Question

Multiply the following polynomials
$\left( {5 - {d^2} + 2d} \right)\left( {2d - 1} \right)$

Answer 1

Hint: Here, we are required to multiply two polynomials. We would multiply the brackets by multiplying each variable present in one bracket by the other present in another bracket. After opening the brackets and multiplying the variables, we would find the required answer which we could verify by assuming any value of the given variable $d$.

Complete step-by-step answer:
A polynomial is a mathematical expression which consists of variables and coefficients.
In this question, we are required to multiply the two given polynomials.
Now, as we know,
When we multiply two brackets, then, every element of the first bracket is multiplied by every element present in the second bracket.
Hence, the given polynomials are:
$\left( {5 - {d^2} + 2d} \right)\left( {2d - 1} \right)$
Multiplying them,
$ = 2d\left( {5 - {d^2} + 2d} \right) - 1\left( {5 - {d^2} + 2d} \right)$
Here, we have shown that the two elements present in the second bracket have to be multiplied by the elements in the third bracket.
Also, as we can see, there is a ‘minus’ sign outside the second bracket.
Hence, all the signs inside the second bracket would get reversed.
In other words, a ‘minus’ would become a ‘plus’ and vice-versa.
Now, opening the brackets and multiplying, we get,
$ = \left( {2d} \right)\left( 5 \right) - \left( {2d} \right)\left( {{d^2}} \right) + \left( {2d} \right)\left( {2d} \right) - 5 + {d^2} - 2d$
Solving further,
$ = 10d - 2{d^3} + 4{d^2} - 5 + {d^2} - 2d$
Now, solving the like variables,
$ = - 2{d^3} + \left( {4{d^2} + {d^2}} \right) + \left( {10d - 2d} \right) - 5$
$ = - 2{d^3} + 5{d^2} + 8d - 5$
Therefore, when we multiply the polynomials$\left( {5 - {d^2} + 2d} \right)\left( {2d - 1} \right)$, we get
$\left( {5 - {d^2} + 2d} \right)\left( {2d - 1} \right) = - 2{d^3} + 5{d^2} + 8d - 5$
Hence, this is the required answer.

Note: We can also cross check whether our answer is correct or not.
First of all, we would substitute $d$ by any real number.
Now, let $d = 1$
We would substitute this value in both LHS and RHS and if they become equal then our answer is correct.
LHS:
$\left( {5 - {d^2} + 2d} \right)\left( {2d - 1} \right)$
Here, let $d = 1$
$ = \left( {5 - {{\left( 1 \right)}^2} + 2\left( 1 \right)} \right)\left( {2\left( 1 \right) - 1} \right)$
$ = \left( {5 - 1 + 2} \right)\left( {2 - 1} \right)$
Solving further,
$ = \left( 6 \right)\left( 1 \right) = 6$
Now, RHS:
$ - 2{d^3} + 5{d^2} + 8d - 5$
Let, $d = 1$
$ = - 2{\left( 1 \right)^3} + 5{\left( 1 \right)^2} + 8\left( 1 \right) - 5$
$ = - 2\left( 1 \right) + 5\left( 1 \right) + 8 - 5$
Solving further,
$ = - 2 + 5 + 8 - 5 = 6$
Clearly,
LHS$ = $RHS$ = $6
Hence, we have multiplied the two given polynomials correctly.