GSEB Class 8 Maths Solutions Chapter 14 Factorization Exercise 14.3

Question 1. Carry out the following divisions.
(i) \( 28x^4 \div 56x \)
(ii) \( -36y^3 \div 9y^2 \)
(iii) \( 66pq^2r^3 \div 11qr^2 \)
(iv) \( 34x^3y^3z^3 \div 51xy^2z^3 \)
(v) \( 12a^8b^8 \div (-6a^6b^4) \)
Answer:
(i) We start with \( 28x^4 \div 56x \). We can rewrite this as a fraction \( \frac{28 x^{4}}{56 x} \). When we factorize the numerator and denominator, we get \( \frac{2 \times 2 \times 7 \times x \times x \times x \times x}{2 \times 2 \times 2 \times 7 \times x} \). After cancelling out common terms, this simplifies to \( \frac{1}{2} x^{4-1} \), which gives us \( \frac{1}{2}x^3 \). Thus, \( 28x^4 \div 56x = \frac{1}{2}x^3 \).
In simple words: First, we write the division as a fraction. Then, we break down the numbers and variables into their smaller parts. We cross out anything that is the same on the top and bottom. What's left is our simplified answer.
(ii) We start with \( -36y^3 \div 9y^2 \). First, we find the prime factors for both terms: \( -36y^3 = (-1) \times 2 \times 2 \times 3 \times 3 \times y \times y \times y \) and \( 9y^2 = 3 \times 3 \times y \times y \). We can then express the division as \( \frac{-36y^3}{9y^2} \). By substituting the factors, we get \( \frac{(-1) \times 2 \times 2 \times 3 \times 3 \times y \times y \times y}{3 \times 3 \times y \times y} \). After cancelling out the common factors, the expression simplifies to \( \frac{(-1) \times 2 \times 2 \times y}{1} \), which gives us \( -4y \).
In simple words: We break down each part of the problem into its prime factors. Then, we write it as a fraction and cross out matching parts from the top and bottom. The remaining parts form our final answer.
(iii) We begin with \( 66pq^2r^3 \div 11qr^2 \). We first determine the prime factorization of each term: \( 66pq^2r^3 = 2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r \) and \( 11qr^2 = 11 \times q \times r \times r \). We can write this division as a fraction: \( \frac{66pq^2r^3}{11qr^2} \). By replacing the terms with their factors, we have \( \frac{2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r}{11 \times q \times r \times r} \). After cancelling out the common factors, we get \( \frac{2 \times 3 \times p \times q \times r}{1} \), which simplifies to \( 6pqr \).
In simple words: We factorize both expressions. Then, we set them up as a fraction. We cancel out any identical factors from the top and bottom. The remaining parts are combined to give us the final answer.
(iv) We need to divide \( 34x^3y^3z^3 \) by \( 51xy^2z^3 \). We write this as a fraction: \( \frac{34x^3 y^3 z^3}{51xy^2z^3} \). Then, we perform the division of the coefficients and subtract the exponents for each variable. This gives us \( \frac{2 \times 17 \times x^{3} \times y^{3} \times z^{3}}{3 \times 17 \times x \times y^{2} \times z^{3}} \). Simplifying, we get \( \frac{2}{3} x^{3-1} y^{3-2} z^{3-3} \), which becomes \( \frac{2}{3} x^2y^1 z^0 \). Since \( z^0 = 1 \), the final answer is \( \frac{2}{3}x^2y \).
In simple words: We write the division as a fraction. We divide the numbers and subtract the powers for the same letters. If a letter has a power of zero, it becomes 1.
(v) We are given the division \( 12a^8b^8 \div (-6a^6b^4) \). We can write this expression as a fraction: \( \frac{12a^8b^8}{-6a^6b^4} \). We then factorize the coefficients and separate the variables: \( \frac{2 \times 2 \times 3 \times a^8 \times b^8}{-1 \times 2 \times 3 \times a^6 \times b^4} \). After simplifying, we get \( \frac{2}{-1} \times \frac{a^8}{a^6} \times \frac{b^8}{b^4} \). Subtracting the exponents for the variables, this equals \( -2 \times a^{8-6} \times b^{8-4} \), which gives us \( -2a^2b^4 \).
In simple words: We write the problem as a fraction. We divide the numbers and subtract the powers of the matching letters. The negative sign carries through from the denominator.

Exam Tip: Always show your steps when simplifying algebraic expressions to avoid errors, especially when dealing with negative signs and exponents.

Divide The Given Polynomial By The Given Monomial?

