Question

If $\left( {{y^2} - 1} \right)$ is a factor of $a{y^4} + b{y^3} + c{y^2} + dy + e$, then show that $a + c + e = b + d = 0$.

Answer 1

Hint: In the given question, we have to use the factor theorem to prove that the value of the given algebraic expression is zero. A thorough understanding of functions, division algorithms and its applications will be of great significance. We must have an understanding of the factor theorem in order to deal with the problem. According to the factor theorem, if the value of the variable obtained on equating the divisor to zero is a root of the polynomial, then the divisor polynomial is a factor of the dividend polynomial.

Complete step-by-step answer:
In the given question, we are given that $\left( {{y^2} - 1} \right)$ is a factor of the polynomial $a{y^2} + b{y^3} + c{y^2} + dy + e$.
We first equate the divisor polynomial as zero.
So, we get, $\left( {{y^2} - 1} \right) = 0$
Factoring the left side of the equation using the algebraic identity $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$ , we get,
$ \Rightarrow \left( {y - 1} \right)\left( {y + 1} \right) = 0$
Now, either $y - 1 = 0$ or $y + 1 = 0$.
Hence, $y = 1$ or $y = - 1$.
The value of the variable so obtained is put into the dividend function to find the remainder.
Now, putting $y = 1$ into $a{y^4} + b{y^3} + c{y^2} + dy + e$, we get,
$ \Rightarrow a{\left( 1 \right)^4} + b{\left( 1 \right)^3} + c{\left( 1 \right)^2} + d\left( 1 \right) + e$
$ \Rightarrow a + b + c + d + e$
Since the divisor polynomial $\left( {{y^2} - 1} \right)$ is a factor of $a{y^2} + b{y^3} + c{y^2} + dy + e$, we equate the remainder as zero. So, we get,
$ \Rightarrow a + b + c + d + e = 0 - - - - - \left( 1 \right)$
Now, putting $y = - 1$ into $a{y^4} + b{y^3} + c{y^2} + dy + e$, we get,
$ \Rightarrow a{\left( { - 1} \right)^4} + b{\left( { - 1} \right)^3} + c{\left( { - 1} \right)^2} + d\left( { - 1} \right) + e$
$ \Rightarrow a - b + c - d + e$
Equating the remainder to zero, we get,
$ \Rightarrow a - b + c - d + e = 0 - - - - - \left( 2 \right)$
Now, we add both the equations $\left( 1 \right)$ and $\left( 2 \right)$.
$ \Rightarrow a + b + c + d + e + a - b + c - d + e = 0$
$ \Rightarrow 2a + 2c + 2e = 0$
Dividing both sides of the equation by $2$, we get,
$ \Rightarrow a + c + e = 0$
Now, putting in the value of $\left( {a + c + e} \right)$ as zero in equation $\left( 1 \right)$, we get,
 $ \Rightarrow b + d = 0$
Therefore, we conclude that $a + c + e = b + d = 0$.
Hence, proved.

Note: One must know the significance of the factor theorem in order to solve such problems. Substitution of a variable involves putting a certain value in place of the variable. That specified value may be a certain number or even any other variable. We must take care of the calculations while doing such problems. Factor theorem is a theorem linking the factors of a polynomial with the zeros or roots of the polynomial. It is a special case of the remainder theorem where the remainder on dividing the dividend polynomial by the divisor polynomial comes out to be zero.