
How do you simplify $(8{y^3})( - 3{x^2}{y^2})\left( {\dfrac{3}{8}x{y^4}} \right)?$
Answer
557.1k+ views
Hint:Firstly use the commutative property of multiplication to separate or group the similar variables together and constants together in order to easily simplify. Then apply law of indices for multiplication of same base variables or constants to get the final answer.
Commutative property of multiplication is given as $a \times b \times c = a \times c \times b$ , that the order of multiplicands does not matter in multiplication, the answer will always be the same.
Law of indices for multiplication of same base can be given as ${a^m} \times {a^n} = {a^{m + n}}$ , that is in multiplication operation if bases of multiplicands are same then their power will be added together.
Complete step by step solution:
To simplify the expression $(8{y^3})( - 3{x^2}{y^2})\left(
{\dfrac{3}{8}x{y^4}} \right)$, we will first separate the similar variables together and the constants together with the help of commutative property of multiplication which says that in multiplication the the order of multiplicands do not affect the result. So the given expression can be rewritten as follows
$
= (8{y^3})( - 3{x^2}{y^2})\left( {\dfrac{3}{8}x{y^4}} \right) \\
= 8 \times ( - 3) \times \dfrac{3}{8} \times {x^2} \times x \times {y^3} \times {y^2} \times {y^4}
\\
= ( - 9) \times {x^2} \times x \times {y^3} \times {y^2} \times {y^4} \\
$
Now we will use the law of indices for multiplication of same base multiplicands that says powers get added when the bases of multiplicands are equal. So simplifying further we will get
$
= ( - 9) \times {x^2} \times x \times {y^3} \times {y^2} \times {y^4} \\
= ( - 9) \times {x^{2 + 1}} \times {y^{3 + 2 + 4}} \\
= - 9{x^3}{y^9} \\
$
So the simplified form of $(8{y^3})( - 3{x^2}{y^2})\left( {\dfrac{3}{8}x{y^4}} \right)\;{\text{is }}\left( { -9{x^3}{y^9}\;} \right)$
Note: Multiplicand is a quantity which is to be multiplied by the other quantity.
Commutative property is also called the order property and it is only applied in addition and
multiplication operation not in subtraction and division.
There are more laws of indices which are very useful in solving this type of problems, understanding and learning them will be good for a student.
Commutative property of multiplication is given as $a \times b \times c = a \times c \times b$ , that the order of multiplicands does not matter in multiplication, the answer will always be the same.
Law of indices for multiplication of same base can be given as ${a^m} \times {a^n} = {a^{m + n}}$ , that is in multiplication operation if bases of multiplicands are same then their power will be added together.
Complete step by step solution:
To simplify the expression $(8{y^3})( - 3{x^2}{y^2})\left(
{\dfrac{3}{8}x{y^4}} \right)$, we will first separate the similar variables together and the constants together with the help of commutative property of multiplication which says that in multiplication the the order of multiplicands do not affect the result. So the given expression can be rewritten as follows
$
= (8{y^3})( - 3{x^2}{y^2})\left( {\dfrac{3}{8}x{y^4}} \right) \\
= 8 \times ( - 3) \times \dfrac{3}{8} \times {x^2} \times x \times {y^3} \times {y^2} \times {y^4}
\\
= ( - 9) \times {x^2} \times x \times {y^3} \times {y^2} \times {y^4} \\
$
Now we will use the law of indices for multiplication of same base multiplicands that says powers get added when the bases of multiplicands are equal. So simplifying further we will get
$
= ( - 9) \times {x^2} \times x \times {y^3} \times {y^2} \times {y^4} \\
= ( - 9) \times {x^{2 + 1}} \times {y^{3 + 2 + 4}} \\
= - 9{x^3}{y^9} \\
$
So the simplified form of $(8{y^3})( - 3{x^2}{y^2})\left( {\dfrac{3}{8}x{y^4}} \right)\;{\text{is }}\left( { -9{x^3}{y^9}\;} \right)$
Note: Multiplicand is a quantity which is to be multiplied by the other quantity.
Commutative property is also called the order property and it is only applied in addition and
multiplication operation not in subtraction and division.
There are more laws of indices which are very useful in solving this type of problems, understanding and learning them will be good for a student.
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