Question

Factories the given expression and divide them: 45y^3(y^2 + 2y - 48) / 30y^2 (y^2 - 64)​

Answer 1

[{3y(y-6)}/{2(y-8)}]

Explanation:

Given expression is [{45y³(y²+2y-48)}/{30y²(y²-64)}]

Now, in numerator

y² + 2y - 48

= y² + 8x -6x - 48

= y(y + 8) - 6(y + 8)

⇛y² + 2y - 48 = (y-6)(y+8)

Next, in denominator

y² - 64

= y² - 8²

= (y - 8)(y + 8)

Consider,

[{45y³(y²+2y-48)}/{30y²(y²-64)}]

= [{(3*3*3)y³(y-6)(y+8)}/{(2*3*5)y²(y-8)(y+8)}]

= [{3y(y-6)}/{2(y-8)}]

⇛[{45y³(y²+2y-48)}/{30y²(y²-64)}] = [{3y(y-6)}/{2(y-8)}] Ans.

Please let me know if you have any other questions.

Answer 2

`\frac{ {45y}^(3)( {y}^(2) + 2y - 48) }{ {30y}^(2) ( {y}^(2) - 64) }`

• Let us factorise the middle term of the bracket portion in the numerator. And factorise the bracket portion in the denominator by using the identity a² - b² = (a - b)(a + b)

`= \frac{ {45y}^(3)( {y}^(2) + 8y - 6y - 48)}{ {30y}^(2) ( {(y)}^(2) - {(8)}^(2) } ) \\ = \frac{ {45y}^(3)( y(y + 8)- 6(y + 8)}{ {30y}^(2)(y - 8)(y + 8)} \\ = \frac{ {45y}^(3)( y - 6)(y + 8)}{ {30y}^(2)(y - 8)(y + 8)} \\`

• Now cancel out (y + 8) from both denominator and numerator.

`= \frac{45 {y}^(3)(y - 6) }{30 {y}^(2) (y - 8)}`

• Now, divide 45y^3 and 30y^2.

`= (3y(y - 6))/(2(y - 8))`

Answer:

`(3y(y - 6))/(2(y - 8))`

Hope you could understand.

If you have any query, feel free to ask.