Question 1. Carry out the following divisions.
(i) \( (5x^2 - 6x) \div 3x \)
(ii) \( (3y^8 - 4y^6 + 5y^4) \div y^4 \)
(iii) \( 8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3) \div 4x^2y^2z^2 \)
(iv) \( (x^3 + 2x^2 + 3x) \div 2x \)
(v) \( (p^3q^6 - p^6q^3) \div p^3q^3 \)
Answer:
(i) We have \( (5x^2 - 6x) \div 3x \). We can write this as a fraction: \( \frac{5x^2-6x}{3x} \). Factor out \( x \) from the numerator to get \( \frac{x(5x-6)}{3x} \). Now, we can cancel out the common \( x \) term, leaving us with \( \frac{5x-6}{3} \). Therefore, \( (5x^2 - 6x) \div 3x = \frac{1}{3} (5x - 6) \).
In simple words: We write the problem as a fraction. We take out common factors from the top part. Then, we cancel matching factors from the top and bottom. This gives us our simpler answer.
(ii) For \( (3y^8 - 4y^6 + 5y^4) \div y^4 \), we express this as a fraction: \( \frac{3y^8 - 4y^6 + 5y^4}{y^4} \). We can factor out \( y^4 \) from each term in the numerator, resulting in \( \frac{y^4 (3y^4 - 4y^2 + 5)}{y^4} \). Canceling the \( y^4 \) from both the numerator and denominator, we obtain \( 3y^4 - 4y^2 + 5 \).
In simple words: Write as a fraction. Take out the common \( y^4 \) from every part on top. Then, cancel the \( y^4 \) from top and bottom to get the answer.
(iii) We need to divide \( 8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3) \) by \( 4x^2y^2z^2 \). We write this as the fraction \( \frac{8 (x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3)}{4 x^2 y^2z^2} \). Factor out \( x^2y^2z^2 \) from the terms inside the parenthesis in the numerator. This gives us \( \frac{2 \times 4 \times x^2y^2z^2 [x + y + z]}{4 \times x^2y^2z^2} \). By cancelling the common \( 4x^2y^2z^2 \) from both numerator and denominator, we are left with \( 2(x+y+z) \).
In simple words: We write the problem as a fraction. We take out common parts from the longer top expression. Then, we cancel out any matching parts from the top and bottom to get the simple answer.
(iv) We have \( (x^3 + 2x^2 + 3x) \div 2x \). We express this as a fraction: \( \frac{x^3 + 2x^2 + 3x}{2x} \). Factor out \( x \) from the numerator: \( \frac{x (x^2 + 2x + 3)}{2x} \). Now, cancel the common \( x \) term, which results in \( \frac{1}{2} (x^2 + 2x + 3) \).
In simple words: We write the problem as a fraction. We take out the common \( x \) from the top part. Then, we cancel the matching \( x \) from top and bottom. The remaining parts form the simplified answer.
(v) We need to divide \( (p^3q^6 - p^6q^3) \) by \( p^3q^3 \). We write this as a fraction: \( \frac{p^3q^6 - p^6q^3}{p^3q^3} \). We factor out the common term \( p^3q^3 \) from the numerator, giving us \( \frac{p^3q^3[q^3-p^3]}{p^3q^3} \). By cancelling the \( p^3q^3 \) from the numerator and denominator, we are left with \( q^3 - p^3 \).
In simple words: Write as a fraction. Take out the common \( p^3q^3 \) from the top expression. Cancel the matching \( p^3q^3 \) from top and bottom. The remainder is the answer.

Exam Tip: Remember to factor out the greatest common factor (GCF) from both the numerator and the denominator before simplifying to avoid mistakes.

