Answer:
the given expression
is factored as (d+1)(2 d-5)
Explanation:
Here, the given expression is:
![6d^2 - 9d- 15= 0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/oan2ju0f1mhzlbnytyta2blyjwk7oeqta2.png)
Now, simplifying the given expression,we get:
![6d^2 - 9d- 15= 0 \\\implies 3(2d^2 - 3d- 5) = 0\\\implies 2d^2 - 3d- 5 = 0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/r860fve7wxst9q1wx7adn3cud53t06p9sl.png)
Now, we need to split the middle term -3 in such a way, that on addition it given -3 and on multiplication it given -10.
![2d^2 - 3d- 5 = 0\\\implies 2d^2 + 2d- 5d - 5 = 0\\\implies 2d(d+1) -5(d+1) = 0\\\implies (d+1)(2d-5) = 0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/hw1mkin46m4a5tgfglq2xv3tjhtm8iimpm.png)
So, either(d+1) = 0 OR (2d-5) = 0
⇒
![(6d^2 - 9d- 15) = (d+1)(2d-5)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/8u3o3jpi4bc8ljxjswlp65aormhp4e20a1.png)
Hence the given expression
is factored as (d+1)(2 d-5) = 0