OP Malhotra Class 9 Maths Solutions Chapter 4 Factorisation Exercise 4 (G)

Factorise:

Question 1. \( 2x^2 + 3x + 1 \)
Answer:
We need to factorise \( 2x^2 + 3x + 1 \).
First, find two numbers whose product is \( 2 \times 1 = 2 \) and whose sum is \( 3 \). These numbers are \( 2 \) and \( 1 \).
\( \implies 2x^2 + 2x + x + 1 \)
\( \implies 2x (x + 1) + 1 (x + 1) \)
\( \implies (x + 1) (2x + 1) \)
In simple words: We split the middle term \( 3x \) into \( 2x + x \) because \( 2 \times 1 = 2 \) (product of first and last term coefficients) and \( 2 + 1 = 3 \) (middle term coefficient). Then, we group terms and factor out common parts to get the final answer.

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

Question 2. \( 3x^2 + 17x + 10 \)
Answer:
We need to factorise \( 3x^2 + 17x + 10 \).
Find two numbers whose product is \( 3 \times 10 = 30 \) and whose sum is \( 17 \). These numbers are \( 15 \) and \( 2 \).
\( \implies 3x^2 + 15x + 2x + 10 \)
\( \implies 3x (x + 5) + 2 (x + 5) \)
\( \implies (x + 5) (3x + 2) \)
In simple words: To factor this, we find two numbers that multiply to \( 30 \) and add up to \( 17 \). We use these numbers, \( 15 \) and \( 2 \), to split the middle term \( 17x \) into \( 15x + 2x \), then factor by grouping.

🎯 Exam Tip: Remember to always check for a common factor in all terms first before splitting the middle term. If there is one, factor it out.

Question 3. \( 5x^2 + 9x - 2 \)
Answer:
We need to factorise \( 5x^2 + 9x - 2 \).
Find two numbers whose product is \( 5 \times (-2) = -10 \) and whose sum is \( 9 \). These numbers are \( 10 \) and \( -1 \).
\( \implies 5x^2 + 10x - x - 2 \)
\( \implies 5x (x + 2) - 1 (x + 2) \)
\( \implies (x + 2) (5x - 1) \)
In simple words: For factorisation, we look for two numbers that multiply to \( -10 \) and add to \( 9 \). We found \( 10 \) and \( -1 \), so we split \( 9x \) into \( 10x - x \). After grouping, we get the factored form.

🎯 Exam Tip: When the product \( ac \) is negative, the two numbers you find must have opposite signs. The one with the larger absolute value will match the sign of \( b \).

Question 4. \( 6x^2 - 7x - 5 \)
Answer:
We need to factorise \( 6x^2 - 7x - 5 \).
Find two numbers whose product is \( 6 \times (-5) = -30 \) and whose sum is \( -7 \). These numbers are \( -10 \) and \( 3 \).
\( \implies 6x^2 - 10x + 3x - 5 \)
\( \implies 2x (3x - 5) + 1 (3x - 5) \)
\( \implies (3x - 5) (2x + 1) \)
In simple words: We need to split the middle term \( -7x \). We found two numbers, \( -10 \) and \( 3 \), that multiply to \( -30 \) and add to \( -7 \). Then, we group the terms and factor out what is common.

🎯 Exam Tip: Be careful with signs when grouping terms. If you factor out a negative number, the signs inside the parenthesis will change.

Question 5. \( 6y^2 - 17y + 12 \)
Answer:
We need to factorise \( 6y^2 - 17y + 12 \).
Find two numbers whose product is \( 6 \times 12 = 72 \) and whose sum is \( -17 \). These numbers are \( -9 \) and \( -8 \).
\( \implies 6y^2 - 9y - 8y + 12 \)
\( \implies 3y (2y - 3) - 4 (2y - 3) \)
\( \implies (2y - 3) (3y - 4) \)
In simple words: To factor this expression, we split \( -17y \) into \( -9y - 8y \) because \( (-9) \times (-8) \) is \( 72 \) and \( (-9) + (-8) \) is \( -17 \). Then, we group and pull out common factors to get the final answer.

