Samacheer Kalvi Class 11 Maths Solutions Chapter 2 Basic Algebra Exercise 2.9

Chapter 2 Basic Algebra Ex 2.9

Question 1. Resolve the following rational expressions into partial fractions : \( \frac{1}{x^{2}-a^{2}} \)
Answer:
We need to break down the rational expression into simpler partial fractions. First, factorize the denominator:
\( x^2 - a^2 = (x+a)(x-a) \)
Now, we can write the expression as a sum of two simple fractions with unknown constants A and B:
\( \frac{1}{x^2 - a^2} = \frac{A}{x+a} + \frac{B}{x-a} \)
To find A and B, we combine the fractions on the right side:
\( \frac{1}{(x+a)(x-a)} = \frac{A(x-a) + B(x+a)}{(x+a)(x-a)} \)
This means the numerators must be equal:
\( 1 = A(x-a) + B(x+a) \) (Let's call this Equation (1))
To find B, substitute \( x = a \) into Equation (1):
\( 1 = A(a-a) + B(a+a) \)
\( 1 = A(0) + B(2a) \)
\( 1 = 2aB \)
\( \implies B = \frac{1}{2a} \)
Next, to find A, substitute \( x = -a \) into Equation (1):
\( 1 = A(-a-a) + B(-a+a) \)
\( 1 = A(-2a) + B(0) \)
\( 1 = -2aA \)
\( \implies A = -\frac{1}{2a} \)
Now, substitute the values of A and B back into the partial fraction form:
\( \frac{1}{x^2 - a^2} = \frac{-\frac{1}{2a}}{x+a} + \frac{\frac{1}{2a}}{x-a} \)
This can be rewritten for clarity:
\( \frac{1}{x^2 - a^2} = \frac{1}{2a(x-a)} - \frac{1}{2a(x+a)} \)
In simple words: We broke the complicated fraction into two simpler ones. We found the numbers for each simple fraction by carefully putting in specific values for x, which helped us solve for the unknowns. This method is often used to make complex fractions easier to work with, especially in calculus.

🎯 Exam Tip: Always fully factorize the denominator first. For distinct linear factors, use the substitution method to quickly find the constants, as it simplifies the equation significantly.

Question 2. Resolve \( \frac{3x+1}{(x-2)(x+1)} \) into partial fractions.
Answer:
The denominator is already factored into distinct linear factors \( (x-2) \) and \( (x+1) \). So, we can write the partial fraction decomposition as:
\( \frac{3x+1}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1} \)
To combine the terms on the right side, we find a common denominator:
\( \frac{3x+1}{(x-2)(x+1)} = \frac{A(x+1) + B(x-2)}{(x-2)(x+1)} \)
Equating the numerators, we get:
\( 3x+1 = A(x+1) + B(x-2) \) (Let's call this Equation (1))
To find A, substitute \( x = 2 \) into Equation (1):
\( 3(2)+1 = A(2+1) + B(2-2) \)
\( 6+1 = A(3) + B(0) \)
\( 7 = 3A \)
\( \implies A = \frac{7}{3} \)
Next, to find B, substitute \( x = -1 \) into Equation (1):
\( 3(-1)+1 = A(-1+1) + B(-1-2) \)
\( -3+1 = A(0) + B(-3) \)
\( -2 = -3B \)
\( \implies B = \frac{2}{3} \)
Now, substitute the values of A and B back into the partial fraction form:
\( \frac{3x+1}{(x-2)(x+1)} = \frac{\frac{7}{3}}{x-2} + \frac{\frac{2}{3}}{x+1} \)
This can be written more cleanly as:
\( \frac{3x+1}{(x-2)(x+1)} = \frac{7}{3(x-2)} + \frac{2}{3(x+1)} \)
In simple words: We took a fraction with two different factors in the bottom part and separated it into two easier fractions. We found the numbers on top of these new fractions by plugging in specific values for x, which helps us quickly find the unknown values. This is a powerful way to break down complex expressions.

🎯 Exam Tip: When dealing with distinct linear factors, choosing the roots of the factors as substitution values for x is the quickest way to solve for the constants.

