Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# If $\dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = f\left( x \right) + \dfrac{A}{{x - 2}} + \dfrac{B}{{x - 3}}$, then $f\left( x \right) =$(a) $x - 1$ (b) $x + 1$ (c) $x$ (d) $x + 2$

Last updated date: 11th Sep 2024
Total views: 419.1k
Views today: 11.19k
Verified
419.1k+ views
Hint: Here, we need to find the value of $f\left( x \right)$. We will simplify the left hand side using a long division method. Then, we will use the division algorithm and rewrite the expression. Finally, we will compare the equation obtained with the given equation to find the value of $f\left( x \right)$.

Formula Used:
The division algorithm states that if $p\left( x \right)$ and $g\left( x \right)$ are two polynomials where $g\left( x \right) \ne 0$, then there are two polynomials $q\left( x \right)$ and $r\left( x \right)$ such that $p\left( x \right) = q\left( x \right) \times g\left( x \right) + r\left( x \right)$. Here, $p\left( x \right)$ is the dividend, $g\left( x \right)$ is the divisor, $q\left( x \right)$ is the quotient, and $r\left( x \right)$ is the remainder.

First, we will use a long division method to divide ${x^3} - 6{x^2} + 10x - 2$ by ${x^2} - 5x + 6$.
Therefore, we get