🎯 Exam Tip: When both the product \( ac \) and the sum \( b \) are positive, the two numbers used for splitting will both be positive. If \( b \) is negative and \( ac \) is positive, both numbers will be negative.

Question 6. \( 8y^2 - 2y - 1 \)
Answer:
We need to factorise \( 8y^2 - 2y - 1 \).
Find two numbers whose product is \( 8 \times (-1) = -8 \) and whose sum is \( -2 \). These numbers are \( -4 \) and \( 2 \).
\( \implies 8y^2 - 4y + 2y - 1 \)
\( \implies 4y (2y - 1) + 1 (2y - 1) \)
\( \implies (2y - 1) (4y + 1) \)
In simple words: We need to split the middle term \( -2y \). We found \( -4 \) and \( 2 \) are the numbers that multiply to \( -8 \) and add to \( -2 \). Then we group the terms and factor out the common factors.

🎯 Exam Tip: Always double-check your factoring by multiplying the two binomials back together to ensure you get the original expression. This helps catch any sign errors.

Question 7. \( 18bx^2 + 18bx - 20b \)
Answer:
We need to factorise \( 18bx^2 + 18bx - 20b \).
First, factor out the common term \( 2b \).
\( \implies 2b (9x^2 + 9x - 10) \)
Now, factor the quadratic expression inside the parenthesis. Find two numbers whose product is \( 9 \times (-10) = -90 \) and whose sum is \( 9 \). These numbers are \( 15 \) and \( -6 \).
\( \implies 2b [9x^2 + 15x - 6x - 10] \)
\( \implies 2b [3x (3x + 5) - 2 (3x + 5)] \)
\( \implies 2b (3x + 5) (3x - 2) \)
In simple words: First, take out the common factor \( 2b \) from all terms. Then, focus on the remaining quadratic expression. Split its middle term by finding two numbers that multiply to \( -90 \) and add to \( 9 \), which are \( 15 \) and \( -6 \). Finally, group and factor to get the complete answer.

🎯 Exam Tip: Always look for a Greatest Common Factor (GCF) at the very beginning of any factorisation problem. Factoring out the GCF simplifies the remaining expression and makes further steps easier.

Question 8. \( 14x^2 - 60xy + 16y^2 \)
Answer:
We need to factorise \( 14x^2 - 60xy + 16y^2 \).
First, factor out the common term \( 2 \).
\( \implies 2 [7x^2 - 30xy + 8y^2] \)
Now, factor the quadratic expression inside the brackets. Find two numbers whose product is \( 7 \times 8 = 56 \) and whose sum is \( -30 \). These numbers are \( -28 \) and \( -2 \).
\( \implies 2 [7x^2 - 28xy - 2xy + 8y^2] \)
\( \implies 2 [7x (x - 4y) - 2y (x - 4y)] \)
\( \implies 2 (x - 4y) (7x - 2y) \)
In simple words: First, pull out the common factor \( 2 \). Then, inside the bracket, find two numbers that multiply to \( 56 \) and add to \( -30 \) to split the middle term \( -30xy \). The numbers are \( -28 \) and \( -2 \). Finally, group the terms and factor them out.

🎯 Exam Tip: When factorising expressions with two variables, the splitting numbers will still be based on the product of the first and last coefficients and the middle coefficient, but the terms will include both variables (e.g., \( xy \)).

Question 9. \( 30x^2 + 103xy - 7y^2 \)
Answer:
We need to factorise \( 30x^2 + 103xy - 7y^2 \).
Find two numbers whose product is \( 30 \times (-7) = -210 \) and whose sum is \( 103 \). These numbers are \( 105 \) and \( -2 \).
\( \implies 30x^2 + 105xy - 2xy - 7y^2 \)
\( \implies 15x (2x + 7y) - y (2x + 7y) \)
\( \implies (2x + 7y) (15x - y) \)
In simple words: To factor this, we need to split the middle term \( 103xy \). We find two numbers, \( 105 \) and \( -2 \), that multiply to \( -210 \) and add to \( 103 \). We use these to split the middle term, then group the terms and factor out common parts.