Question 3. Resolve \( \frac{x}{(x^2+1)(x-1)(x+2)} \) into partial fractions.
Answer:
The denominator has one irreducible quadratic factor \( (x^2+1) \) and two distinct linear factors \( (x-1) \) and \( (x+2) \). So, the partial fraction decomposition will be:
\( \frac{x}{(x^2+1)(x-1)(x+2)} = \frac{Ax+B}{x^2+1} + \frac{C}{x-1} + \frac{D}{x+2} \)
Combine the terms on the right side:
\( x = (Ax+B)(x-1)(x+2) + C(x^2+1)(x+2) + D(x^2+1)(x-1) \) (Let's call this Equation (1))
First, find C by substituting \( x = 1 \) into Equation (1):
\( 1 = (A(1)+B)(1-1)(1+2) + C(1^2+1)(1+2) + D(1^2+1)(1-1) \)
\( 1 = (A+B)(0)(3) + C(2)(3) + D(2)(0) \)
\( 1 = 6C \)
\( \implies C = \frac{1}{6} \)
Next, find D by substituting \( x = -2 \) into Equation (1):
\( -2 = (A(-2)+B)(-2-1)(-2+2) + C((-2)^2+1)(-2+2) + D((-2)^2+1)(-2-1) \)
\( -2 = (-2A+B)(-3)(0) + C(5)(0) + D(5)(-3) \)
\( -2 = -15D \)
\( \implies D = \frac{2}{15} \)
Now, find B by substituting \( x = 0 \) into Equation (1):
\( 0 = (A(0)+B)(0-1)(0+2) + C(0^2+1)(0+2) + D(0^2+1)(0-1) \)
\( 0 = B(-1)(2) + C(1)(2) + D(1)(-1) \)
\( 0 = -2B + 2C - D \)
Substitute the values of C and D:
\( 0 = -2B + 2\left(\frac{1}{6}\right) - \frac{2}{15} \)
\( 0 = -2B + \frac{1}{3} - \frac{2}{15} \)
To solve for B, find a common denominator for the fractions:
\( 0 = -2B + \frac{5}{15} - \frac{2}{15} \)
\( 0 = -2B + \frac{3}{15} \)
\( 0 = -2B + \frac{1}{5} \)
\( \implies 2B = \frac{1}{5} \)
\( \implies B = \frac{1}{10} \)
Finally, find A by comparing the coefficients of \( x^3 \) on both sides of Equation (1). Expand the right side:
\( x = (Ax+B)(x^2+x-2) + C(x^3+2x^2+x+2) + D(x^3-x^2+x-1) \)
The \( x^3 \) terms are \( Ax^3 \), \( Cx^3 \), and \( Dx^3 \). The left side has no \( x^3 \) term, so its coefficient is 0.
\( 0 = A + C + D \)
Substitute the values of C and D:
\( 0 = A + \frac{1}{6} + \frac{2}{15} \)
\( A = -\frac{1}{6} - \frac{2}{15} \)
Find a common denominator (30):
\( A = -\frac{5}{30} - \frac{4}{30} \)
\( A = -\frac{9}{30} \)
\( \implies A = -\frac{3}{10} \)
Now, substitute the values of A, B, C, and D back into the partial fraction form:
\( \frac{x}{(x^2+1)(x-1)(x+2)} = \frac{-\frac{3}{10}x + \frac{1}{10}}{x^2+1} + \frac{\frac{1}{6}}{x-1} + \frac{\frac{2}{15}}{x+2} \)
We can simplify the first term:
\( \frac{x}{(x^2+1)(x-1)(x+2)} = \frac{1-3x}{10(x^2+1)} + \frac{1}{6(x-1)} + \frac{2}{15(x+2)} \)
In simple words: This problem had different kinds of factors in the bottom part. We used a special form for the quadratic factor and then found all the unknown numbers by putting in some easy values for x and also by looking at the numbers next to the highest powers of x. This helps us break down a complex expression into much simpler parts.

🎯 Exam Tip: When dealing with irreducible quadratic factors, remember to use a linear term \( Ax+B \) as the numerator. Combining the substitution method with coefficient comparison is often necessary for these more complex problems.

