Factorisation using identities | Factorization | Assessment | English | Grade 8

Quiz
•
Mathematics
•
8th Grade
•
Hard
Standards-aligned
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6 questions
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1.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Factorised form of p⁴ + 2p²q² + q⁴ is __________.
(p² - q²)²
(p² + q²)²
(p² - q²)
(p² + q²)
Answer explanation
The given expression is of the form a² + 2ab + b², where a = p², b = q² and 2ab = 2p²q² So, using the identity a² + 2ab + b² = (a + b)², we can write, p⁴ + 2p²q² + q⁴ = (p²)² + 2p²q² + (p²)² = (p² + q²)² So, option 2 is correct. Alternatively, p⁴ + 2p²q² + q⁴ = p⁴ + p²q² + p²q² + q⁴ = p² (p² + q²) + q²(p² + q²) = (p² + q²)(p² + q²) = (p²+q²)²
Tags
CCSS.HSA.APR.C.4
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Factorised form of 45p⁴ + 30p³q + 5p²q² is ___________.
5p²(3p - q)²
5p(3p + q)²
5p²(3p + q)
5p²(3p + q)²
Answer explanation
45p⁴ + 30p³q + 5p²q² = 5p²(9p² + 6pq + q²) = 5p²[(3p)² + 2(3p)(q) + (q²)] {Using the identity (a + b)²= a² + 2ab + b², where a = 3p, b = q} = 5p²(3p + q)² So, option 4 is correct.
Tags
CCSS.HSA.APR.C.4
3.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Factorised form of 144k² - 81 is __________.
(12k + 9)(12k + 9)
(12k - 9)(12k - 9)
(12k + 9)(12k - 9)
(12k + 9)
Answer explanation
Option 3 is correct as by identity (a + b)(a – b) = a²–b², we can write, 144k² - 81 = (12k)²-(9)² = (12k - 9)(12k + 9)
Tags
CCSS.HSA.APR.C.4
4.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Factorised form of 7a⁴ - 7b⁴ is _________.
7(a² + b²)(a + b)(a - b)
7(a² - b²)(a + b)(a - b)
7(a⁴ + b⁴)(a² - b²)
(a² - b²)(a + b)(a - b)
Answer explanation
7p⁴ - 7q⁴ = 7(p⁴ - q⁴) = 7[(p²)² - (q²)²] = 7[(p² + q²)(p² - q²)] {Applying identity a²–b² = (a + b)(a – b), where a = p² and b = q²} = 7[(p² + q²)(p + q)(p - q)] {Applying identity a²–b² = (a + b)(a – b), where a = p and b = q} So, option 1 is correct.
Tags
CCSS.HSA.APR.C.4
5.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Factorised form of t² - 15 is _________.
(t + √15)(t + √15)
(t - √15)(t + √15)
(t - √15)(t - √15)
(√t - √15)(√t + √15)
Answer explanation
Option 2 is correct as by identity (a + b)(a – b) = a² – b², we can write, t² - 15 = t² - (√15)² = (t - √15)(t + √15)
Tags
CCSS.HSA.APR.C.4
6.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
If the area of a square is given by the expression 64k² - 112k + 49, then what will be its side?
(8k + 7)
(8k - 7)²
(8k + 7)²
(8k - 7)
Answer explanation
By using identity (a - b)²= a² - 2ab + b² and, taking a = 8k, b = 7, 2ab = 2(8k)(7) 64k² - 112k + 49 = (8k - 7)² Since the area of a square = side² and area of square = 64k² - 112k + 49 = (8k - 7)² Therefore, the side of the square = 8k - 7 So, option 4 is correct.
Tags
CCSS.HSA.APR.C.4
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