Algebraic Identities and Factorisation Test - 1

(z² + 4) (z² − 5)

= (z²)² + [4 + (− 5)] z² + 4 × (− 5) [(x + a) (x + b) = x² + (a + b) x + ab]

= z⁴ + (4 − 5) z² − 20

The given expression is –16y²+ 7x²+ 8xy²+ 6y – 14x – 3xy.

It can be factorized by regrouping the terms.

This can be done as

–16y²+ 7x²+ 8xy²+ 6y – 14x – 3xy = –16y²+ 8xy² + 7x²– 14x+ 6y – 3xy

= 8y²(–2 + x) + 7x(x– 2) – 3y(–2 + x)

= 8y²(x– 2) + 7x(x– 2) – 3y(x– 2)

= (x – 2)(8y²+ 7x – 3y)

Thus, the given expression can be factorised as (x– 2)(8y²+ 7x – 3y).

It is known that: (a − b − c)² = a² + b² + c² − 2ab + 2bc − 2ac

∴(a − b − c)² = a² + b² + c² + 2 (bc − ac − ab)

⇒ a² + b² + c² = (a − b − c)² − 2 (bc − ac − ab)

= 49 − 2(12) [(a − b −c)² = 49, bc −ac − ab = 12]

= 25

We can write 706 × 695 as (700 + 6)(700 − 5).

This expression is of the form (x + a) (x + b), where x = 700, a = 6, and b = − 5.

∴ 706 × 695 = (700 + 6) (700 − 5)

= (700)² + (6 − 5) × 700 + 6 × (− 5)

[(x + a) (x + b) = x² + (a + b) x + ab]

⇒ 706 × 695 = (700)² + 700 − 6 × 5