Question 4. Resolve \( \frac{x}{(x-1)^3} \) into partial fractions.
Answer:
The denominator has a repeated linear factor \( (x-1)^3 \). For repeated factors, we need a term for each power up to the highest power. So, the partial fraction decomposition is:
\( \frac{x}{(x-1)^3} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3} \)
Multiply both sides by \( (x-1)^3 \) to clear the denominators:
\( x = A(x-1)^2 + B(x-1) + C \) (Let's call this Equation (1))
To find C, substitute \( x = 1 \) into Equation (1):
\( 1 = A(1-1)^2 + B(1-1) + C \)
\( 1 = A(0) + B(0) + C \)
\( \implies C = 1 \)
Next, to find A, we can compare the coefficients of \( x^2 \) on both sides of Equation (1). Expand the right side:
\( x = A(x^2-2x+1) + B(x-1) + C \)
\( x = Ax^2 - 2Ax + A + Bx - B + C \)
The coefficient of \( x^2 \) on the left side is 0. On the right side, the coefficient of \( x^2 \) is A.
So, \( 0 = A \)
\( \implies A = 0 \)
Finally, to find B, substitute \( x = 0 \) into Equation (1) and use the known values of A and C:
\( 0 = A(0-1)^2 + B(0-1) + C \)
\( 0 = A(1) + B(-1) + C \)
\( 0 = A - B + C \)
Substitute \( A=0 \) and \( C=1 \):
\( 0 = 0 - B + 1 \)
\( \implies B = 1 \)
Now, substitute the values of A, B, and C back into the partial fraction form:
\( \frac{x}{(x-1)^3} = \frac{0}{x-1} + \frac{1}{(x-1)^2} + \frac{1}{(x-1)^3} \)
The term with A is zero, so it simplifies to:
\( \frac{x}{(x-1)^3} = \frac{1}{(x-1)^2} + \frac{1}{(x-1)^3} \)
In simple words: When the bottom part of a fraction has the same factor multiplied many times (like \( (x-1) \) three times), we break it into parts, each with a different power of that factor. We then find the numbers for each part using simple substitutions and by matching the terms of x on both sides. This makes the original fraction much simpler.

🎯 Exam Tip: For repeated linear factors like \( (x-r)^n \), remember to include terms for all powers from 1 to n: \( \frac{A_1}{x-r} + \frac{A_2}{(x-r)^2} + \dots + \frac{A_n}{(x-r)^n} \).

Question 5. Resolve \( \frac{1}{x^{4}-1} \) into partial fractions.
Answer:
First, factorize the denominator completely. Recognize \( x^4-1 \) as a difference of squares:
\( x^4-1 = (x^2)^2 - 1^2 = (x^2-1)(x^2+1) \)
Further factorize \( (x^2-1) \):
\( x^2-1 = (x-1)(x+1) \)
So, the complete factorization is: \( x^4-1 = (x-1)(x+1)(x^2+1) \)
This gives us two distinct linear factors and one irreducible quadratic factor. The partial fraction decomposition is:
\( \frac{1}{(x-1)(x+1)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1} \)
Multiply both sides by \( (x-1)(x+1)(x^2+1) \) to clear the denominators:
\( 1 = A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx+D)(x-1)(x+1) \) (Let's call this Equation (1))
To find A, substitute \( x = 1 \) into Equation (1):
\( 1 = A(1+1)(1^2+1) + B(1-1)(1^2+1) + (C(1)+D)(1-1)(1+1) \)
\( 1 = A(2)(2) + B(0)(2) + (C+D)(0)(2) \)
\( 1 = 4A \)
\( \implies A = \frac{1}{4} \)
To find B, substitute \( x = -1 \) into Equation (1):
\( 1 = A(-1+1)((-1)^2+1) + B(-1-1)((-1)^2+1) + (C(-1)+D)(-1-1)(-1+1) \)
\( 1 = A(0)(2) + B(-2)(2) + (-C+D)(-2)(0) \)
\( 1 = -4B \)
\( \implies B = -\frac{1}{4} \)
To find D, substitute \( x = 0 \) into Equation (1):
\( 1 = A(0+1)(0^2+1) + B(0-1)(0^2+1) + (C(0)+D)(0-1)(0+1) \)
\( 1 = A(1)(1) + B(-1)(1) + D(-1)(1) \)
\( 1 = A - B - D \)
Substitute the values of A and B:
\( 1 = \frac{1}{4} - \left(-\frac{1}{4}\right) - D \)
\( 1 = \frac{1}{4} + \frac{1}{4} - D \)
\( 1 = \frac{2}{4} - D \)
\( 1 = \frac{1}{2} - D \)
\( D = \frac{1}{2} - 1 \)
\( \implies D = -\frac{1}{2} \)
To find C, compare the coefficients of \( x^3 \) on both sides of Equation (1). Expand the terms that produce \( x^3 \):
From \( A(x+1)(x^2+1) = A(x^3+x^2+x+1) \), the \( x^3 \) term is \( Ax^3 \).
From \( B(x-1)(x^2+1) = B(x^3-x^2+x-1) \), the \( x^3 \) term is \( Bx^3 \).
From \( (Cx+D)(x-1)(x+1) = (Cx+D)(x^2-1) = Cx^3-Cx+Dx^2-D \), the \( x^3 \) term is \( Cx^3 \).
The coefficient of \( x^3 \) on the left side is 0.
So, \( 0 = A + B + C \)
Substitute the values of A and B:
\( 0 = \frac{1}{4} + \left(-\frac{1}{4}\right) + C \)
\( 0 = 0 + C \)
\( \implies C = 0 \)
Substitute A, B, C, and D back into the partial fraction form:
\( \frac{1}{x^4-1} = \frac{\frac{1}{4}}{x-1} + \frac{-\frac{1}{4}}{x+1} + \frac{0x-\frac{1}{2}}{x^2+1} \)
This simplifies to:
\( \frac{1}{x^4-1} = \frac{1}{4(x-1)} - \frac{1}{4(x+1)} - \frac{1}{2(x^2+1)} \)
In simple words: We first broke down the bottom part of the fraction into simpler pieces, including a quadratic part that could not be factored more. Then, we used a mix of putting in easy numbers for x and matching the terms of x to find all the missing numbers for our simplified fractions. This helps us work with very complex fractions more easily.

