Factorisation Test

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Factorisation Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

One of the factors of 4(x + y)(3a − b)+6(x + y)(2b − 3a) is

A
(2b − 3a)

B
(3a − b)

C
(4a − 3b)

D
(−3a + 4b)

Solution

We have, 4(x + y)(3a − b) + 6(x + y)(2b − 3a) = 2(x + y)[ 2 (3a − b)+ 3(2b − 3a)] = 2(x + y)[6a − 2b + 6b − 9a] = 2(x + y)[−3a + 4b]
Question 2
1 / -0

Factorise (2x + 3y)² − 5(2x + 3y) - 14.

A
4(2x+3y)(x+y−2)

B
4(2x+3y)(x+y+2)

C
(2x−3y+7)(2x−3y+2)

D
(2x+3y−7)(2x+3y+2)

Solution

We have, (2x + 3y)² − 5 (2x + 3y) − 14 = (2x + 3y)² − 7 (2x + 3y) + 2(2x + 3y) − 14 = (2x + 3y)(2x + 3y − 7) + 2(2x + 3y − 7) = (2x + 3y + 2)(2x + 3y − 7)
Question 3
1 / -0

Which of the following is the factor of 12(a² + 7a)² − 8(a² + 7a)(2a − 1) − 15(2a − 1)²?

(i) (2a² + 8a + 3)

(ii) (6a² + 52a − 5)

(iii) (3a + 5)

A
Only (i)

B
Both (i) and (ii)

C
Only (ii)

D
All (i), (ii) and (iii)

Solution

12(a² + 7a)² − 8(a² + 7a)(2a − 1) − 15(2a − 1)² = 12(a² + 7a)² − 18(a² + 7a)(2a − 1) +10(a² + 7a)(2a − 1) − 15 (2a − 1)² = 6(a² + 7a)[2(a² + 7a) − 3 (2a − 1)] +5 (2a − 1)[2a² + 7a) − 3(2a − 1)] = (6a² + 42a)(2a² + 8a + 3) +(10a − 5)(2a² + 8a + 3) = (6a² + 42a + 10a − 5)(2a² + 8a + 3) = (6a² + 52a − 5)(2a² + 8a + 3)
Question 4
1 / -0

Which of the following statements is CORRECT?

A
The factors of an expression are always either algebraic variable or algebraic expression.

B
An irreducible factor is a factor that cannot be expressed further as a product of factors.

C
Every binomial expression can be factorised into two monomial expression.

D
The process of writing a given expression as the product of two or more factors is called multiplication of factors.

Question 5

1 / -0

Match the expression given in Column-I to one of their factors given in Column-II.

Column - I	Column - II
P. 9x² + 24x + 16	(i) (2x − 4)
Q. 25x² + 30x + 9	(ii) (4x + 1)
R. 40x² + 14x + 1	(iii) (5x + 3)
S. 4x² − 16x + 16	(iv) (3x + 4)

P→(iv); Q→(iii); R→(ii); S→(i)

P→(iii): Q→(i); R→(iv); S→(ii)

P→(ii); Q→(i); R→(iv): S→(iii)

P→(iv); Q→(iii); R→(i); S→(ii)

Solution

9x² + 24x + 16 = (3x)² + 2(3x)(4) + (4)² =(3x + 4)² = (3x + 4)(3x + 4) Q. We have,

25x² + 30x + 9 = (5x)² + 2(5x) (3) + (3)² =(5x + 3)² = (5x + 3)(5x + 3) R. We have, 40x² + 14x + 1 = 40x² + 10x + 4x + 1

=10x(4x + 1) + 1(4x + 1) = (10x + 1)(4x + 1) S. We have, 4x² − 16x + 16 = (2x)² − 2(2x)(4) + (4)² = (2x − 4)² = (2x − 4)(2x − 4)

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Re-Attempt Weekly Quiz Competition

Factorisation Test - 6

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Factorisation Test - 6

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Question 1

(2b − 3a)

(3a − b)

(4a − 3b)

(−3a + 4b)

Solution

Question 2

4(2x+3y)(x+y−2)

4(2x+3y)(x+y+2)

(2x−3y+7)(2x−3y+2)

(2x+3y−7)(2x+3y+2)

Solution

Question 3

Only (i)

Both (i) and (ii)

Only (ii)

All (i), (ii) and (iii)

Solution

Question 4

The factors of an expression are always either algebraic variable or algebraic expression.

An irreducible factor is a factor that cannot be expressed further as a product of factors.

Every binomial expression can be factorised into two monomial expression.

The process of writing a given expression as the product of two or more factors is called multiplication of factors.

Question 5

P→(iv); Q→(iii); R→(ii); S→(i)

P→(iii): Q→(i); R→(iv); S→(ii)

P→(ii); Q→(i); R→(iv): S→(iii)

P→(iv); Q→(iii); R→(i); S→(ii)

Solution

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