Question 3. Work out the following divisions:
1. \( (10x-25) \div 5 \)
2. \( (10x-25) \div (2x-5) \)
3. \( 10y(6y + 21) \div 5(2y + 7) \)
4. \( 9x^2y^2(3z-24) \div 27xy(z-8) \)
5. \( 96abc(3a-12)(5b-30) \div 144(a-4)(b-6) \)
Answer:
1. We start with \( (10x-25) \div 5 \). First, factorize \( 10x-25 \) to get \( 5(2x-5) \). So the division becomes \( \frac{5(2x-5)}{5} \). We can cancel out the common factor of 5, leaving us with \( (2x-5) \). Hence, \( (10x-25) \div 5 = 2x-5 \).
In simple words: First, we factor out a common number from the expression. Then, we write it as a fraction and cancel out the matching number from the top and bottom.
2. We are given \( (10x-25) \div (2x-5) \). We start by factoring \( 10x-25 \), which gives \( 5(2x-5) \). So the expression becomes \( \frac{5(2x-5)}{2x-5} \). Now, we can cancel the common factor \( (2x-5) \), which leaves us with 5. Therefore, \( (10x-25) \div (2x-5) = 5 \).
In simple words: We factorize the top expression. Then, we write it as a fraction and cancel the matching terms from both top and bottom. The remaining number is our answer.
3. We need to divide \( 10y(6y + 21) \) by \( 5(2y + 7) \). First, factorize \( 6y+21 \) to get \( 3(2y+7) \). So the expression becomes \( \frac{10y \times 3(2y+7)}{5(2y+7)} \). We can cancel out the common factor \( (2y+7) \) and simplify the numerical coefficients \( \frac{10 \times 3}{5} \), which gives \( 2 \times y \times 3 \). This simplifies to \( 6y \). Therefore, \( 10y(6y + 21) \div 5(2y + 7) = 6y \).
In simple words: We factorize one of the expressions. Then, we set it up as a fraction. We cancel matching terms and simplify the numbers. The result is the answer.
4. We are asked to divide \( 9x^2y^2(3z-24) \) by \( 27xy(z-8) \). First, factorize \( 3z-24 \) as \( 3(z-8) \). The division is then \( \frac{9x^2y^2 \times 3(z-8)}{27xy(z-8)} \). We can write the prime factors: \( \frac{3 \times 3 \times x^2 \times y^2 \times 3 \times (z-8)}{3 \times 3 \times 3 \times x \times y \times (z-8)} \). Cancel the common terms, which includes \( (z-8) \), the numerical factors, and one \( x \) and one \( y \). This leaves us with \( \frac{x^2}{x} \times \frac{y^2}{y} \). Subtracting the exponents, we get \( x^{2-1}y^{2-1} \), which simplifies to \( xy \).
In simple words: Factorize the expression with 3z-24. Then, write the problem as a fraction and cancel out all the common factors. Subtract powers of same letters to get the final answer.
5. We are asked to divide \( 96abc(3a-12)(5b-30) \) by \( 144(a-4)(b-6) \). First, factorize the terms: \( 3a-12 = 3(a-4) \) and \( 5b-30 = 5(b-6) \). Also, prime factorize \( 96 = 2^5 \times 3 \) and \( 144 = 2^4 \times 3^2 \).
Now substitute these into the division: \( \frac{96abc \times 3(a-4) \times 5(b-6)}{144(a-4)(b-6)} \).
This becomes \( \frac{2^5 \times 3 \times abc \times 3 \times 5 \times (a-4) \times (b-6)}{2^4 \times 3^2 \times (a-4) \times (b-6)} \).
Cancel out the common factors: \( (a-4) \), \( (b-6) \), \( 2^4 \), \( 3^2 \).
Remaining factors are \( (2 \times 5) \times abc = 10abc \).
Thus, the expression simplifies to \( 10abc \).
In simple words: We factorize all parts of the expression, including numbers and polynomials. Then, we write the problem as a fraction and cancel out all matching terms from the top and bottom. What remains is our final, simpler answer.

Exam Tip: Always fully factorize all polynomial expressions and constants before attempting to simplify the division. This makes cancellation much clearer.