🎯 Exam Tip: Always completely factorize the denominator first, identifying linear and irreducible quadratic factors. For \( x^4-1 \), remember the factorization pattern for differences of squares twice.

Question 6. Resolve \( \frac{(x-1)^2}{x^3+x} \) into partial fractions.
Answer:
First, factorize the denominator. We can take out a common factor of \( x \):
\( x^3+x = x(x^2+1) \)
The factor \( (x^2+1) \) is an irreducible quadratic factor. So, the partial fraction decomposition will be:
\( \frac{(x-1)^2}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1} \)
Multiply both sides by \( x(x^2+1) \) to clear the denominators:
\( (x-1)^2 = A(x^2+1) + (Bx+C)x \) (Let's call this Equation (1))
To find A, substitute \( x = 0 \) into Equation (1):
\( (0-1)^2 = A(0^2+1) + (B(0)+C)(0) \)
\( (-1)^2 = A(1) + 0 \)
\( 1 = A \)
\( \implies A = 1 \)
Next, to find B, compare the coefficients of \( x^2 \) on both sides of Equation (1). Expand the right side:
\( (x-1)^2 = Ax^2+A + Bx^2+Cx \)
\( x^2-2x+1 = (A+B)x^2 + Cx + A \)
The coefficient of \( x^2 \) on the left side is 1. On the right side, it is \( (A+B) \).
So, \( 1 = A + B \)
Substitute the value of A:
\( 1 = 1 + B \)
\( \implies B = 0 \)
Finally, to find C, compare the coefficients of \( x \) on both sides of Equation (1).
The coefficient of \( x \) on the left side is -2. On the right side, it is C.
So, \( -2 = C \)
\( \implies C = -2 \)
Alternatively, we could substitute \( x = 1 \) into Equation (1) to find C (though comparing coefficients is usually faster when many are already found):
\( (1-1)^2 = A(1^2+1) + (B(1)+C)(1) \)
\( 0 = A(2) + B + C \)
Substitute \( A=1 \) and \( B=0 \):
\( 0 = 1(2) + 0 + C \)
\( 0 = 2 + C \)
\( \implies C = -2 \)
Now, substitute the values of A, B, and C back into the partial fraction form:
\( \frac{(x-1)^2}{x^3+x} = \frac{1}{x} + \frac{0x-2}{x^2+1} \)
This simplifies to:
\( \frac{(x-1)^2}{x^3+x} = \frac{1}{x} - \frac{2}{x^2+1} \)
In simple words: We first found the factors of the bottom part of the fraction, which included a basic 'x' and a quadratic part that doesn't easily break down. Then, we used a specific setup for these kinds of factors and found the missing numbers by putting in some easy values for x and by matching the terms of x on both sides. This process helps us simplify the fraction into easier parts.

