Question

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Answer 1

Answer:

iii) z2-4z-12

= z2 - 6z-2z-12

= z(z-6)-2(z-6)

= (z-2) (z-6)

Answer 2

Question:-

Factorize:-

Solutions:-

(i) a² - 2ab + b² - c²

→ We know,

(a - b) = a² - 2ab + b²

Hence,

a² - 2ab + b² = (a - b)²

(a - b)² - c²

Now,

Applying a² - b² = (a + b)(a - b)

Therefore,

= {(a - b) + c}{(a - b) - c}

= (a - b + c)(a - b - c) (Ans)

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(ii) 16x⁴ - 625y⁴

→ = (4x²)² - (25y²)²

Applying a² - b² = (a + b)(a - b)

= (4x² + 25y²)(4x² - 25y²)

= (4x² + 25y²){(2x)² - (5y)²}

Applying a² - b² = (a + b)(a - b) again,

= (4x² + 25y²){(2x + 5y)(2x - 5y)}

= (4x² + 25y²)(2x + 5y)(2x - 5y) (Ans)

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(iii) z² - 4z - 12

→ By splitting the middle term,

= z² - 6z + 2z - 12

= z(z - 6) + 2(z - 6)

= (z - 6)(z + 2) (Ans)

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(iv) x⁴ - (x - z)⁴

→ (x²)² - {(x - z)²}²

Applying a² - b² = (a + b)(a - b)

= {x² + (x - z)²}{x² - (x - z)²}

Using (a - b)² = a² + b² - 2ab

And applying a² - b² = (a + b)(a - b)

= {x² + (x² + z² - 2xz)}[{x - (x - z)}{x + (x - z)}]

= {x² + x² + z² - 2xz}[{x - x + z}{x + x - z}]

= (2x² + z² - 2xz)(z)(2x - z) (Ans)

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Some other algebraic identities:-

• (a + b)² = a² + b² + 2ab
• (a + b)³ = a³ + b³ + 3a²b + 3ab²
• (a - b)³ = a³ - b³ - 3a²b + 3ab²
• a³ - b³ = (a - b)(a² + ab + b²)

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