OP Malhotra Class 9 Maths Solutions Chapter 4 Factorisation Chapter Test

Factorise :

Question 1. \( 8x^2y^3 - x^5 \)
Answer: We start by finding the common factor in the expression \( 8x^2y^3 - x^5 \).
We see that \( x^2 \) is common to both terms.
\( 8x^2y^3 - x^5 = x^2(8y^3 - x^3) \)
Now, we recognize that \( 8y^3 \) can be written as \( (2y)^3 \). So, the expression inside the bracket is a difference of cubes, which follows the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \).
Here, \( a = 2y \) and \( b = x \).
\( x^2[(2y)^3 - (x)^3] \)
\( = x^2(2y - x) ((2y)^2 + (2y)(x) + x^2) \)
\( = x^2(2y - x) (4y^2 + 2xy + x^2) \)
In simple words: First, take out the common part, which is \( x^2 \). Then, use the special rule for subtracting cubes to break down the rest of the expression. This gives you the fully factored form.

🎯 Exam Tip: Always look for common factors first before applying other factorization identities. Knowing the \( a^3 - b^3 \) identity is crucial for solving such problems quickly.

Question 2. \( x^2 + \frac { 1 }{ x^2 } + 2 - 2x - \frac { 2 }{ x } \)
Answer: Let's rearrange the terms to group similar parts together for easier factorization.
The given expression is \( x^2 + \frac{1}{x^2} + 2 - 2x - \frac{2}{x} \)
We notice that \( x^2 + \frac{1}{x^2} + 2 \) is a perfect square, specifically \( (x + \frac{1}{x})^2 \).
And \( -2x - \frac{2}{x} \) can be factored as \( -2(x + \frac{1}{x}) \).
So, we can rewrite the expression as:
\( = \left(x+\frac{1}{x}\right)^2 - 2\left(x+\frac{1}{x}\right) \)
Now, let \( A = (x + \frac{1}{x}) \). The expression becomes \( A^2 - 2A \).
We can factor out \( A \) from this: \( A(A - 2) \).
Substitute \( (x + \frac{1}{x}) \) back for \( A \):
\( = \left(x+\frac{1}{x}\right)\left(x+\frac{1}{x}-2\right) \)
In simple words: Group the terms to find a perfect square and a common factor. Then, take out the common term \( (x + \frac{1}{x}) \) to simplify the expression further. This helps in factoring the whole problem.

🎯 Exam Tip: Recognizing identities like \( (a+b)^2 = a^2+2ab+b^2 \) and common factors is key. Sometimes, rearranging terms makes the pattern clearer.

Question 3. \( 2x^2 - x - 6 \)
Answer: To factor the quadratic expression \( 2x^2 - x - 6 \), we use the splitting the middle term method. We need to find two numbers that multiply to \( (2 \times -6) = -12 \) and add up to the middle term coefficient, which is \( -1 \).
The two numbers are \( -4 \) and \( 3 \), because \( -4 \times 3 = -12 \) and \( -4 + 3 = -1 \).
So, we rewrite the middle term \( -x \) as \( -4x + 3x \).
\( 2x^2 - x - 6 \)
\( = 2x^2 - 4x + 3x - 6 \)
Now, we group the terms and factor out common factors from each group:
\( = 2x(x - 2) + 3(x - 2) \)
We see that \( (x - 2) \) is a common factor in both terms. So, we factor it out:
\( = (x - 2)(2x + 3) \)
In simple words: To factor this, find two numbers that multiply to \( -12 \) and add up to \( -1 \). Then, split the middle term using these numbers, group the terms, and factor out what's common to get the final answer.

🎯 Exam Tip: When splitting the middle term for a quadratic \( ax^2 + bx + c \), always look for two numbers that multiply to \( ac \) and add to \( b \).

Question 4. \( a^3 - 0.216 \)
Answer: We need to factor the expression \( a^3 - 0.216 \).
We recognize that this is a difference of cubes because \( a^3 \) is a cube and \( 0.216 \) can be written as \( (0.6)^3 \) since \( 0.6 \times 0.6 \times 0.6 = 0.216 \).
So, the expression is in the form \( a^3 - b^3 \), where \( b = 0.6 \).
The formula for difference of cubes is \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \).
Applying this formula:
\( a^3 - 0.216 = (a)^3 - (0.6)^3 \)
\( = (a - 0.6)(a^2 + a(0.6) + (0.6)^2) \)
\( = (a - 0.6)(a^2 + 0.6a + 0.36) \)
In simple words: This problem asks us to factor a "cube minus a cube". We first find out what number, when multiplied by itself three times, gives 0.216 (which is 0.6). Then we use the special formula for \( a^3 - b^3 \) to get the answer.

🎯 Exam Tip: Memorize the difference of cubes formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \) and be able to identify perfect cubes, including decimals.

