Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.4

Question 1. Reduce each of the following rational expression to its lowest form.
(i) \( \frac{x^{2}-1}{x^{2}+x} \)
(ii) \( \frac{x^{2}-11 x+18}{x^{2}-4 x+4} \)
(iii) \( \frac{9 x^{2}+81 x}{x^{3}+8 x^{2}-9 x} \)
(iv) \( \frac{p^{2}-3 p-40}{2 p^{3}-24 p^{2}+64 p} \)
Answer:
(i) To reduce the expression \( \frac{x^{2}-1}{x^{2}+x} \):
First, we factor the numerator using the difference of squares formula, \( a^2 - b^2 = (a-b)(a+b) \):
\( x^{2}-1 = (x+1)(x-1) \)
Next, we factor the denominator by taking out the common factor \( x \):
\( x^{2}+x = x(x+1) \)
Now, we rewrite the expression with the factored forms:
\( \frac{x^{2}-1}{x^{2}+x} = \frac{(x+1)(x-1)}{x(x+1)} \)
We can cancel out the common factor \( (x+1) \) from both the numerator and the denominator, as long as \( x \neq -1 \).
So, the simplified form is \( \frac{x-1}{x} \).

(ii) To reduce the expression \( \frac{x^{2}-11 x+18}{x^{2}-4 x+4} \):
First, we factor the quadratic numerator \( x^{2}-11x+18 \). We look for two numbers that multiply to 18 and add up to -11. These numbers are -9 and -2.
\( x^{2}-11x+18 = (x-9)(x-2) \)
Next, we factor the quadratic denominator \( x^{2}-4x+4 \). We look for two numbers that multiply to 4 and add up to -4. These numbers are -2 and -2. This is also a perfect square trinomial, \( (a-b)^2 = a^2 - 2ab + b^2 \).
\( x^{2}-4x+4 = (x-2)(x-2) \)
Now, we write the expression using the factored forms:
\( \frac{x^{2}-11x+18}{x^{2}-4x+4} = \frac{(x-9)(x-2)}{(x-2)(x-2)} \)
We can cancel out the common factor \( (x-2) \) from both the numerator and the denominator, provided \( x \neq 2 \).
So, the simplified form is \( \frac{x-9}{x-2} \).

(iii) To reduce the expression \( \frac{9 x^{2}+81 x}{x^{3}+8 x^{2}-9 x} \):
First, we factor the numerator \( 9x^{2}+81x \). We take out the common factor \( 9x \):
\( 9x^{2}+81x = 9x(x+9) \)
Next, we factor the denominator \( x^{3}+8x^{2}-9x \). We start by taking out the common factor \( x \):
\( x^{3}+8x^{2}-9x = x(x^{2}+8x-9) \)
Now, we factor the quadratic expression inside the parenthesis, \( x^{2}+8x-9 \). We look for two numbers that multiply to -9 and add up to 8. These numbers are 9 and -1.
\( x^{2}+8x-9 = (x+9)(x-1) \)
So, the fully factored denominator is \( x(x+9)(x-1) \).
Now, we write the expression with the factored forms:
\( \frac{9x^{2}+81x}{x^{3}+8x^{2}-9x} = \frac{9x(x+9)}{x(x+9)(x-1)} \)
We can cancel out the common factors \( x \) and \( (x+9) \) from both the numerator and the denominator, provided \( x \neq 0 \) and \( x \neq -9 \).
So, the simplified form is \( \frac{9}{x-1} \).

(iv) To reduce the expression \( \frac{p^{2}-3 p-40}{2 p^{3}-24 p^{2}+64 p} \):
First, we factor the quadratic numerator \( p^{2}-3p-40 \). We look for two numbers that multiply to -40 and add up to -3. These numbers are -8 and 5.
\( p^{2}-3p-40 = (p-8)(p+5) \)
Next, we factor the denominator \( 2p^{3}-24p^{2}+64p \). We start by taking out the common factor \( 2p \):
\( 2p^{3}-24p^{2}+64p = 2p(p^{2}-12p+32) \)
Now, we factor the quadratic expression inside the parenthesis, \( p^{2}-12p+32 \). We look for two numbers that multiply to 32 and add up to -12. These numbers are -8 and -4.
\( p^{2}-12p+32 = (p-8)(p-4) \)
So, the fully factored denominator is \( 2p(p-8)(p-4) \).
Now, we write the expression with the factored forms:
\( \frac{p^{2}-3p-40}{2p^{3}-24p^{2}+64p} = \frac{(p-8)(p+5)}{2p(p-8)(p-4)} \)
We can cancel out the common factor \( (p-8) \) from both the numerator and the denominator, provided \( p \neq 8 \).
So, the simplified form is \( \frac{p+5}{2p(p-4)} \).
In simple words: To simplify these math problems, first find the parts that can be multiplied together to make the top and bottom expressions (this is called factoring). Then, look for any same parts on both the top and the bottom, and cross them out. This makes the fraction simpler.

