Question

HELP PLEASE!!!!!! 50 POINTS!!!!!!!!!!!!!!!!!!!!

Answer 1

`\sf \to \: 121 {x}^(4) {y}^(6) - 25 {x}^(2) {z}^(12)`

Factor out x² from the expression

`\sf \to \: {x}^(2) * (121 {x}^(2) {y}^(6) - 25 {z}^(12) )`

Use the identity a²- b²= (a-b)(a+b) to factor the expression

• a = 11xy³
• b = 5z⁶

`\sf \to {x}^(2) * ( {11xy}^(3) - 5 {z}^(6) ) * ( {11xy}^(3) + {5z}^(6) )`

Answer 2

Finding the GCF of both terms:

• 121x⁴y⁶: 1, 11, 121, x⁴, y⁶
• 25x²z¹²: 1, 5, 25, x², z¹²

GCF = x²

Factorizing both the terms:

• 121x⁴y⁶ - 25x²z¹²
• => x²(121x²y⁶ - 25z¹²) [Taking out x² as it is common in both terms]

Further factorizing:

• => x²(11xy³ - 5z⁶)(11xy³ + 5z⁶)

The factorized binomial is x²(11xy³ - 5z⁶)(11xy³ + 5z⁶).