🎯 Exam Tip: If the numbers for splitting are large, break down the product \( ac \) into its prime factors to help find possible pairs more easily.

Question 10. \( 12x^2 - 29xy + 14y^2 \)
Answer:
We need to factorise \( 12x^2 - 29xy + 14y^2 \).
Find two numbers whose product is \( 12 \times 14 = 168 \) and whose sum is \( -29 \). These numbers are \( -8 \) and \( -21 \).
\( \implies 12x^2 - 8xy - 21xy + 14y^2 \)
\( \implies 4x (3x - 2y) - 7y (3x - 2y) \)
\( \implies (3x - 2y) (4x - 7y) \)
In simple words: We split the middle term \( -29xy \) into \( -8xy - 21xy \) because \( (-8) \times (-21) = 168 \) and \( (-8) + (-21) = -29 \). Then, we group the terms and factor out the common parts to get the final factored form.

🎯 Exam Tip: Remember to include the variable \( y \) when splitting the middle term \( -29xy \) into \( -8xy \) and \( -21xy \) to maintain the correct expression structure.

Question 11. \( 15 + p (7 - 2p) \)
Answer:
We need to factorise \( 15 + p (7 - 2p) \).
First, expand the expression:
\( \implies 15 + 7p - 2p^2 \)
Rearrange into standard quadratic form \( ap^2 + bp + c \):
\( \implies -2p^2 + 7p + 15 \)
Now, find two numbers whose product is \( (-2) \times 15 = -30 \) and whose sum is \( 7 \). These numbers are \( 10 \) and \( -3 \).
\( \implies 15 + 10p - 3p - 2p^2 \)
\( \implies 5 (3 + 2p) - p (3 + 2p) \)
\( \implies (3 + 2p) (5 - p) \)
In simple words: First, expand the expression and rearrange it. Then, find two numbers that multiply to \( -30 \) and add to \( 7 \). These numbers are \( 10 \) and \( -3 \). We use them to split the middle term, then factor by grouping.

🎯 Exam Tip: Always rearrange quadratic expressions into standard form \( ax^2 + bx + c \) before attempting to factor by splitting the middle term. This makes it easier to identify \( a, b, \) and \( c \).

Question 12. \( 2 - y(7 - 5y) \)
Answer:
We need to factorise \( 2 - y(7 - 5y) \).
First, expand the expression:
\( \implies 2 - 7y + 5y^2 \)
Rearrange into standard quadratic form \( ay^2 + by + c \):
\( \implies 5y^2 - 7y + 2 \)
Find two numbers whose product is \( 5 \times 2 = 10 \) and whose sum is \( -7 \). These numbers are \( -5 \) and \( -2 \).
\( \implies 5y^2 - 5y - 2y + 2 \)
\( \implies 5y (y - 1) - 2 (y - 1) \)
\( \implies (y - 1) (5y - 2) \)
In simple words: First, expand the given expression to get a standard quadratic equation. Then, find two numbers that multiply to \( 10 \) and add up to \( -7 \), which are \( -5 \) and \( -2 \). Use these to split the middle term and factor by grouping.

🎯 Exam Tip: Be careful with the signs when expanding and combining terms. A negative sign in front of a parenthesis can change all the signs inside.

Question 13. \( x(2x + 5) - 3 \)
Answer:
We need to factorise \( x(2x + 5) - 3 \).
First, expand the expression:
\( \implies 2x^2 + 5x - 3 \)
Find two numbers whose product is \( 2 \times (-3) = -6 \) and whose sum is \( 5 \). These numbers are \( 6 \) and \( -1 \).
\( \implies 2x^2 + 6x - x - 3 \)
\( \implies 2x (x + 3) - 1 (x + 3) \)
\( \implies (x + 3) (2x - 1) \)
In simple words: Expand the expression to turn it into a quadratic form. Then, find two numbers that multiply to \( -6 \) and add to \( 5 \), which are \( 6 \) and \( -1 \). Use these to split the middle term \( 5x \), and then factor by grouping.