Question 4. Divide as directed.
1. \( 5(2x + 1)(3x + 5) \div (2x + 1) \)
2. \( 26xy(x + 5)(y - 4) \div 13x(y - 4) \)
3. \( 52pqr(p + q)(q + r)(r + p) \div 104pq(q + r)(r + p) \)
4. \( 20(y + 4)(y^2 + 5y + 3) \div 5(y + 4) \)
5. \( x(x + 1)(x + 2)(x + 3) \div x(x + 1) \)
Answer:
1. We are given \( 5(2x + 1)(3x + 5) \div (2x + 1) \). We can write this as a fraction: \( \frac{5(2x+1)(3x+5)}{(2x+1)} \). We observe that \( (2x+1) \) is a common factor in both the numerator and the denominator. Cancelling this factor, we are left with \( 5(3x+5) \).
In simple words: We write the division as a fraction. Then, we find matching terms in the top and bottom and cancel them out. The leftover part is our answer.
2. We need to divide \( 26xy(x + 5)(y - 4) \) by \( 13x(y - 4) \). We express this as the fraction \( \frac{26xy(x+5)(y-4)}{13x(y-4)} \). We can cancel common factors: \( (y-4) \), \( x \), and \( \frac{26}{13} \). After cancellation, we are left with \( 2y(x+5) \).
In simple words: Write the problem as a fraction. Cancel out all matching letters and terms from the top and bottom. Also, simplify the numbers.
3. We are asked to divide \( 52pqr(p + q)(q + r)(r + p) \) by \( 104pq(q + r)(r + p) \). We write this as a fraction: \( \frac{52pqr(p+q)(q+r)(r+p)}{104pq(q+r)(r+p)} \). We notice common factors like \( (q+r) \), \( (r+p) \), \( p \), and \( q \). Also, \( \frac{52}{104} \) simplifies to \( \frac{1}{2} \). After cancelling these factors, we are left with \( \frac{1}{2}r(p+q) \).
In simple words: Set the problem as a fraction. Find and cancel out all identical parts from the top and bottom. Simplify the numbers as well.
4. We need to divide \( 20(y + 4)(y^2 + 5y + 3) \) by \( 5(y + 4) \). We express this as the fraction \( \frac{20(y+4)(y^2+5y+3)}{5(y+4)} \). We observe that \( (y+4) \) is a common factor. Also, \( \frac{20}{5} = 4 \). Cancelling \( (y+4) \) and simplifying the numerical part, we get \( 4(y^2+5y+3) \).
In simple words: Write as a fraction. Cancel the matching parts in the top and bottom. Simplify the numbers, and the remaining expression is the answer.
5. We are asked to divide \( x(x + 1)(x + 2)(x + 3) \) by \( x(x + 1) \). We write this as a fraction: \( \frac{x(x+1)(x+2)(x+3)}{x(x+1)} \). We can clearly see that \( x \) and \( (x+1) \) are common factors in both the numerator and denominator. Cancelling these common factors, we are left with \( (x+2)(x+3) \).
In simple words: We set up the problem as a fraction. Then, we cancel out the identical factors from the top and bottom. What's left over is the answer.

Exam Tip: When dividing polynomials, always look for common factors in both the numerator and denominator to simplify the expression efficiently.

