Multiply \[ - 3{x^2}{y^2} \] by \[ - 5x{y^2} \]
Answer
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Hint: To solve this kind of question, have some knowledge about the basic rules of exponentiation. That is we need to know about the product of exponentials with the same base. We can multiply the constant term by a simple multiplication method. We know multiplying two negative numbers makes a positive number.
Complete step-by-step answer:
Let us see the meaning of above given problem,
In general if n is a positive integer then \[{x^n} = { \left \{ {x \times x \times x \times x........ \times x} \right \}_{n{ \text{ times}}}} \] , we have multiplied \[x \] ,
\[n \] -times.
Now we have \[{x^2} \] , that is \[x \] is multiplied 2 times, and \[{y^2} \] , \[y \] is multiplied 2 times.
Given: Multiply \[ - 3{x^2}{y^2} \] by \[ - 5x{y^2} \]
\[ = \left( { - 3{x^2}{y^2}} \right) \times \left( { - 5x{y^2}} \right) \]
If we have an exponential with the same base, we simply add the exponents.
That is, \[{x^a} \times {x^b} = {x^{a + b}} \] , rearranging above we get:
\[ = ( - 3) \times ( - 5) \times {x^2} \times x \times {y^2} \times {y^2} \]
As we know multiplying two negative numbers makes a positive number.
Regrouping with like terms,
\[ \Rightarrow 15 \times ({x^2} \times x) \times ({y^2} \times {y^2}) \]
Know that, \[x \] can be written as \[{x^1} \] , which is the same for any variable and constant. (Example \[5 = {5^1} \] )
Applying the product exponential with the same base for \[x \] and \[y \] .
We get,
\[ \Rightarrow 15 \times {x^{2 + 1}} \times {y^{2 + 2}} \]
\[ \Rightarrow 15 \times {x^3} \times {y^4} \]
We do not see any exponent with the same base. Only one constant is there. So we stop here.
Hence, the required answer is \[15{x^3}{y^4} \] .
So, the correct answer is “\[15{x^3}{y^4} \]”.
Note: There are nine basic rules for exponentiation. Namely Product, Quotient, Power of power, Power of a product, power of one, Power of zero, power of negative, Change sign of exponents and Fractional exponents. Remembering these we can solve these types of questions. Also note that two like signs make positive signs. Two unlike signs make a negative sign.
Complete step-by-step answer:
Let us see the meaning of above given problem,
In general if n is a positive integer then \[{x^n} = { \left \{ {x \times x \times x \times x........ \times x} \right \}_{n{ \text{ times}}}} \] , we have multiplied \[x \] ,
\[n \] -times.
Now we have \[{x^2} \] , that is \[x \] is multiplied 2 times, and \[{y^2} \] , \[y \] is multiplied 2 times.
Given: Multiply \[ - 3{x^2}{y^2} \] by \[ - 5x{y^2} \]
\[ = \left( { - 3{x^2}{y^2}} \right) \times \left( { - 5x{y^2}} \right) \]
If we have an exponential with the same base, we simply add the exponents.
That is, \[{x^a} \times {x^b} = {x^{a + b}} \] , rearranging above we get:
\[ = ( - 3) \times ( - 5) \times {x^2} \times x \times {y^2} \times {y^2} \]
As we know multiplying two negative numbers makes a positive number.
Regrouping with like terms,
\[ \Rightarrow 15 \times ({x^2} \times x) \times ({y^2} \times {y^2}) \]
Know that, \[x \] can be written as \[{x^1} \] , which is the same for any variable and constant. (Example \[5 = {5^1} \] )
Applying the product exponential with the same base for \[x \] and \[y \] .
We get,
\[ \Rightarrow 15 \times {x^{2 + 1}} \times {y^{2 + 2}} \]
\[ \Rightarrow 15 \times {x^3} \times {y^4} \]
We do not see any exponent with the same base. Only one constant is there. So we stop here.
Hence, the required answer is \[15{x^3}{y^4} \] .
So, the correct answer is “\[15{x^3}{y^4} \]”.
Note: There are nine basic rules for exponentiation. Namely Product, Quotient, Power of power, Power of a product, power of one, Power of zero, power of negative, Change sign of exponents and Fractional exponents. Remembering these we can solve these types of questions. Also note that two like signs make positive signs. Two unlike signs make a negative sign.
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