We can observe that when ${x^3} - 6{x^2} + 10x - 2$ is divided by ${x^2} - 5x + 6$ using long division method, the quotient is $x - 1$, and the remainder is $- x + 4$.
Now, we will use the division algorithm.
The division algorithm states that if $p\left( x \right)$ and $g\left( x \right)$ are two polynomials where $g\left( x \right) \ne 0$, then there are two polynomials $q\left( x \right)$ and $r\left( x \right)$ such that $p\left( x \right) = q\left( x \right) \times g\left( x \right) + r\left( x \right)$. Here, $p\left( x \right)$ is the dividend, $g\left( x \right)$ is the divisor, $q\left( x \right)$ is the quotient, and $r\left( x \right)$ is the remainder.
Thus, we get
$p\left( x \right) = {x^3} - 6{x^2} + 10x - 2$
$q\left( x \right) = x - 1$
$r\left( x \right) = - x + 4$
$g\left( x \right) = {x^2} - 5x + 6$
Substituting $p\left( x \right) = {x^3} - 6{x^2} + 10x - 2$, $q\left( x \right) = x - 1$, $r\left( x \right) = - x + 4$, and $g\left( x \right) = {x^2} - 5x + 6$ in the division algorithm, we get
$\Rightarrow {x^3} - 6{x^2} + 10x - 2 = \left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right) + \left( { - x + 4} \right)$
Dividing both sides by ${x^2} - 5x + 6$, we get
$\Rightarrow \dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 + \left( {\dfrac{{ - x + 4}}{{{x^2} - 5x + 6}}} \right)$
Rewriting the expression, we get
$\Rightarrow \dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 - \left( {\dfrac{{x - 4}}{{{x^2} - 5x + 6}}} \right)$
Factoring the denominator, we get
$\begin{array}{l} \Rightarrow \dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 - \left( {\dfrac{{x - 4}}{{{x^2} - 3x - 2x + 6}}} \right)\\ \Rightarrow \dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 - \left[ {\dfrac{{x - 4}}{{x\left( {x - 3} \right) - 2\left( {x - 3} \right)}}} \right]\\ \Rightarrow \dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 - \left[ {\dfrac{{x - 4}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}} \right]\end{array}$
Multiplying and dividing the fraction by 2, we get
$\begin{array}{l} \Rightarrow \dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 - \dfrac{2}{2}\left[ {\dfrac{{x - 4}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}} \right]\\ \Rightarrow \dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 - \dfrac{1}{2}\left[ {\dfrac{{2\left( {x - 4} \right)}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}} \right]\end{array}$
Multiplying the above terms using the distributive property, we get
$\Rightarrow \dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 - \dfrac{1}{2}\left[ {\dfrac{{2x - 8}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}} \right]$
Rewriting the numerator, we get
$\begin{array}{l} \Rightarrow \dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 - \dfrac{1}{2}\left[ {\dfrac{{x + x - 3 - 3 - 2}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}} \right]\\ \Rightarrow \dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 - \dfrac{1}{2}\left[ {\dfrac{{x - 3 + x - 2 - 3}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}} \right]\end{array}$
Separating the fractions with the same denominator, we get
$\Rightarrow \dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 - \dfrac{1}{2}\left[ {\dfrac{{x - 3}}{{\left( {x - 3} \right)\left( {x - 2} \right)}} + \dfrac{{x - 2}}{{\left( {x - 3} \right)\left( {x - 2} \right)}} - \dfrac{3}{{\left( {x - 3} \right)\left( {x - 2} \right)}}} \right]$
Simplifying the expressions, we get
$\Rightarrow \dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 - \dfrac{1}{2}\left[ {\dfrac{1}{{x - 2}} + \dfrac{1}{{x - 3}} - \dfrac{3}{{\left( {x - 3} \right)\left( {x - 2} \right)}}} \right]$
Multiplying the above terms using the distributive property, we get$\Rightarrow \dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 - \dfrac{1}{{2\left( {x - 2} \right)}} - \dfrac{1}{{2\left( {x - 3} \right)}} + \dfrac{3}{{2\left( {x - 3} \right)\left( {x - 2} \right)}}$
Rearranging the terms of the expression, we get
$\Rightarrow \dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 + \dfrac{3}{{2\left( {x - 3} \right)\left( {x - 2} \right)}} + \dfrac{{ - \dfrac{1}{2}}}{{\left( {x - 2} \right)}} + \dfrac{{ - \dfrac{1}{2}}}{{\left( {x - 3} \right)}}$
It is given that $\dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = f\left( x \right) + \dfrac{A}{{x - 2}} + \dfrac{B}{{x - 3}}$.
Here, $A$ and $B$ are constants.
Therefore, from the equations $\dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = x - 1 + \dfrac{3}{{2\left( {x - 3} \right)\left( {x - 2} \right)}} + \dfrac{{ - \dfrac{1}{2}}}{{\left( {x - 2} \right)}} + \dfrac{{ - \dfrac{1}{2}}}{{\left( {x - 3} \right)}}$ and $\dfrac{{{x^3} - 6{x^2} + 10x - 2}}{{{x^2} - 5x + 6}} = f\left( x \right) + \dfrac{A}{{x - 2}} + \dfrac{B}{{x - 3}}$, we get
$\Rightarrow x - 1 + \dfrac{3}{{2\left( {x - 3} \right)\left( {x - 2} \right)}} + \dfrac{{ - \dfrac{1}{2}}}{{\left( {x - 2} \right)}} + \dfrac{{ - \dfrac{1}{2}}}{{\left( {x - 3} \right)}} = f\left( x \right) + \dfrac{A}{{x - 2}} + \dfrac{B}{{x - 3}}$
Comparing the terms of the expressions, we get
$x - 1 + \dfrac{3}{{2\left( {x - 3} \right)\left( {x - 2} \right)}} = f\left( x \right)$
$\Rightarrow - \dfrac{1}{2} = A$
$\Rightarrow - \dfrac{1}{2} = B$
Therefore, we get the value of $f\left( x \right)$ as $x - 1 + \dfrac{3}{{2\left( {x - 3} \right)\left( {x - 2} \right)}}$.
Thus, none of the given options are correct.

Note: We have used the distributive law of multiplication to multiply some expressions in the solution. The distributive law of multiplication states that $a\left( {b + c} \right) = a \cdot b + a \cdot c$. Here for finding the value of $f\left( x \right)$ it is important for us to find the value of A and B.