Chapter 8 Brain Teasers | Polynomials | Class-8 DAV Secondary Mathematics

Are you looking for complete, step-by-step verified solutions for DAV Class 8 Maths Chapter 8 Brain Teasers (Polynomials)? Here you will find clear, fully typed, teacher-verified solutions for all questions, covering all three textbook questions: dividing by a monomial, long division with quotient and remainder, and checking whether one polynomial is a factor of another, plus extra HOTS and value-based practice.

Key Concepts: Polynomials & Division Algorithm

Unit 8 Brain Teasers: Step-by-Step Solutions

Q.1. Divide:

(i) -39x⁴ by √13 x²
Solution:
= (-39x⁴) / (√13 x²)
= (-39 / √13) × x^{4 – 2}
Rationalize denominator: -39√13 / 13 = -3√13.
Result: -3√13 x².

(ii) √125 y² by 5y²
Solution:
√125 = √(25 × 5) = 5√5.
= (5√5 y²) / (5 y²) = √5.

(iii) 49p³ by -7√7 p²
Solution:
= (49p³) / (-7√7 p²)
= [49 / (-7√7)] × p = [-7 / √7] × p = -√7 p.

(iv) 14z⁶ by (14/3) z³ (Clarification & Verified Solution)
Solution:
= 14z⁶ ÷ [(14/3) z³]
= [14 × (3/14)] × z^{6 – 3}
= 1 × 3 × z³
= 3z³.
Explanation: The coefficient 14 divided by 14/3 gives 14 × 3/14 = 3, and z⁶/z³ = z³. Hence the answer is 3z³.

(v) 6x³ – 4x² + 8x by (2/3) x
Solution:
Divide each term by (2/3)x, which is equivalent to multiplying by 3 / (2x):
= [6x³ × 3/(2x)] – [4x² × 3/(2x)] + [8x × 3/(2x)]
= 3(3x²) – 2(3x) + 4(3)
= 9x² – 6x + 12.

(vi) y² – 5y + 1 by -(1/3) y
Solution:
Multiply each term by -3/y:
= [y² × (-3/y)] – [5y × (-3/y)] + [1 × (-3/y)]
= -3y + 15 – 3/y.

(vii) 18a³ – 12a² + 9a by 3a
Solution:
= (18a³ / 3a) – (12a² / 3a) + (9a / 3a)
= 6a² – 4a + 3.

Q.2. Divide and write down the quotient and remainder. Also check your answer.

(i) 4x² + 7x − 11 by 2x − 4
Solution:
4x² ÷ 2x = 2x; 2x(2x − 4) = 4x² − 8x; subtract: 15x − 11
15x ÷ 2x = 15/2; (15/2)(2x − 4) = 15x − 30; subtract: −11 + 30 = 19
Quotient = 2x + 15/2, Remainder = 19
Check: (2x − 4)(2x + 15/2) + 19 = 4x² + 15x − 8x − 30 + 19 = 4x² + 7x − 11 ✓

(ii) 125 − 225x + 135x² − 27x³ by 5 − 3x
Solution:
Write both in descending powers: (−27x³ + 135x² − 225x + 125) ÷ (−3x + 5)
−27x³ ÷ (−3x) = 9x²; 9x²(−3x + 5) = −27x³ + 45x²; subtract: 90x² − 225x
90x² ÷ (−3x) = −30x; −30x(−3x + 5) = 90x² − 150x; subtract: −75x + 125
−75x ÷ (−3x) = 25; 25(−3x + 5) = −75x + 125; subtract: 0
Quotient = 9x² − 30x + 25 = (3x − 5)², Remainder = 0
Check: the dividend is (5 − 3x)³, so dividing by (5 − 3x) leaves (5 − 3x)² ✓

(iii) y³ + 5y² + 12y + 9 by y + 2
Solution:
y³ ÷ y = y²; y²(y + 2) = y³ + 2y²; subtract: 3y² + 12y
3y² ÷ y = 3y; 3y(y + 2) = 3y² + 6y; subtract: 6y + 9
6y ÷ y = 6; 6(y + 2) = 6y + 12; subtract: 9 − 12 = −3
Quotient = y² + 3y + 6, Remainder = −3
Check: (y + 2)(y² + 3y + 6) − 3 = y³ + 5y² + 12y + 12 − 3 = y³ + 5y² + 12y + 9 ✓

(iv) 8p³ − 729 − 108p² + 486p by 2p − 9
Solution:
Descending powers: 8p³ − 108p² + 486p − 729
8p³ ÷ 2p = 4p²; 4p²(2p − 9) = 8p³ − 36p²; subtract: −72p² + 486p
−72p² ÷ 2p = −36p; −36p(2p − 9) = −72p² + 324p; subtract: 162p − 729
162p ÷ 2p = 81; 81(2p − 9) = 162p − 729; subtract: 0
Quotient = 4p² − 36p + 81 = (2p − 9)², Remainder = 0
Check: (2p − 9)³ = 8p³ − 108p² + 486p − 729 ✓