🎯 Exam Tip: Remember to simplify the numerator \( (x-1)^2 \) to \( x^2-2x+1 \) before comparing coefficients, to ensure accurate extraction of the coefficients of \( x^2 \), \( x \), and the constant term.

Question 7. Resolve \( \frac{x^2+x+1}{x^2-5x+6} \) into partial fractions.
Answer:
In this expression, the degree of the numerator (2) is equal to the degree of the denominator (2). When this happens, it's an improper rational function, and we must perform polynomial long division first.
First, factorize the denominator: \( x^2-5x+6 = (x-2)(x-3) \)
Now, perform the long division:
\[ \begin{array}{r} 1 \\ x^2-5x+6 \overline{) x^2+x+1} \\ \underline{-(x^2-5x+6)} \\ 6x-5 \end{array} \]
So, the original expression can be written as:
\( \frac{x^2+x+1}{x^2-5x+6} = 1 + \frac{6x-5}{x^2-5x+6} \) (Let's call this Equation (1))
Now, we need to resolve the proper fraction \( \frac{6x-5}{x^2-5x+6} \) into partial fractions. The denominator is \( (x-2)(x-3) \).
\( \frac{6x-5}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3} \)
Multiply both sides by \( (x-2)(x-3) \):
\( 6x-5 = A(x-3) + B(x-2) \) (Let's call this Equation (2))
To find B, substitute \( x = 3 \) into Equation (2):
\( 6(3)-5 = A(3-3) + B(3-2) \)
\( 18-5 = A(0) + B(1) \)
\( 13 = B \)
\( \implies B = 13 \)
To find A, substitute \( x = 2 \) into Equation (2):
\( 6(2)-5 = A(2-3) + B(2-2) \)
\( 12-5 = A(-1) + B(0) \)
\( 7 = -A \)
\( \implies A = -7 \)
Substitute the values of A and B back into the partial fraction form for the remainder term:
\( \frac{6x-5}{x^2-5x+6} = \frac{-7}{x-2} + \frac{13}{x-3} \)
Finally, combine this with the result of the long division (Equation (1)):
\( \frac{x^2+x+1}{x^2-5x+6} = 1 + \frac{-7}{x-2} + \frac{13}{x-3} \)
This can be written as:
\( \frac{x^2+x+1}{x^2-5x+6} = 1 - \frac{7}{x-2} + \frac{13}{x-3} \)
In simple words: First, we divided the top by the bottom because they had the same highest power of x. This gave us a whole number part and a new fraction. Then, we broke the new fraction into two simpler pieces using the factors of its bottom part. We found the missing numbers for these pieces by plugging in easy values for x. This is how we simplify fractions where the top is as complex as the bottom.

🎯 Exam Tip: Always check the degrees of the numerator and denominator. If the numerator's degree is equal to or greater than the denominator's, perform polynomial long division first to obtain a proper fraction.