Question 5. Factorise the expressions and divide them as directed.
1. \( (y^2 + 7y + 10) \div (y + 5) \)
2. \( (m^2 - 14m - 32) \div (m + 2) \)
3. \( (5p^2 - 25p + 20) \div (p - 1) \)
4. \( 4yz(z^2 + 6z-16) \div 2y(z + 8) \)
5. \( 5pq(p^2 - q^2) \div 2p(p + q) \)
6. \( 12xy(9x^2 - 16y^2) \div 4xy(3x + 4y) \)
7. \( 39y^3(50y^2 - 98) \div 26y^2(5y + 7) \)
Answer:
1. We need to divide \( (y^2 + 7y + 10) \) by \( (y + 5) \). First, we factorize the quadratic expression \( y^2 + 7y + 10 \). We split the middle term \( 7y \) into \( 2y + 5y \), such that their product \( (2y)(5y) = 10y^2 \) matches the product of the first and last terms \( (y^2)(10) = 10y^2 \).
So, \( y^2 + 7y + 10 = y^2 + 2y + 5y + 10 \).
Factor by grouping: \( y(y + 2) + 5(y + 2) \).
This gives us \( (y + 2)(y + 5) \).
Now, the division becomes \( \frac{(y+2)(y+5)}{(y+5)} \). We cancel the common factor \( (y+5) \), leaving us with \( (y+2) \).
In simple words: First, we factorize the top expression by splitting the middle term. This creates two groups that can be factored. Then, we write the division as a fraction and cancel out the common terms from the top and bottom.
2. We need to divide \( (m^2 - 14m - 32) \) by \( (m + 2) \). First, we factorize the quadratic expression \( m^2 - 14m - 32 \). We look for two numbers that multiply to -32 and add up to -14. These numbers are -16 and 2.
So, \( m^2 - 14m - 32 = m^2 - 16m + 2m - 32 \).
Factor by grouping: \( m(m - 16) + 2(m - 16) \).
This gives us \( (m - 16)(m + 2) \).
Now, the division is \( \frac{(m-16)(m+2)}{(m+2)} \). We cancel the common factor \( (m+2) \), resulting in \( (m-16) \).
In simple words: We factorize the top expression by finding two numbers that multiply to the last term and add to the middle term. Then, we write the division as a fraction. We cancel matching parts from the top and bottom.
3. We are asked to divide \( (5p^2 - 25p + 20) \) by \( (p - 1) \). First, factor out the common numerical factor 5 from the first expression: \( 5(p^2 - 5p + 4) \).
Now, factorize the quadratic expression \( p^2 - 5p + 4 \). We need two numbers that multiply to 4 and add up to -5. These numbers are -1 and -4.
So, \( p^2 - 5p + 4 = p^2 - p - 4p + 4 \).
Factor by grouping: \( p(p - 1) - 4(p - 1) \).
This yields \( (p - 1)(p - 4) \).
Therefore, \( 5p^2 - 25p + 20 = 5(p - 1)(p - 4) \).
The division becomes \( \frac{5(p-1)(p-4)}{(p-1)} \). We cancel the common factor \( (p-1) \), which leaves us with \( 5(p-4) \).
In simple words: First, we factor out the common number. Then, we factorize the remaining quadratic expression. After that, we write the division as a fraction and cancel out the matching terms from the top and bottom.
4. We need to divide \( 4yz(z^2 + 6z - 16) \) by \( 2y(z + 8) \). First, we factorize the quadratic expression \( z^2 + 6z - 16 \). We find two numbers that multiply to -16 and add up to 6. These numbers are 8 and -2.
So, \( z^2 + 6z - 16 = z^2 + 8z - 2z - 16 \).
Factor by grouping: \( z(z + 8) - 2(z + 8) \).
This gives us \( (z + 8)(z - 2) \).
Now, the division becomes \( \frac{4yz(z+8)(z-2)}{2y(z+8)} \).
We cancel common factors: \( (z+8) \) and \( y \). Also, \( \frac{4}{2} = 2 \). This leaves us with \( 2z(z-2) \).
In simple words: First, we factorize the quadratic part. Then, we write the whole problem as a fraction. We cancel out common numbers, letters, and terms from the top and bottom parts.
5. We are asked to divide \( 5pq(p^2 - q^2) \) by \( 2p(p + q) \). First, we factorize \( p^2 - q^2 \) using the difference of squares formula \( a^2 - b^2 = (a - b)(a + b) \). So, \( p^2 - q^2 = (p - q)(p + q) \).
Now, the division becomes \( \frac{5pq(p-q)(p+q)}{2p(p+q)} \).
We cancel common factors: \( p \) and \( (p+q) \). This leaves us with \( \frac{5q(p-q)}{2} \).
In simple words: We factorize the \(p^2-q^2\) term using a special formula. Then, we write the division as a fraction. We cancel out matching factors from the top and bottom.
6. We need to divide \( 12xy(9x^2 - 16y^2) \) by \( 4xy(3x + 4y) \). First, we factorize \( 9x^2 - 16y^2 \) using the difference of squares formula. \( 9x^2 = (3x)^2 \) and \( 16y^2 = (4y)^2 \). So, \( (3x)^2 - (4y)^2 = (3x - 4y)(3x + 4y) \).
Now, the division becomes \( \frac{12xy(3x-4y)(3x+4y)}{4xy(3x+4y)} \).
We cancel common factors: \( x \), \( y \), and \( (3x+4y) \). Also, \( \frac{12}{4} = 3 \). This leaves us with \( 3(3x-4y) \).
In simple words: Factorize the part with \(9x^2-16y^2\) using the difference of squares rule. Then, set up the problem as a fraction. Cancel out all common numbers, letters, and expressions from the top and bottom.
7. We are asked to divide \( 39y^3(50y^2 - 98) \) by \( 26y^2(5y + 7) \). First, we factorize \( 50y^2 - 98 \). Take out the common factor 2: \( 2(25y^2 - 49) \).
Now, factorize \( 25y^2 - 49 \) using the difference of squares formula, since \( 25y^2 = (5y)^2 \) and \( 49 = 7^2 \).
So, \( 2(25y^2 - 49) = 2((5y)^2 - 7^2) = 2(5y - 7)(5y + 7) \).
Therefore, \( 39y^3(50y^2 - 98) = 39y^3 \times 2(5y - 7)(5y + 7) \).
The division becomes \( \frac{39y^3 \times 2(5y-7)(5y+7)}{26y^2(5y+7)} \).
We cancel common factors: \( (5y+7) \) and \( y^2 \).
Also, simplify the coefficients: \( \frac{39 \times 2}{26} = \frac{78}{26} = 3 \).
This leaves us with \( 3y(5y-7) \).
In simple words: Factor out the common number from the first expression. Then, use the difference of squares rule to factorize the remaining part. Set up the problem as a fraction and cancel out all matching numbers, letters, and expressions from the top and bottom.

Exam Tip: Always look for common factors and apply appropriate factorization identities (like difference of squares or splitting the middle term) to simplify polynomials before division.