# Find the value of $a{\text{ and }}b$ so that $\left( {x + 1} \right){\text{ and }}\left( {x - 1} \right)$ are factors of ${x^4} + a{x^3} - 3{x^2} + 2x + b$

Last updated date: 27th Mar 2023

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Answer

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Hint- Use the property of roots or the factors of the equation in order to solve the problem. Find the roots with the help of factors of equations and then put the value of the roots in the function to obtain simple equations in terms of unknown variables.

Complete step-by-step solution -

Given equation: ${x^4} + a{x^3} - 3{x^2} + 2x + b$

Also we know that $\left( {x + 1} \right){\text{ and }}\left( {x - 1} \right)$ are the factors of the given equation.

Let us find the roots of the equation with the help of the factors.

Roots will be given by equating the factors of the equation to zero.

$

\Rightarrow \left( {x + 1} \right) = 0\& \left( {x - 1} \right) = 0 \\

\Rightarrow x = - 1\& x = 1 \\

$

So, we have two roots of the equation.

Let $f\left( x \right) = {x^4} + a{x^3} - 3{x^2} + 2x + b$

For the roots $x = - 1\& x = 1$ the value of the function is zero.

Taking $x = - 1$

\[

f\left( {{x_1}} \right) = 0{\text{ }}\left[ {{\text{where }}{x_1} = - 1} \right] \\

\Rightarrow f\left( {{x_1}} \right) = {x_1}^4 + a{x_1}^3 - 3{x_1}^2 + 2{x_1} + b = 0 \\

\Rightarrow f\left( { - 1} \right) = {\left( { - 1} \right)^4} + a{\left( { - 1} \right)^3} - 3{\left( { - 1} \right)^2} + 2\left( { - 1} \right) + b = 0 \\

\Rightarrow f\left( { - 1} \right) = 1 - a - 3 - 2 + b = 0 \\

\Rightarrow - a + b = - 1 + 3 + 2 \\

\Rightarrow - a + b = 4...............(1) \\

\]

We have one equation in two unknown variables from the above root.

Similarly taking the other root and solving

Taking $x = 1$

\[

f\left( {{x_2}} \right) = 0{\text{ }}\left[ {{\text{where }}{x_2} = 1} \right] \\

\Rightarrow f\left( {{x_2}} \right) = {x_2}^4 + a{x_2}^3 - 3{x_2}^2 + 2{x_2} + b = 0 \\

\Rightarrow f\left( 1 \right) = {\left( 1 \right)^4} + a{\left( 1 \right)^3} - 3{\left( 1 \right)^2} + 2\left( 1 \right) + b = 0 \\

\Rightarrow f\left( 1 \right) = 1 + a - 3 + 2 + b = 0 \\

\Rightarrow a + b = - 1 + 3 - 2 \\

\Rightarrow a + b = 0 \\

\Rightarrow a = - b...............(2) \\

\]

Now we have one more equation in the same unknown variable.

Solving equation (1) and (2) in order to find the value of $a{\text{ and }}b$

Putting the equation (2) in equation (1)

\[

\Rightarrow - a + b = 4...............(1) \\

\Rightarrow - \left( { - b} \right) + b = 4{\text{ }}\left[ {\because a = - b\left( {{\text{from equation (2)}}} \right)} \right] \\

\Rightarrow 2b = 4 \\

\Rightarrow b = 2 \\

\]

In order to solve for $a$ , putting the value of $b$ obtained above in equation (2)

\[

\Rightarrow a = - b = - \left( 2 \right) \\

\Rightarrow a = - 2 \\

\]

Hence, the values of unknown variables are $a = - 2\& b = 2$ .

Note- In order to solve such a question as above there is no need to solve the degree 4 polynomial in order to find all the 4 roots of the equation. Questions like these can be solved by simple manipulation and by the use of property of the roots. Relationship between the factors of the equation and the roots of the equation is very important and must be remembered.

Complete step-by-step solution -

Given equation: ${x^4} + a{x^3} - 3{x^2} + 2x + b$

Also we know that $\left( {x + 1} \right){\text{ and }}\left( {x - 1} \right)$ are the factors of the given equation.

Let us find the roots of the equation with the help of the factors.

Roots will be given by equating the factors of the equation to zero.

$

\Rightarrow \left( {x + 1} \right) = 0\& \left( {x - 1} \right) = 0 \\

\Rightarrow x = - 1\& x = 1 \\

$

So, we have two roots of the equation.

Let $f\left( x \right) = {x^4} + a{x^3} - 3{x^2} + 2x + b$

For the roots $x = - 1\& x = 1$ the value of the function is zero.

Taking $x = - 1$

\[

f\left( {{x_1}} \right) = 0{\text{ }}\left[ {{\text{where }}{x_1} = - 1} \right] \\

\Rightarrow f\left( {{x_1}} \right) = {x_1}^4 + a{x_1}^3 - 3{x_1}^2 + 2{x_1} + b = 0 \\

\Rightarrow f\left( { - 1} \right) = {\left( { - 1} \right)^4} + a{\left( { - 1} \right)^3} - 3{\left( { - 1} \right)^2} + 2\left( { - 1} \right) + b = 0 \\

\Rightarrow f\left( { - 1} \right) = 1 - a - 3 - 2 + b = 0 \\

\Rightarrow - a + b = - 1 + 3 + 2 \\

\Rightarrow - a + b = 4...............(1) \\

\]

We have one equation in two unknown variables from the above root.

Similarly taking the other root and solving

Taking $x = 1$

\[

f\left( {{x_2}} \right) = 0{\text{ }}\left[ {{\text{where }}{x_2} = 1} \right] \\

\Rightarrow f\left( {{x_2}} \right) = {x_2}^4 + a{x_2}^3 - 3{x_2}^2 + 2{x_2} + b = 0 \\

\Rightarrow f\left( 1 \right) = {\left( 1 \right)^4} + a{\left( 1 \right)^3} - 3{\left( 1 \right)^2} + 2\left( 1 \right) + b = 0 \\

\Rightarrow f\left( 1 \right) = 1 + a - 3 + 2 + b = 0 \\

\Rightarrow a + b = - 1 + 3 - 2 \\

\Rightarrow a + b = 0 \\

\Rightarrow a = - b...............(2) \\

\]

Now we have one more equation in the same unknown variable.

Solving equation (1) and (2) in order to find the value of $a{\text{ and }}b$

Putting the equation (2) in equation (1)

\[

\Rightarrow - a + b = 4...............(1) \\

\Rightarrow - \left( { - b} \right) + b = 4{\text{ }}\left[ {\because a = - b\left( {{\text{from equation (2)}}} \right)} \right] \\

\Rightarrow 2b = 4 \\

\Rightarrow b = 2 \\

\]

In order to solve for $a$ , putting the value of $b$ obtained above in equation (2)

\[

\Rightarrow a = - b = - \left( 2 \right) \\

\Rightarrow a = - 2 \\

\]

Hence, the values of unknown variables are $a = - 2\& b = 2$ .

Note- In order to solve such a question as above there is no need to solve the degree 4 polynomial in order to find all the 4 roots of the equation. Questions like these can be solved by simple manipulation and by the use of property of the roots. Relationship between the factors of the equation and the roots of the equation is very important and must be remembered.

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