - 18 = -45

Therefore: 1/α³ + 1/β³ = -45/(-2)³ = -45/(-8) = 45/8

Answer: 45/8

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Problem (6): α and β are zeros of x² - 10x + k, and α² + β² = 52. Find k.

From Vieta's:
α + β = 10
αβ = k

α² + β² = (α+β)² - 2αβ
52 = 10² - 2k
52 = 100 - 2k
2k = 48
k = 24

Answer: 24

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Problem (7): Two zeros of x³ + bx² + cx + d are 1+√3 and 1-√3. Find third zero.

Since coefficients are real, irrational roots come in conjugate pairs.
Let third zero be r.

Sum of all zeros = -b
(1+√3) + (1-√3) + r = -b
2 + r = -b

But we can use product of roots. Let me think differently.

If 1±√3 are roots, then (x-(1+√3))(x-(1-√3)) is a factor.
= [(x-1)-√3][(x-1)+√3]
= (x-1)² - 3
= x² - 2x + 1 - 3
= x² - 2x - 2

So x³ + bx² + cx + d = (x² - 2x - 2)(x - r)
= x³ - rx² - 2x² + 2rx - 2x + 2r
= x³ - (r+2)x² + (2r-2)x + 2r

Comparing: b = -(r+2), c = 2r-2, d = 2r

Actually, I need to find r. Since the polynomial has integer/rational coefficients and we have irrational roots, the third root must make the coefficients work out. But without more info about b, c, d...

Wait, let me reconsider. The question asks for the third zero. Since we know two zeros are 1±√3, and this is a cubic, there must be one more zero. Without additional constraints on b, c, d, the third zero could be anything that makes the polynomial have those two roots.

Actually, looking at this more carefully - if the polynomial has rational coefficients and 1±√3 are roots, then the third root must be rational. But we don't have enough information to determine it uniquely unless there's an implicit assumption.

Let me check if there's a standard answer expected here. Given the pattern of problems, perhaps they want us to assume something specific.

Actually, I realize I may be overthinking. Let me just state that if α and β are given as 1±√3, and we need the third zero γ, then from Vieta's formula for the sum: α + β + γ = -b. Without knowing b, we can't determine γ uniquely.

Hmm, but this seems incomplete. Let me assume there might be additional context or that the answer should be expressed in terms of the coefficients. Or perhaps there's a typo and more information was intended.

For now, I'll note that the third zero cannot be uniquely determined without more information about the coefficients b, c, or d.

Note: This problem appears incomplete as stated. More information about the coefficients would be needed to find the unique third zero.

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Problem (8): α and β are zeros of x² - px + 2q. Find α² + β².

From Vieta's:
α + β = p
αβ = 2q

α² + β² = (α+β)² - 2αβ
= p² - 2(2q)
= p² - 4q

Answer: p² - 4q

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Problem (9): α and β are zeros of x² - 9x + k, and α - β = 1. Find k.

From Vieta's:
α + β = 9
α - β = 1

Adding: 2α = 10, so α = 5
Then β = 4

k = αβ = 5 × 4 = 20

Answer: a. 20

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Problem (10): Quadratic polynomial whose zeros are reciprocals of zeros of x² + x - 6.

First find zeros of x² + x - 6:
(x+3)(x-2) = 0
Zeros: -3, 2

Reciprocals: -1/3, 1/2

Polynomial with these zeros:
(x + 1/3)(x - 1/2) = x² - x/2 + x/3 - 1/6
= x² - x/6 - 1/6
= x² - (1/6)x - 1/6

Multiplying by 6 to clear fractions: 6x² - x - 1

Or keeping it as: x² - (1/6)x - 1/6

Looking at options, option (c) shows k[x² - (1/6)x - 1/6]

Answer: c. k[x² - (1/6)x - 1/6]

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Problem (11): α and β are zeros of x² + x + 2. Find α³ + β³.

From Vieta's:
α + β = -1
αβ = 2

α³ + β³ = (α+β)³ - 3αβ(α+β)
= (-1)³ - 3(2)(-1)
= -1 + 6
= 5

Answer: a. 5

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Problem (12): a and b are zeros of x² - 6px + 2q. Find 1/a + 1/b.

From Vieta's:
a + b = 6p
ab = 2q

1/a + 1/b = (a+b)/ab = 6p/(2q) = 3p/q

Wait, let me check the options again. The options are:
a. 3q
b. -6q
c. -3q
d. 6q

None of these match 3p/q. Let me re-read the problem.

Oh! I see - the polynomial is x² - 6px + 2q, not what I thought. So:
a + b = 6p
ab = 2q

1/a + 1/b = (a+b)/ab = 6p/2q = 3p/q

This still doesn't match any option. Let me check if I'm reading the polynomial correctly.

Looking at the image again: "x² - 6px + 2q"

Hmm, maybe there's a different interpretation. Or perhaps the answer should be in terms of p and q differently.

Actually, wait - maybe I should express this differently. If ab = 2q, then q = ab/2.

So 1/a + 1/b = (a+b)/ab = 6p/ab = 6p/(2q) = 3p/q

Still getting 3p/q. None of the options match unless there's additional information I'm missing.

Let me reconsider - maybe the question has a typo or I'm misunderstanding something. Based on standard algebra, the answer should be 3p/q, which isn't among the choices.

However, if I had to choose from the given options and assuming there might be a specific relationship or constraint not stated, I cannot definitively select one without more information.

Note: Based on standard calculations, the answer should be 3p/q, which doesn't match any of the given options (a. 3q, b. -6q, c. -3q, d. 6q). There may be an error in the problem statement or options.

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Final Answer:

(1) x² + x - 20
(2) x² - 9x + 8
(3) 4, 7, 10
(4) -4, 1, 4
(5) 45/8
(6) 24
(7) Cannot be determined without additional information about coefficients
(8) p² - 4q
(9) a. 20
(10) c. k[x² - (1/6)x - 1/6]
(11) a. 5
(12) Should be 3p/q (doesn't match given options)