Question 5. \( 6x^2y - xy - 2y \)
Answer: First, we look for common factors in all terms of the expression \( 6x^2y - xy - 2y \).
We can see that \( y \) is common to all three terms.
Factoring out \( y \):
\( = y[6x^2 - x - 2] \)
Now, we need to factor the quadratic expression inside the bracket: \( 6x^2 - x - 2 \). We use the splitting the middle term method. We need two numbers that multiply to \( (6 \times -2) = -12 \) and add up to \( -1 \).
The two numbers are \( -4 \) and \( 3 \), because \( -4 \times 3 = -12 \) and \( -4 + 3 = -1 \).
So, we rewrite the middle term \( -x \) as \( -4x + 3x \).
\( = y[6x^2 - 4x + 3x - 2] \)
Next, group the terms inside the bracket and factor out common factors from each group:
\( = y[2x(3x - 2) + 1(3x - 2)] \)
Now, \( (3x - 2) \) is a common factor inside the bracket:
\( = y(3x - 2)(2x + 1) \)
In simple words: First, pull out the common letter \( y \) from all parts. Then, focus on the remaining part that looks like \( 6x^2 - x - 2 \). Find two numbers that multiply to \( -12 \) and add to \( -1 \), then split the middle term and factor by grouping to get the final answer.

🎯 Exam Tip: Always factor out any common monomial (like 'y' in this case) from all terms before attempting other factorization methods on the remaining polynomial.

Question 6. \( (x^2 - 3x)^2 - 8(x^2 - 3x) - 20 \)
Answer: To factor the expression \( (x^2 - 3x)^2 - 8(x^2 - 3x) - 20 \), we can use a substitution to simplify it into a standard quadratic form.
Let \( y = x^2 - 3x \).
Substituting \( y \) into the expression, we get:
\( y^2 - 8y - 20 \)
Now we need to factor this quadratic expression. We look for two numbers that multiply to \( -20 \) and add up to \( -8 \).
These numbers are \( -10 \) and \( 2 \), because \( -10 \times 2 = -20 \) and \( -10 + 2 = -8 \).
So, we split the middle term \( -8y \) into \( -10y + 2y \):
\( = y^2 - 10y + 2y - 20 \)
Group the terms and factor out common factors:
\( = y(y - 10) + 2(y - 10) \)
Factor out the common binomial \( (y - 10) \):
\( = (y - 10)(y + 2) \)
Now, substitute back \( y = x^2 - 3x \):
\( = (x^2 - 3x - 10)(x^2 - 3x + 2) \)
We need to factor each of these quadratic expressions further if possible.
For \( (x^2 - 3x - 10) \): We look for two numbers that multiply to \( -10 \) and add to \( -3 \). These are \( -5 \) and \( 2 \).
So, \( x^2 - 3x - 10 = (x - 5)(x + 2) \).
For \( (x^2 - 3x + 2) \): We look for two numbers that multiply to \( 2 \) and add to \( -3 \). These are \( -1 \) and \( -2 \).
So, \( x^2 - 3x + 2 = (x - 1)(x - 2) \).
Combining these, the fully factored expression is:
\( = (x - 5)(x + 2)(x - 1)(x - 2) \)
In simple words: First, replace the repeating part \( (x^2 - 3x) \) with a simpler letter like \( y \). This turns the problem into a regular quadratic. Factor that quadratic, then put \( (x^2 - 3x) \) back in place of \( y \). Finally, factor those two new quadratic expressions to get the full answer.

🎯 Exam Tip: When you see a repeated expression like \( (x^2 - 3x) \), use substitution to simplify the problem into a standard quadratic form. Remember to substitute back and continue factoring the resulting expressions.

Multiple Choice Questions

Question 7. One of the factors of \( (x - 1) - (x^2 - 1) \) is
(a) x + 4
(b) x + 1
(c) x - 1
(d) x + 4
Answer: (c) x - 1
We are given the expression \( (x - 1) - (x^2 - 1) \).
First, we recognize that \( (x^2 - 1) \) is a difference of squares, which factors as \( (x - 1)(x + 1) \).
So, substitute this back into the expression:
\( (x - 1) - (x + 1)(x - 1) \)
Now, we can see that \( (x - 1) \) is a common factor in both terms.
Factor out \( (x - 1) \):
\( = (x - 1)[1 - (x + 1)] \)
\( = (x - 1)[1 - x - 1] \)
\( = (x - 1)[-x] \)
\( = -x(x - 1) \)
So, \( (x - 1) \) is one of its factors.
In simple words: The problem asks for one part that the expression can be divided by. First, break down \( (x^2 - 1) \) into \( (x - 1)(x + 1) \). Then, notice that \( (x - 1) \) is in both big parts of the expression, so you can pull it out as a common factor.

🎯 Exam Tip: Always look for common factors first. Also, recognize standard algebraic identities like the difference of squares \( (a^2 - b^2) = (a - b)(a + b) \) as they simplify expressions greatly.

