Question

If ${x^2} - 1$ is factor of $a{x^4} + b{x^3} + c{x^2} + dx + e$ , then show that $a + c + e = b + d = 0$

Answer 1

Hint:
As the factor of the equation is given, we can find the roots of the equations by equating the factor to zero. Then we can apply the roots to the polynomial to obtain 2 equations. Then we can add the equations and subtract the equations separately to obtain the required relation.

Complete step by step solution:
Let $f\left( x \right) = a{x^4} + b{x^3} + c{x^2} + dx + e$
It is given that ${x^2} - 1$ is a factor of this polynomial.
We know that the roots of the factor will also be the roots of the equation. So, we can find the roots of the factor. For that we equate it to zero.
 $ \Rightarrow {x^2} - 1 = 0$
 $ \Rightarrow {x^2} = 1$
On taking the square root, we get,
 $ \Rightarrow x = \pm 1$
Thus 1 and -1 are the roots of the factor. So, they will be the roots of the equations.
When $x = 1$ , we get,
 $ \Rightarrow f\left( 1 \right) = 0$
 $ \Rightarrow a \times {1^4} + b \times {1^3} + c \times {1^2} + d \times 1 + e = 0$
 $ \Rightarrow a + b + c + d + e = 0$ ….. (1)
When $x = - 1$ , we get,
 $ \Rightarrow f\left( { - 1} \right) = 0$
 $ \Rightarrow a \times {\left( { - 1} \right)^4} + b \times {\left( { - 1} \right)^3} + c \times {\left( { - 1} \right)^2} + d \times \left( { - 1} \right) + e = 0$
 $ \Rightarrow a - b + c - d + e = 0$ ….. (2)
Now we can add equations (1) and (2),
 $
   \Rightarrow a + b + c + d + e = 0 \\
  \underline {\left( + \right)a - b + c - d + e = 0} \\
  2a + 0 + 2c + 0 + 2e = 0 \\
 $
On dividing throughout with 2, we get,
 $ \Rightarrow a + c + e = 0$ … (3)
Now we can subtract equation (2) from equation (1),
 $
   \Rightarrow a + b + c + d + e = 0 \\
  \underline {\left( - \right)a - b + c - d + e = 0} \\
  0 + 2b + 0 + 2d + 0 = 0 \\
 $
On dividing throughout with 2, we get,
 $ \Rightarrow b + d = 0$ … (4).
 On comparing equations (3) and (4), we get,
 $ \Rightarrow a + c + e = b + d = 0$
Thus, we have proved the required result.

Note:
We know that roots of the polynomial are the values of the variable at which the value of the expression will become zero. If one of the factors is equal to zero, then the whole expression will become zero. So, we used the roots of the factor and took it as the root of the expression. While taking the square root of 1, we must take both the negative and positive roots. Then we must substitute both the values in the polynomial and equate it to zero.