Question

If \[{{x}^{2}}-1\] is a factor of \[a{{x}^{4}}+\text{ }b{{x}^{3}}+\text{ }c{{x}^{2}}+\text{ }dx\text{ }+\text{ }e\] , then:-
a)a + c + e = b + d
b)a + b + e = c + d
c)a + b + c = d + e
d)b + c + d = a + e

Answer 1

Hint:-The roots of an expression are also the roots of the polynomial of which that expression is a factor of.
This is very important information. Without this information, we will not be able to solve this question.
Also, an important formula which may be used in solving this question is:-
\[{{a}^{2}}-{{b}^{2}}=\text{ }\left( a\text{ }+\text{ }b \right)\text{ }\left( a-b \right)\]

Complete step-by-step answer:
As mentioned in the question, we have to find the correct option from the four options provided above.
As mentioned in the hint provided above, we can write \[{{x}^{2}}-1\] as (x + 1) (x – 1).
\[{{x}^{2}}\text{-1 }=\text{ }\left( x\text{ }+\text{ }1 \right)\text{ }\left( x-1 \right)\]
So, now, after the above explanation, we can say that ‘1’ and ‘-1’ are the roots of the expression \[{{x}^{2}}-1\] because, on putting ‘1’ or ‘-1’ as the value of ‘x’, we will get 0.
Now, as mentioned in the hint provided above, the roots of an expression are also the roots of the polynomial of which that expression is a factor of.
So, therefore, ‘1’ and ‘-1’ are also the roots of the expression \[a{{x}^{4}}+\text{ }b{{x}^{3}}+\text{ }c{{x}^{2}}+\text{ }dx\text{ }+\text{ }e\] .
So, if we will put ‘1’ or ‘-1’ as the value of ‘x’ in the expression \[a{{x}^{4}}+\text{ }b{{x}^{3}}+\text{ }c{{x}^{2}}+\text{ }dx\text{ }+\text{ }e\] , then the result that we will get will be 0.
So, let us put ‘-1’ as the value of ‘x’ in the expression \[a{{x}^{4}}+\text{ }b{{x}^{3}}+\text{ }c{{x}^{2}}+\text{ }dx\text{ }+\text{ }e\] .
\[\begin{align}
  & a~\times -{{(1)}^{4}}+\text{ }b~\times -{{(1)}^{3}}+\text{ }c~\times -{{(1)}^{2}}+\text{ }d\times ~-1\text{ }+\text{ }e\text{ }=\text{ }0 \\
 & a~1\text{ }+\text{ }b~\left( -1 \right)\text{ }+\text{ }c~1\text{ }+\text{ }d~\left( -1 \right)\text{ }+\text{ }e\text{ }=\text{ }0 \\
\end{align}\]
$\Rightarrow$ a + (-b) + c + (-d) + e = 0
$\Rightarrow$ a – b + c – d + e = 0
$\Rightarrow$ a + c + e = b + d
The only option that matches with the above explanation is the option (a) a + c + e = b + d.
Therefore, the answer of this question is (a) a + c + e = b + d.

Note:-The formulas mentioned above in the hint are very important.
The two formulas are:-
The roots of an expression are also the roots of the polynomial of which that expression is a factor of.
\[{{a}^{2}}-{{b}^{2}}=\text{ }\left( a\text{ }+\text{ }b \right)\text{ }\left( a-b \right)\]
These are the two most important formulas that are used in the solution of this question.
If the students do not know these two formulas, then he or she will not be able to solve this question.
Also, one must do all the calculations in this question very carefully.