Question 8. If \( \frac{x}{y}+\frac{y}{x} = -1 \) (x, y ≠ 0), then the value of \( x^3 - y^3 \) is
(a) 1
(b) -1
(c) \( \frac { 1 }{ 2 } \)
(d) 0
Answer: (d) 0
We are given the condition \( \frac{x}{y}+\frac{y}{x} = -1 \).
To simplify this, find a common denominator, which is \( xy \).
\( \frac{x^2+y^2}{xy} = -1 \)
Multiply both sides by \( xy \):
\( x^2 + y^2 = -xy \)
Bring \( -xy \) to the left side:
\( x^2 + y^2 + xy = 0 \)
Now, we need to find the value of \( x^3 - y^3 \).
The formula for the difference of cubes is \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \).
From our initial simplification, we found that \( x^2 + xy + y^2 = 0 \).
Substitute this value into the difference of cubes formula:
\( x^3 - y^3 = (x - y)(0) \)
Any number multiplied by zero is zero.
\( = 0 \)
In simple words: We are given an equation that can be simplified to \( x^2 + xy + y^2 = 0 \). We know that \( x^3 - y^3 \) can be written as \( (x - y) \) multiplied by \( (x^2 + xy + y^2) \). Since this second part is zero, the whole expression becomes zero.

🎯 Exam Tip: Always simplify the given conditions first. Recognizing that \( x^2 + xy + y^2 \) is a factor of \( x^3 - y^3 \) is the key to solving this type of problem efficiently.

Question 9. The product of \( \left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right) \)
(a) \( x^4 + \frac{1}{x^4} \)
(b) \( x^3 + \frac { 1 }{ x^3 } - 2 \)
(c) \( x^4 - \frac{1}{x^4} \)
(d) \( x^2 + \frac { 1 }{ x^2 } + 2 \)
Answer: (c) \( x^4 - \frac{1}{x^4} \)
We need to find the product of \( \left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right) \).
First, consider the first two terms: \( \left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right) \).
This is in the form of the difference of squares identity: \( (a - b)(a + b) = a^2 - b^2 \).
Here, \( a = x \) and \( b = \frac{1}{x} \).
So, \( \left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right) = x^2 - \left(\frac{1}{x}\right)^2 = x^2 - \frac{1}{x^2} \).
Now, substitute this result back into the original product:
\( = \left(x^2-\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}\right) \)
Again, this is in the form of the difference of squares: \( (A - B)(A + B) = A^2 - B^2 \).
Here, \( A = x^2 \) and \( B = \frac{1}{x^2} \).
So, \( \left(x^2-\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}\right) = (x^2)^2 - \left(\frac{1}{x^2}\right)^2 \)
\( = x^{2 \times 2} - \frac{1}{x^{2 \times 2}} \)
\( = x^4 - \frac{1}{x^4} \)
In simple words: This problem uses the "difference of squares" rule two times. First, combine \( (x - \frac{1}{x}) \) and \( (x + \frac{1}{x}) \) to get \( (x^2 - \frac{1}{x^2}) \). Then, combine this new term with \( (x^2 + \frac{1}{x^2}) \) using the same rule to reach the final answer.

🎯 Exam Tip: The difference of squares identity \( (a-b)(a+b)=a^2-b^2 \) is very powerful and often applied multiple times in factorization problems. Always look for this pattern.

Question 10. If x – 2y= 11 and xy = 8, then the value of \( x^3 – 8y^3 \) is
(a) 1860
(b) 1600
(c) 1859
(d) 2000
Answer: (c) 1859
We are given two equations:
1) \( x - 2y = 11 \)
2) \( xy = 8 \)
We need to find the value of \( x^3 - 8y^3 \).
Notice that \( 8y^3 \) can be written as \( (2y)^3 \). So, the expression we need to find is \( x^3 - (2y)^3 \).
This is in the form of \( a^3 - b^3 \), where \( a = x \) and \( b = 2y \).
We know the identity \( (a - b)^3 = a^3 - b^3 - 3ab(a - b) \).
Rearranging this to solve for \( a^3 - b^3 \):
\( a^3 - b^3 = (a - b)^3 + 3ab(a - b) \)
Substitute \( a = x \) and \( b = 2y \):
\( x^3 - (2y)^3 = (x - 2y)^3 + 3(x)(2y)(x - 2y) \)
Now, substitute the given values \( x - 2y = 11 \) and \( xy = 8 \):
\( = (11)^3 + 3(2xy)(11) \)
\( = 1331 + 6(xy)(11) \)
\( = 1331 + 6(8)(11) \)
\( = 1331 + 48 \times 11 \)
\( = 1331 + 528 \)
\( = 1859 \)
Therefore, the value of \( x^3 - 8y^3 \) is 1859.
In simple words: We want to find the value of \( x^3 - (2y)^3 \). We use the special formula for \( a^3 - b^3 \), which can be found by expanding \( (a - b)^3 \). We then put in the given values for \( (x - 2y) \) and \( xy \) to calculate the final answer.

🎯 Exam Tip: Recognize that \( 8y^3 \) is \( (2y)^3 \), allowing you to use the difference of cubes identity or related cubic expansion formulas. Always be careful with the coefficients when substituting.