
What is the product of $\left( {2a + 5b} \right)$ and $\left( {a - 3b} \right)$?
(A) $2{a^2} - 2ab - 15{b^2}$
(B) $2{a^2} - ab - 15{b^2}$
(C) $2{a^2} + 2ab - 15{b^2}$
(D) $2{a^2} + 11ab - 15{b^2}$
Answer
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Hint: We have multiplication of two polynomials that are $\left( {2a + 5b} \right)$ and $\left( {a - 3b} \right)$. Both the polynomials have degree one and have two terms each. Hence, both polynomials are called binomials and linear polynomials. To multiply polynomials, first multiply each term in one polynomial by each term in the other polynomial using distributive law. Then, simplify the resulting polynomial by adding or subtracting the like terms.
Complete answer:
So, we have to calculate the product of $\left( {2a + 5b} \right)$ and $\left( {a - 3b} \right)$.
So, we have, $\left( {2a + 5b} \right)\left( {a - 3b} \right)$
In the first polynomial, we have two terms and in the second polynomial also we have two terms. Multiplying the first term of a polynomial with second polynomial and then the second term with second polynomial, we have,
$2a\left( {a - 3b} \right) + 5b\left( {a - 3b} \right)$
Opening the brackets and multiplying, we have,
$ = 2{a^2} - 6ab + 5ab - 15{b^2}$
Adding the like terms, we have,
$ = 2{a^2} - ab - 15{b^2}$
Thus we have, $\left( {2a + 5b} \right)\left( {a - 3b} \right) = 2{a^2} - ab - 15{b^2}$.
Therefore, the product of two binomial polynomials $\left( {2a + 5b} \right)$ and $\left( {a - 3b} \right)$ is equal to $2{a^2} - ab - 15{b^2}$.
Hence, option (B) is the correct answer.
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 multiplication of any two given polynomials irrespective of the degree and the number of terms.
Complete answer:
So, we have to calculate the product of $\left( {2a + 5b} \right)$ and $\left( {a - 3b} \right)$.
So, we have, $\left( {2a + 5b} \right)\left( {a - 3b} \right)$
In the first polynomial, we have two terms and in the second polynomial also we have two terms. Multiplying the first term of a polynomial with second polynomial and then the second term with second polynomial, we have,
$2a\left( {a - 3b} \right) + 5b\left( {a - 3b} \right)$
Opening the brackets and multiplying, we have,
$ = 2{a^2} - 6ab + 5ab - 15{b^2}$
Adding the like terms, we have,
$ = 2{a^2} - ab - 15{b^2}$
Thus we have, $\left( {2a + 5b} \right)\left( {a - 3b} \right) = 2{a^2} - ab - 15{b^2}$.
Therefore, the product of two binomial polynomials $\left( {2a + 5b} \right)$ and $\left( {a - 3b} \right)$ is equal to $2{a^2} - ab - 15{b^2}$.
Hence, option (B) is the correct answer.
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 multiplication of any two given polynomials irrespective of the degree and the number of terms.
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