Factorisation Test

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Factorisation Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

The factors of (1 − 6z − 9z²) are

A
(1+3z)(1−3z)

B
(z+3)²

C
(3z−1)²

D
(z−3)²

Solution

(c): 1 − 6z + 9z² or 9z² − 6z + 1 or (3z)² − 2 × 3z × 1 + (1)² or (3z − 1)².
Question 2
1 / -0

Factorize a³ − 2√2b³ is

A
(a+√2b)(a² − √2ab + 2b²)

B
(a − √2b)(a² + √2ab + 2b²)

C
(a − √2b)(a² − √2ab + 2b²)

D
(a − √2b)(a² + √2ab + b²)

Solution

(b): a³ − 3√2b³ = a³ − (√2b)³ Then use formula, a³ − b³ = (a − b)(a² + ab + b²).
Question 3
1 / -0

Factorize 16x² − 25y²

A
(4x−5y)²

B
(4x+5y)²

C
4x+5y / 4x−5y

D
(4x+5y)(4x−5y)

Solution

(d): 16x² − 25y² = (4x)² − (5y)² = (4x + 5y)(4x − 5y).
Question 4
1 / -0

Factorize 36x³y − 60x²yz

A
12x²y(3x−5z)

B
12x²y(3x+5z)

C
12x²(5z−3x)

D
6x²y(6x−2z)

Solution

(a): 36x³y − 60x²yz = 12x²y[3x − 5z].
Question 5
1 / -0

Which of the following is equal to x³(a − 2b) + x²(a − 2b).

A
x²(a−2b)(x+1)

B
(x²−1)(a−2b)

C
(x²+1)(a−2b)(x+1)

D
x(a−2b)(x−1)

Solution

(a): x³(a − 2b) + x²(a − 2b) = x²(a − 2b)[x + 1]

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Factorisation Test - 5

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Factorisation Test - 5

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Question 1

(1+3z)(1−3z)

(z+3)2

(3z−1)2

(z−3)2

Solution

Question 2

(a+√2b)(a2 − √2ab + 2b2)

(a − √2b)(a2 + √2ab + 2b2)

(a − √2b)(a2 − √2ab + 2b2)

(a − √2b)(a2 + √2ab + b2)

Solution

Question 3

(4x−5y)2

(4x+5y)2

4x+5y / 4x−5y

(4x+5y)(4x−5y)

Solution

Question 4

12x2y(3x−5z)

12x2y(3x+5z)

12x2(5z−3x)

6x2y(6x−2z)

Solution

Question 5

x2(a−2b)(x+1)

(x2−1)(a−2b)

(x2+1)(a−2b)(x+1)

x(a−2b)(x−1)

Solution

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(z+3)²

(3z−1)²

(z−3)²

(a+√2b)(a² − √2ab + 2b²)

(a − √2b)(a² + √2ab + 2b²)

(a − √2b)(a² − √2ab + 2b²)

(a − √2b)(a² + √2ab + b²)

(4x−5y)²

(4x+5y)²

12x²y(3x−5z)

12x²y(3x+5z)

12x²(5z−3x)

6x²y(6x−2z)

x²(a−2b)(x+1)

(x²−1)(a−2b)

(x²+1)(a−2b)(x+1)