Answer
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Hint: We will first use the distributive property and divide the given factor and then open the bracket up by multiplying respective required terms.
Complete step-by-step answer:
We are given that we have to find the product $\left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right)$.
Now, we will use the fact of Distributive property which states that: $(u + v)y = uy + vy$.
Here, we will just replace u = 2a, v = -9 and $y = 3{a^2} + 4a - 4$.
Thus, we will get the following result:-
$ \Rightarrow \left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right) = 2a\left( {3{a^2} + 4a - 4} \right) - 9\left( {3{a^2} + 4a - 4} \right)$
Let $I = 2a\left( {3{a^2} + 4a - 4} \right)$ and $J = 9\left( {3{a^2} + 4a - 4} \right)$
Putting this in above expression, we will then obtain:
$ \Rightarrow \left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right) = I - J$ ………………….(1)
Now, we will use the fact that: $x\left( {u + v} \right) = xu + xv$
Now, we will use the above formula for finding I:-
$ \Rightarrow I = 2a\left( {3{a^2} + 4a - 4} \right)$
$ \Rightarrow I = 2a \times 3{a^2} + 2a \times 4a - 2a \times 4$
Simplifying the multiplication calculations in the above expression’s right hand side, we will then obtain:-
$ \Rightarrow I = 6{a^3} + 8{a^2} - 8a$ …………..(2)
Now, we will use the $x\left( {u + v} \right) = xu + xv$ for finding J:-
$ \Rightarrow J = 9\left( {3{a^2} + 4a - 4} \right)$
$ \Rightarrow J = 9 \times 3{a^2} + 9 \times 4a - 9 \times 4$
Simplifying the multiplication calculations in the above expression’s right hand side, we will then obtain:-
$ \Rightarrow J = 27{a^3} + 36a - 36$ …………..(3)
Putting (2) and (3) in equation (1), we will then obtain:-
$ \Rightarrow \left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right) = \left( {6{a^3} + 8{a^2} - 8a} \right) - \left( {27{a^3} + 36a - 36} \right)$
Opening the bracket on the right hand side to obtain the following expression:-
$ \Rightarrow \left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right) = 6{a^3} + 8{a^2} - 8a - 27{a^3} - 36a + 36$
Clubbing the like terms on the right hand side to obtain the following expression:-
$ \Rightarrow \left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right) = \left( {6{a^3} - 27{a^3}} \right) + 8{a^2} - \left( {8a + 36a} \right) + 36$
Simplifying the calculations inside the brackets to obtain the following expression:-
$ \Rightarrow \left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right) = - 21{a^3} + 8{a^2} - 44a + 36$
Hence, the answer is $\left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right) = - 21{a^3} + 8{a^2} - 44a + 36$.
Note:
The students must note that we have used the distributive property and associative property of real numbers and thus got to a conclusion.
Associative property of real numbers say that we can club any random real numbers in any order.
And students must also note that we have used the BODMAS rule in it as well when multiplying the terms before addition. Because BODMAS stands for doing the calculations in the order as follows: bracket of division multiplication addition and subtraction in the same order.
Complete step-by-step answer:
We are given that we have to find the product $\left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right)$.
Now, we will use the fact of Distributive property which states that: $(u + v)y = uy + vy$.
Here, we will just replace u = 2a, v = -9 and $y = 3{a^2} + 4a - 4$.
Thus, we will get the following result:-
$ \Rightarrow \left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right) = 2a\left( {3{a^2} + 4a - 4} \right) - 9\left( {3{a^2} + 4a - 4} \right)$
Let $I = 2a\left( {3{a^2} + 4a - 4} \right)$ and $J = 9\left( {3{a^2} + 4a - 4} \right)$
Putting this in above expression, we will then obtain:
$ \Rightarrow \left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right) = I - J$ ………………….(1)
Now, we will use the fact that: $x\left( {u + v} \right) = xu + xv$
Now, we will use the above formula for finding I:-
$ \Rightarrow I = 2a\left( {3{a^2} + 4a - 4} \right)$
$ \Rightarrow I = 2a \times 3{a^2} + 2a \times 4a - 2a \times 4$
Simplifying the multiplication calculations in the above expression’s right hand side, we will then obtain:-
$ \Rightarrow I = 6{a^3} + 8{a^2} - 8a$ …………..(2)
Now, we will use the $x\left( {u + v} \right) = xu + xv$ for finding J:-
$ \Rightarrow J = 9\left( {3{a^2} + 4a - 4} \right)$
$ \Rightarrow J = 9 \times 3{a^2} + 9 \times 4a - 9 \times 4$
Simplifying the multiplication calculations in the above expression’s right hand side, we will then obtain:-
$ \Rightarrow J = 27{a^3} + 36a - 36$ …………..(3)
Putting (2) and (3) in equation (1), we will then obtain:-
$ \Rightarrow \left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right) = \left( {6{a^3} + 8{a^2} - 8a} \right) - \left( {27{a^3} + 36a - 36} \right)$
Opening the bracket on the right hand side to obtain the following expression:-
$ \Rightarrow \left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right) = 6{a^3} + 8{a^2} - 8a - 27{a^3} - 36a + 36$
Clubbing the like terms on the right hand side to obtain the following expression:-
$ \Rightarrow \left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right) = \left( {6{a^3} - 27{a^3}} \right) + 8{a^2} - \left( {8a + 36a} \right) + 36$
Simplifying the calculations inside the brackets to obtain the following expression:-
$ \Rightarrow \left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right) = - 21{a^3} + 8{a^2} - 44a + 36$
Hence, the answer is $\left( {2a - 9} \right)\left( {3{a^2} + 4a - 4} \right) = - 21{a^3} + 8{a^2} - 44a + 36$.
Note:
The students must note that we have used the distributive property and associative property of real numbers and thus got to a conclusion.
Associative property of real numbers say that we can club any random real numbers in any order.
And students must also note that we have used the BODMAS rule in it as well when multiplying the terms before addition. Because BODMAS stands for doing the calculations in the order as follows: bracket of division multiplication addition and subtraction in the same order.
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