Question

How do you find the product of $\left( 3d+3 \right)\left( 2{{d}^{2}}+5d-2 \right)$ ?

Answer 1

Hint: Product of two bracket terms require only the patience of applying the distributive property. Here, we multiply the term outside the bracket with the terms inside the bracket and then adding up the products of the process. We repeat this process on and on and then finally simplify the entire expression to arrive at the final answer.

Complete step by step solution:
The given product that we are requested to find in the given problem is,
$\left( 3d+3 \right)\left( 2{{d}^{2}}+5d-2 \right)$
To solve this, there is no other way than to use the distributive property of multiplication. The distributive property states that the product of two expressions, for example \[a\left( x+y+z \right)\] can be written as $ax+ay+az$ which is done just simply by multiplying the term outside the bracket with the terms inside the bracket and then adding up the products of the process.
In our problem, the product of two expressions is,
$\left( 3d+3 \right)\left( 2{{d}^{2}}+5d-2 \right)$
Here, we can take $a=3d+3,x=2{{d}^{2}},y=5d,z=-2$ . The individual terms of the distributive property having been assigned, we can carry out the distributive property as,
$\Rightarrow \left( 3d+3 \right)2{{d}^{2}}+\left( 3d+3 \right)5d-\left( 3d+3 \right)2$
Upon performing the previous distributive property, we are know again left with quite a many bracket expression. So, we again have to apply the distributive property one by one to the expressions, and then we can arrive at the final result. In order to solve the first expression, $a=2{{d}^{2}},x=3d,y=3$ . Applying the distributive property, we get,
$\Rightarrow \left( 6{{d}^{3}}+6{{d}^{2}} \right)+\left( 3d+3 \right)5d-\left( 3d+3 \right)2$
In order to solve the second expression, $a=5d,x=3d,y=3$ . Applying the distributive property, we get,
$\Rightarrow \left( 6{{d}^{3}}+6{{d}^{2}} \right)+\left( 15{{d}^{2}}+15d \right)-\left( 3d+3 \right)2$
In order to solve the first expression, $a=2,x=3d,y=3$ . Applying the distributive property, we get,
$\Rightarrow \left( 6{{d}^{3}}+6{{d}^{2}} \right)+\left( 15{{d}^{2}}+15d \right)-\left( 6d+6 \right)$
Opening up the brackets, we get,
$\Rightarrow 6{{d}^{3}}+21{{d}^{2}}+9d-6$

Therefore, we can conclude that the final answer to the product is $6{{d}^{3}}+21{{d}^{2}}+9d-6$

Note: Distributive property are very easy on one hand, but too prone to mistakes on the other. The only disadvantage of the distributive property is that as the number of terms increases, the chances of committing a mistake increases. So, we must be very cautious over here as the problem involves two terms in the first bracket and three terms in the second one. Also, we must take care of the sines and the degrees.