Question 8. Resolve \( \frac{x^3+2x+1}{x^2+5x+6} \) into partial fractions.
Answer:
The degree of the numerator (3) is greater than the degree of the denominator (2). This is an improper rational function, so we must perform polynomial long division first.
First, factorize the denominator: \( x^2+5x+6 = (x+2)(x+3) \)
Now, perform the long division:
\[ \begin{array}{r} x-5 \\ x^2+5x+6 \overline{) x^3+0x^2+2x+1} \\ \underline{-(x^3+5x^2+6x)} \\ -5x^2-4x+1 \\ \underline{-(-5x^2-25x-30)} \\ 21x+31 \end{array} \]
So, the original expression can be written as:
\( \frac{x^3+2x+1}{x^2+5x+6} = x-5 + \frac{21x+31}{x^2+5x+6} \) (Let's call this Equation (1))
Now, we need to resolve the proper fraction \( \frac{21x+31}{x^2+5x+6} \) into partial fractions. The denominator is \( (x+2)(x+3) \).
\( \frac{21x+31}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3} \)
Multiply both sides by \( (x+2)(x+3) \):
\( 21x+31 = A(x+3) + B(x+2) \) (Let's call this Equation (2))
To find B, substitute \( x = -3 \) into Equation (2):
\( 21(-3)+31 = A(-3+3) + B(-3+2) \)
\( -63+31 = A(0) + B(-1) \)
\( -32 = -B \)
\( \implies B = 32 \)
To find A, substitute \( x = -2 \) into Equation (2):
\( 21(-2)+31 = A(-2+3) + B(-2+2) \)
\( -42+31 = A(1) + B(0) \)
\( -11 = A \)
\( \implies A = -11 \)
Substitute the values of A and B back into the partial fraction form for the remainder term:
\( \frac{21x+31}{x^2+5x+6} = \frac{-11}{x+2} + \frac{32}{x+3} \)
Finally, combine this with the result of the long division (Equation (1)):
\( \frac{x^3+2x+1}{x^2+5x+6} = x-5 + \frac{-11}{x+2} + \frac{32}{x+3} \)
This can be written as:
\( \frac{x^3+2x+1}{x^2+5x+6} = x-5 - \frac{11}{x+2} + \frac{32}{x+3} \)
In simple words: Because the top part of the fraction had a higher power of x than the bottom part, we first divided them, which gave us a simple polynomial and a new, simpler fraction. Then, we broke this new fraction into two even simpler pieces. We found the numbers for these pieces by plugging in values that would make parts of the equation zero. This is how we simplify fractions where the top is more complex than the bottom.

🎯 Exam Tip: When the numerator's degree is higher, always perform polynomial long division first. The polynomial part is as much a part of the final answer as the partial fractions of the remainder.

Question 9. Resolve \( \frac{x+12}{(x+1)^2(x-2)} \) into partial fractions.
Answer:
The denominator has a repeated linear factor \( (x+1)^2 \) and a distinct linear factor \( (x-2) \). For the repeated factor, we need a term for each power up to the highest. So, the partial fraction decomposition is:
\( \frac{x+12}{(x+1)^2(x-2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-2} \)
Multiply both sides by \( (x+1)^2(x-2) \) to clear the denominators:
\( x+12 = A(x+1)(x-2) + B(x-2) + C(x+1)^2 \) (Let's call this Equation (1))
To find C, substitute \( x = 2 \) into Equation (1):
\( 2+12 = A(2+1)(2-2) + B(2-2) + C(2+1)^2 \)
\( 14 = A(3)(0) + B(0) + C(3)^2 \)
\( 14 = 9C \)
\( \implies C = \frac{14}{9} \)
To find B, substitute \( x = -1 \) into Equation (1):
\( -1+12 = A(-1+1)(-1-2) + B(-1-2) + C(-1+1)^2 \)
\( 11 = A(0)(-3) + B(-3) + C(0)^2 \)
\( 11 = -3B \)
\( \implies B = -\frac{11}{3} \)
To find A, substitute \( x = 0 \) into Equation (1):
\( 0+12 = A(0+1)(0-2) + B(0-2) + C(0+1)^2 \)
\( 12 = A(1)(-2) + B(-2) + C(1)^2 \)
\( 12 = -2A - 2B + C \)
Substitute the known values of B and C:
\( 12 = -2A - 2\left(-\frac{11}{3}\right) + \frac{14}{9} \)
\( 12 = -2A + \frac{22}{3} + \frac{14}{9} \)
To clear the fractions, multiply the entire equation by 9:
\( 12 \times 9 = -2A \times 9 + \frac{22}{3} \times 9 + \frac{14}{9} \times 9 \)
\( 108 = -18A + 66 + 14 \)
\( 108 = -18A + 80 \)
\( 18A = 80 - 108 \)
\( 18A = -28 \)
\( A = -\frac{28}{18} \)
\( \implies A = -\frac{14}{9} \)
Substitute the values of A, B, and C back into the partial fraction form:
\( \frac{x+12}{(x+1)^2(x-2)} = \frac{-\frac{14}{9}}{x+1} + \frac{-\frac{11}{3}}{(x+1)^2} + \frac{\frac{14}{9}}{x-2} \)
This can be written as:
\( \frac{x+12}{(x+1)^2(x-2)} = -\frac{14}{9(x+1)} - \frac{11}{3(x+1)^2} + \frac{14}{9(x-2)} \)
In simple words: This fraction had a factor that appeared twice in the bottom part, along with another simple factor. We set it up with different terms for each power of the repeated factor and then found the numbers on top of each new fraction. We did this by plugging in special values for x that made parts of the equation disappear, and also by using general values and solving the equations.