🎯 Exam Tip: When factoring by grouping, ensure that the expressions inside the two parentheses are identical. If they are not, recheck your splitting and factoring steps.

Question 14. \( 1 - 18y - 63y^2 \)
Answer:
We need to factorise \( 1 - 18y - 63y^2 \).
Find two numbers whose product is \( 1 \times (-63) = -63 \) and whose sum is \( -18 \). These numbers are \( -21 \) and \( 3 \).
\( \implies 1 - 21y + 3y - 63y^2 \)
\( \implies 1 (1 - 21y) + 3y (1 - 21y) \)
\( \implies (1 - 21y) (1 + 3y) \)
In simple words: We split the middle term \( -18y \) into \( -21y + 3y \) because \( (-21) \times 3 = -63 \) (product of first and last term coefficients) and \( (-21) + 3 = -18 \) (middle term coefficient). Then, we group the terms and factor out the common factors.

🎯 Exam Tip: When the leading coefficient (of \( y^2 \)) is negative, you can either factor out \( -1 \) first or proceed directly, making sure to handle the negative product \( ac \) correctly.

Question 15. \( 8a^3b - 10a^2b^2 - 12ab^3 \)
Answer:
We need to factorise \( 8a^3b - 10a^2b^2 - 12ab^3 \).
First, find the greatest common factor (GCF) for all terms. The GCF is \( 2ab \).
\( \implies 2ab (4a^2 - 5ab - 6b^2) \)
Now, factor the quadratic expression inside the parenthesis. Find two numbers whose product is \( 4 \times (-6) = -24 \) and whose sum is \( -5 \). These numbers are \( -8 \) and \( 3 \).
\( \implies 2ab [4a^2 - 8ab + 3ab - 6b^2] \)
\( \implies 2ab [4a (a - 2b) + 3b (a - 2b)] \)
\( \implies 2ab (a - 2b) (4a + 3b) \)
In simple words: First, take out the common factor \( 2ab \) from all parts of the expression. Then, focus on the remaining quadratic part. Find two numbers that multiply to \( -24 \) and add to \( -5 \). These numbers are \( -8 \) and \( 3 \). Use them to split the middle term, then group and factor.

🎯 Exam Tip: Always remember to look for the GCF first in any factorisation problem, especially when there are multiple variables and higher powers. This simplifies the expression greatly.

Question 16. \( 2 (a + b)^2 - 5 (a + b) - 3 \)
Answer:
We need to factorise \( 2 (a + b)^2 - 5 (a + b) - 3 \).
Let \( x = (a + b) \). Substitute \( x \) into the expression:
\( \implies 2x^2 - 5x - 3 \)
Now, factor this quadratic expression. Find two numbers whose product is \( 2 \times (-3) = -6 \) and whose sum is \( -5 \). These numbers are \( -6 \) and \( 1 \).
\( \implies 2x^2 - 6x + x - 3 \)
\( \implies 2x (x - 3) + 1 (x - 3) \)
\( \implies (x - 3) (2x + 1) \)
Finally, substitute back \( x = (a + b) \):
\( \implies (a + b - 3) (2(a + b) + 1) \)
\( \implies (a + b - 3) (2a + 2b + 1) \)
In simple words: We can make this expression simpler by replacing \( (a + b) \) with a single variable, like \( x \). This turns it into a standard quadratic equation. Factor this simpler equation by splitting the middle term. Once factored, put \( (a + b) \) back in place of \( x \).

🎯 Exam Tip: For expressions with repeated complex terms like \( (a+b) \), using substitution (e.g., let \( x = a+b \)) can greatly simplify the factorisation process. Remember to substitute back the original term in the final step.