🎯 Exam Tip: Always fully factor both the numerator and the denominator before canceling any terms. This helps prevent errors and ensures the expression is in its lowest form.

Question 2. Find the excluded values, if any of the following expressions.
(i) \( \frac{y}{y^{2}-25} \)
(ii) \( \frac{t}{t^{2}-5 t+6} \)
(iii) \( \frac{x^{2}+6x+8}{x^{2}+x-2} \)
(iv) \( \frac{x^{3}-27}{x^{3}+x^{2}-6 x} \)
Answer:
(i) For the expression \( \frac{y}{y^{2}-25} \), the excluded values are those that make the denominator equal to zero, because division by zero is not allowed.
So, we set the denominator to zero:
\( y^{2}-25 = 0 \)
We can write 25 as \( 5^2 \), so this is a difference of squares:
\( y^{2}-5^{2} = 0 \)
Factoring the difference of squares gives:
\( (y+5)(y-5) = 0 \)
This means either \( y+5=0 \) or \( y-5=0 \).
If \( y+5=0 \), then \( y=-5 \).
If \( y-5=0 \), then \( y=5 \).
Therefore, the excluded values for this expression are -5 and 5.

(ii) For the expression \( \frac{t}{t^{2}-5 t+6} \), we find the values that make the denominator zero.
Set the denominator to zero:
\( t^{2}-5t+6 = 0 \)
To factor this quadratic expression, we look for two numbers that multiply to 6 and add up to -5. These numbers are -3 and -2.
\( (t-3)(t-2) = 0 \)
This means either \( t-3=0 \) or \( t-2=0 \).
If \( t-3=0 \), then \( t=3 \).
If \( t-2=0 \), then \( t=2 \).
Therefore, the excluded values for this expression are 2 and 3.

(iii) For the expression \( \frac{x^{2}+6x+8}{x^{2}+x-2} \), we identify values that make the denominator zero.
Set the denominator to zero:
\( x^{2}+x-2 = 0 \)
To factor this quadratic expression, we look for two numbers that multiply to -2 and add up to 1. These numbers are 2 and -1.
\( (x+2)(x-1) = 0 \)
This means either \( x+2=0 \) or \( x-1=0 \).
If \( x+2=0 \), then \( x=-2 \).
If \( x-1=0 \), then \( x=1 \).
However, when simplifying the expression \( \frac{(x+4)(x+2)}{(x+2)(x-1)} \) to \( \frac{x+4}{x-1} \), the term \( (x+2) \) is cancelled. The standard approach for identifying excluded values generally considers the original denominator. Thus, the actual excluded values are 1 and -2. But if we consider the simplified rational expression, the denominator is \( (x-1) \). So we set this to zero:
\( x-1=0 \)
\( x=1 \)
Therefore, the excluded value based on the simplified form is 1.

(iv) For the expression \( \frac{x^{3}-27}{x^{3}+x^{2}-6 x} \), we find the values that make the denominator zero.
Set the denominator to zero:
\( x^{3}+x^{2}-6x = 0 \)
First, take out the common factor \( x \):
\( x(x^{2}+x-6) = 0 \)
Next, factor the quadratic expression inside the parenthesis, \( x^{2}+x-6 \). We look for two numbers that multiply to -6 and add up to 1. These numbers are 3 and -2.
\( x(x+3)(x-2) = 0 \)
This means one of the factors must be zero:
\( x=0 \)
or \( x+3=0 \implies x=-3 \)
or \( x-2=0 \implies x=2 \)
Therefore, the excluded values for this expression are 0, -3, and 2.
In simple words: Excluded values are the numbers that would make the bottom part of a fraction equal to zero. When the bottom is zero, the fraction doesn't make sense. To find these numbers, set the bottom part of the fraction to zero and solve for the variable. These solutions are the excluded values.

🎯 Exam Tip: Remember to factor the denominator completely to find all possible values that would make it zero. These are your excluded values.