🎯 Exam Tip: For repeated factors like \( (x+1)^2 \), ensure you include both \( \frac{A}{x+1} \) and \( \frac{B}{(x+1)^2} \) in your partial fraction setup to account for all possible terms.

Question 10. Resolve \( \frac{6x^2-x+1}{x^3+x^2+x+1} \) into partial fractions.
Answer:
First, factorize the denominator completely. We can group terms:
\( x^3+x^2+x+1 = x^2(x+1) + 1(x+1) = (x^2+1)(x+1) \)
This gives one irreducible quadratic factor \( (x^2+1) \) and one distinct linear factor \( (x+1) \). The partial fraction decomposition is:
\( \frac{6x^2-x+1}{(x^2+1)(x+1)} = \frac{Ax+B}{x^2+1} + \frac{C}{x+1} \)
Multiply both sides by \( (x^2+1)(x+1) \) to clear the denominators:
\( 6x^2-x+1 = (Ax+B)(x+1) + C(x^2+1) \) (Let's call this Equation (1))
To find C, substitute \( x = -1 \) into Equation (1):
\( 6(-1)^2 - (-1) + 1 = (A(-1)+B)(-1+1) + C((-1)^2+1) \)
\( 6(1) + 1 + 1 = (-A+B)(0) + C(1+1) \)
\( 8 = 2C \)
\( \implies C = 4 \)
To find B, substitute \( x = 0 \) into Equation (1):
\( 6(0)^2 - (0) + 1 = (A(0)+B)(0+1) + C(0^2+1) \)
\( 1 = B(1) + C(1) \)
\( 1 = B + C \)
Substitute the value of C:
\( 1 = B + 4 \)
\( \implies B = -3 \)
To find A, compare the coefficients of \( x^2 \) on both sides of Equation (1). Expand the terms that produce \( x^2 \):
From \( (Ax+B)(x+1) = Ax^2+Ax+Bx+B \), the \( x^2 \) term is \( Ax^2 \).
From \( C(x^2+1) \), the \( x^2 \) term is \( Cx^2 \).
The coefficient of \( x^2 \) on the left side is 6.
So, \( 6 = A + C \)
Substitute the value of C:
\( 6 = A + 4 \)
\( \implies A = 2 \)
Substitute the values of A, B, and C back into the partial fraction form:
\( \frac{6x^2-x+1}{x^3+x^2+x+1} = \frac{2x-3}{x^2+1} + \frac{4}{x+1} \)
In simple words: We first factored the bottom of the fraction by grouping terms. This gave us a simple linear factor and a quadratic factor that couldn't be broken down further. Then, we set up the fraction in a special way for these factors and found the unknown numbers by plugging in easy values for x and by matching the terms with x squared on both sides. This simplifies the fraction into easier-to-manage parts.

🎯 Exam Tip: Always fully factorize the denominator first, which might involve grouping terms. Remember the \( Ax+B \) form for irreducible quadratic factors.