(v) q⁴ + 3q² − 4 by q² − 1
Solution:
q⁴ ÷ q² = q²; q²(q² − 1) = q⁴ − q²; subtract: 4q² − 4
4q² ÷ q² = 4; 4(q² − 1) = 4q² − 4; subtract: 0
Quotient = q² + 4, Remainder = 0
Check: (q² − 1)(q² + 4) = q⁴ + 4q² − q² − 4 = q⁴ + 3q² − 4 ✓

(vi) p⁴ + (63/4)p² − 5 by p² − 1/4
Solution:
p⁴ ÷ p² = p²; p²(p² − 1/4) = p⁴ − (1/4)p²; subtract: (63/4 + 1/4)p² − 5 = 16p² − 5
16p² ÷ p² = 16; 16(p² − 1/4) = 16p² − 4; subtract: −5 + 4 = −1
Quotient = p² + 16, Remainder = −1
Check: (p² − 1/4)(p² + 16) − 1 = p⁴ + (63/4)p² − 4 − 1 = p⁴ + (63/4)p² − 5 ✓

Q.3. Find whether or not the first polynomial is a factor of the second polynomial.

(i) 3x + 2, 3x⁴ + 5x³ − x² + 13x + 10
Solution:
Divide 3x⁴ + 5x³ − x² + 13x + 10 by 3x + 2:
3x⁴ ÷ 3x = x³; subtract 3x⁴ + 2x³ → 3x³ − x²
3x³ ÷ 3x = x²; subtract 3x³ + 2x² → −3x² + 13x
−3x² ÷ 3x = −x; subtract −3x² − 2x → 15x + 10
15x ÷ 3x = 5; subtract 15x + 10 → 0
Quotient = x³ + x² − x + 5, Remainder = 0
The remainder is 0, so 3x + 2 is a factor.

(ii) x² + 1, x⁴ − 3x³ − 4x² + 3x + 2
Solution:
x⁴ ÷ x² = x²; subtract x⁴ + x² → −3x³ − 5x² + 3x
−3x³ ÷ x² = −3x; subtract −3x³ − 3x → −5x² + 6x + 2
−5x² ÷ x² = −5; subtract −5x² − 5 → 6x + 7
Quotient = x² − 3x − 5, Remainder = 6x + 7
The remainder is not 0, so x² + 1 is not a factor.

HOTS (Higher Order Thinking Skills) Questions

Extra practice: the HOTS and value-based questions below are not printed in the textbook. Q.1 to Q.3 above are the textbook Brain Teasers.

HOTS 1: If the polynomial x⁴ + 2x³ + 8x² + 12x + 18 is divided by another polynomial (x² + 5), the remainder comes out to be (px + q). Find the values of p and q.
Solution:
Divide x⁴ + 2x³ + 8x² + 12x + 18 by x² + 5:
1. x⁴ / x² = x² → x²(x² + 5) = x⁴ + 5x².
Subtracting gives: 2x³ + 3x² + 12x + 18.
2. 2x³ / x² = 2x → 2x(x² + 5) = 2x³ + 10x.
Subtracting gives: 3x² + 2x + 18.
3. 3x² / x² = 3 → 3(x² + 5) = 3x² + 15.
Subtracting gives remainder: 2x + 3.

Comparing with px + q:
p = 2 and q = 3.

Value Based Questions (VBQs)

Question: A school decided to construct an eco-friendly playground. The area of the rectangular playground is represented by the polynomial (6x² + 11x – 35) sq. metres. If the width of the ground is (2x + 7) metres:

(a) Find the polynomial representing the length of the playground.
Solution:
Length = Area / Width = (6x² + 11x – 35) ÷ (2x + 7)
Factorise numerator: 6 × (-35) = -210. 21 – 10 = 11.
6x² + 21x – 10x – 35 = 3x(2x + 7) – 5(2x + 7) = (2x + 7)(3x – 5).
Dividing by (2x + 7) gives:
Length = (3x – 5) metres.

(b) If x = 10, find the actual dimensions and perimeter of the playground.
Solution:
Width = 2(10) + 7 = 27 m.
Length = 3(10) – 5 = 25 m.
Perimeter = 2(25 + 27) = 2(52) = 104 metres.

(c) What value is promoted by providing open sports grounds for children?
Solution:
Physical fitness, mental agility, camaraderie, sportsmanship, and holistic personality development.

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Answer 2 (iv):

Arrange in descending powers: 8p³ − 108p² + 486p − 729.

This is (2p − 9)³, because (2p − 9)³ = 8p³ − 3(4p²)(9) + 3(2p)(81) − 729 = 8p³ − 108p² + 486p − 729.

So dividing by (2p − 9): Quotient = 4p² − 36p + 81 = (2p − 9)², Remainder = 0.

Check: (2p − 9)(4p² − 36p + 81) + 0 = 8p³ − 108p² + 486p − 729 ✓