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

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

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

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

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

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

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