Question 11. Resolve \( \frac{2x^2+5x-11}{x^2+2x-3} \) into partial fractions.
Answer:
The degree of the numerator (2) is equal to the degree of the denominator (2). This is an improper rational function, so we must perform polynomial long division first.
First, factorize the denominator: \( x^2+2x-3 = (x-1)(x+3) \)
Now, perform the long division:
\[ \begin{array}{r} 2 \\ x^2+2x-3 \overline{) 2x^2+5x-11} \\ \underline{-(2x^2+4x-6)} \\ x-5 \end{array} \]
So, the original expression can be written as:
\( \frac{2x^2+5x-11}{x^2+2x-3} = 2 + \frac{x-5}{x^2+2x-3} \) (Let's call this Equation (1))
Now, we need to resolve the proper fraction \( \frac{x-5}{x^2+2x-3} \) into partial fractions. The denominator is \( (x-1)(x+3) \).
\( \frac{x-5}{(x-1)(x+3)} = \frac{A}{x-1} + \frac{B}{x+3} \)
Multiply both sides by \( (x-1)(x+3) \):
\( x-5 = A(x+3) + B(x-1) \) (Let's call this Equation (2))
To find A, substitute \( x = 1 \) into Equation (2):
\( 1-5 = A(1+3) + B(1-1) \)
\( -4 = A(4) + B(0) \)
\( -4 = 4A \)
\( \implies A = -1 \)
To find B, substitute \( x = -3 \) into Equation (2):
\( -3-5 = A(-3+3) + B(-3-1) \)
\( -8 = A(0) + B(-4) \)
\( -8 = -4B \)
\( \implies B = 2 \)
Substitute the values of A and B back into the partial fraction form for the remainder term:
\( \frac{x-5}{x^2+2x-3} = \frac{-1}{x-1} + \frac{2}{x+3} \)
Finally, combine this with the result of the long division (Equation (1)):
\( \frac{2x^2+5x-11}{x^2+2x-3} = 2 + \frac{-1}{x-1} + \frac{2}{x+3} \)
This can be written as:
\( \frac{2x^2+5x-11}{x^2+2x-3} = 2 - \frac{1}{x-1} + \frac{2}{x+3} \)
In simple words: Since the highest power of x on top was the same as on the bottom, we first divided the fractions to get a whole number and a simpler fraction. Then, we broke this simpler fraction into two parts, using the factors of its bottom part. We found the numbers for these parts by plugging in specific values for x. This process helps us simplify fractions where the top part is as complicated as the bottom part.

🎯 Exam Tip: Always remember to check if the rational expression is improper. If it is, polynomial long division is a mandatory first step before applying partial fraction decomposition techniques.

Question 12. Resolve the following rational expression into partial fractions: \( \frac{7+x}{(1+x)\left(1+x^{2}\right)} \)
Answer:
To resolve the given rational expression into partial fractions, we set up the equation as follows:
\( \frac{7+x}{(1+x)\left(1+x^{2}\right)} = \frac{A}{1+x} + \frac{Bx+C}{1+x^{2}} \)
Now, we combine the fractions on the right side by finding a common denominator:
\( \frac{7+x}{(1+x)\left(1+x^{2}\right)} = \frac{A(1+x^{2}) + (Bx+C)(1+x)}{(1+x)(1+x^{2})} \)
Equating the numerators from both sides gives us the main equation:
\( 7+x = A(1+x^{2}) + (Bx+C)(1+x) \) ——— (1)
To find the values of A, B, and C, we substitute specific values for \( x \):
First, substitute \( x = -1 \) into equation (1) to find A:
\( 7 + (-1) = A(1 + (-1)^{2}) + (B(-1)+C)(1+(-1)) \)
\( 6 = A(1 + 1) + (-B+C)(0) \)
\( 6 = 2A \)
\( \implies A = 3 \)
Next, substitute \( x = 0 \) into equation (1) to find C:
\( 7 + 0 = A(1 + 0^{2}) + (B(0)+C)(1+0) \)
\( 7 = A(1) + (C)(1) \)
\( 7 = A + C \)
Since we found \( A = 3 \):
\( 7 = 3 + C \)
\( \implies C = 4 \)
Finally, to find B, we equate the coefficients of \( x^{2} \) in equation (1). Expanding the right side of equation (1):
\( 7+x = A + Ax^{2} + Bx + Bx^{2} + C + Cx \)
Gathering the terms with \( x^{2} \):
\( 7+x = (A+B)x^{2} + (B+C)x + (A+C) \)
Comparing the coefficient of \( x^{2} \) on both sides: the left side has \( 0x^{2} \), so:
\( 0 = A + B \)
Substitute \( A = 3 \):
\( 0 = 3 + B \)
\( \implies B = -3 \)
Therefore, the required partial fraction decomposition is:
\( \frac{7+x}{(1+x)(1+x^{2})} = \frac{3}{1+x} + \frac{-3x+4}{1+x^{2}} \)
In simple words: We split the complex fraction into simpler ones, each with an unknown constant. By putting in specific numbers for \( x \), we found the values for A and C. Then, by comparing the parts with \( x^{2} \), we figured out B. This method helps break down difficult fractions into easier parts for calculations.

🎯 Exam Tip: When dealing with a quadratic factor like \( (1+x^{2}) \), remember to use a linear numerator \( Bx+C \) in the partial fraction setup, not just a constant. Always test some values for x, like 0, 1, -1, or the roots of the factors, to quickly find constants.