
Solve: $4x\left( {3x - 2y} \right) - 2y\left( {3x - 2y} \right)$
Answer
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Hint: In the given question, we are required to simplify the operations on the polynomials provided to us. We will first take the common term outside the bracket and make the expression into the multiplication of two polynomials. Then, we will use the distributive property to open the brackets. We will multiply each term in the first polynomial obtained after opening the bracket with each term in the second polynomial. Then, simplify the resulting polynomial by adding or subtracting the like terms and get to the required answer.
Complete step-by-step answer:
Given, $4x\left( {3x - 2y} \right) - 2y\left( {3x - 2y} \right)$.
So, the given polynomial has two terms. We first take the bracket that is present in both the terms as common. So, we get,
$ \Rightarrow \left( {3x - 2y} \right)\left( {4x - 2y} \right)$
Now, in the first polynomial, we have two terms and in the second polynomial also we have two terms. Multiply the first term of a polynomial with second polynomial and then the second term with second polynomial, we have,
$ \Rightarrow 3x\left( {4x - 2y} \right) - 2y\left( {4x - 2y} \right)$
Opening the brackets and multiplying, we have,
$ \Rightarrow 12{x^2} - 6xy - 8xy + 4{y^2}$
Adding the like terms, we have,
$ \Rightarrow 12{x^2} - 14xy + 4{y^2}$
Thus we have, $4x\left( {3x - 2y} \right) - 2y\left( {3x - 2y} \right) = 12{x^2} - 14xy + 4{y^2}$.
Note: For avoiding mistakes, write the terms in the decreasing order of their exponent. Thus we obtained a polynomial of degree two. Hence, the obtained polynomial is a quadratic polynomial. When we multiply two polynomials of any degree the obtained polynomial must have degree higher than multiplied individual polynomials. We can check this in the given above problem and it satisfies the condition. Be careful with the sign when you open the brackets. Follow the same procedure for any two polynomials.
Complete step-by-step answer:
Given, $4x\left( {3x - 2y} \right) - 2y\left( {3x - 2y} \right)$.
So, the given polynomial has two terms. We first take the bracket that is present in both the terms as common. So, we get,
$ \Rightarrow \left( {3x - 2y} \right)\left( {4x - 2y} \right)$
Now, in the first polynomial, we have two terms and in the second polynomial also we have two terms. Multiply the first term of a polynomial with second polynomial and then the second term with second polynomial, we have,
$ \Rightarrow 3x\left( {4x - 2y} \right) - 2y\left( {4x - 2y} \right)$
Opening the brackets and multiplying, we have,
$ \Rightarrow 12{x^2} - 6xy - 8xy + 4{y^2}$
Adding the like terms, we have,
$ \Rightarrow 12{x^2} - 14xy + 4{y^2}$
Thus we have, $4x\left( {3x - 2y} \right) - 2y\left( {3x - 2y} \right) = 12{x^2} - 14xy + 4{y^2}$.
Note: For avoiding mistakes, write the terms in the decreasing order of their exponent. Thus we obtained a polynomial of degree two. Hence, the obtained polynomial is a quadratic polynomial. When we multiply two polynomials of any degree the obtained polynomial must have degree higher than multiplied individual polynomials. We can check this in the given above problem and it satisfies the condition. Be careful with the sign when you open the brackets. Follow the same procedure for any two polynomials.
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