Respuesta :

Answer:

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

Step-by-step 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.

[tex] \frac{ {45y}^{3}( {y}^{2} + 2y - 48) }{ {30y}^{2} ( {y}^{2} - 64) } [/tex]

  • 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)

[tex] = \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)} \\[/tex]

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

[tex] = \frac{45 {y}^{3}(y - 6) }{30 {y}^{2} (y - 8)} [/tex]

  • Now, divide 45y^3 and 30y^2.

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

Answer:

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

Hope you could understand.

If you have any query, feel free to ask.