RBSE Solutions Class 8 Maths Chapter 10 Factorization Exercise 10.3

Factorization Ex 10.3

Question 1. Carry out the following divisions.
(i) \( 28x^4 \) by \( 56x \)
(ii) \( - 36y^3 \) by \( 9y^2 \)
(iii) \( 34x^3y^3z^3 \) by \( 51xy^2z^3 \)
(iv) \( 12a^8b^8 \) by \( (- 6a^6b^4) \)
Answer:
(i) To divide \( 28x^4 \) by \( 56x \), we can write it as a fraction and simplify:
\( \frac{28x^4}{56x} \)
Now, we break down the numbers and variables into their factors:
\( = \frac{2 \times 2 \times 7 \times x \times x \times x \times x}{2 \times 2 \times 2 \times 7 \times x} \)
We cancel out the common factors from the top and bottom:
\( = \frac{x \times x \times x}{2} \)
\( = \frac{x^3}{2} \)
This can also be written as \( \frac{1}{2}x^3 \). Always simplify numerical coefficients and variable powers separately.

(ii) To divide \( -36y^3 \) by \( 9y^2 \), we set up the division as a fraction:
\( \frac{-36y^3}{9y^2} \)
We factorize the numbers and variables:
\( = \frac{- (2 \times 2 \times 3 \times 3 \times y \times y \times y)}{3 \times 3 \times y \times y} \)
Cancel out the common factors:
\( = - \frac{2 \times 2 \times y}{1} \)
\( = - \frac{4y}{1} \)
\( = -4y \)

(iii) To divide \( 34x^3y^3z^3 \) by \( 51xy^2z^3 \), we express it as a fraction:
\( \frac{34x^3y^3z^3}{51xy^2z^3} \)
We find the common factors for the numbers (34 and 51 share 17) and cancel variables:
\( = \frac{2 \times 17 \times x \times x \times x \times y \times y \times y \times z \times z \times z}{3 \times 17 \times x \times y \times y \times z \times z \times z} \)
After cancelling common terms:
\( = \frac{2 \times x \times x \times y}{3} \)
\( = \frac{2x^2y}{3} \)
This means the simplified expression is two-thirds of \( x^2y \).

(iv) To divide \( 12a^8b^8 \) by \( (-6a^6b^4) \), we write it as a fraction:
\( \frac{12a^8b^8}{-6a^6b^4} \)
We simplify the numerical coefficients and the variables separately by subtracting the exponents for the same base:
\( = - \frac{12}{6} \times a^{8-6} \times b^{8-4} \)
\( = -2a^2b^4 \)
When dividing terms with exponents, remember to subtract the exponent of the divisor from the exponent of the dividend for each variable.
In simple words: To divide these algebraic terms, first divide the numbers, then divide each variable by subtracting their powers. Write it as a fraction and cross out common parts from top and bottom to make it simpler.

🎯 Exam Tip: Always factorize both the numerical coefficients and the variable parts completely to avoid errors when cancelling common terms. Pay close attention to negative signs.

Question 2. Divide the given polynomial by the given monomial.
(i) \( (5x^2 - 6x) \) by \( 3x \)
(ii) \( (x^3 + 2x^2 + 3x) \) by \( 2x \)
(iii) \( (p^3q^6 - p^6q^3) \) by \( p^3q^3 \)
(iv) \( (3x^8 - 4x^6 + 5x^4) \) by \( x^4 \)
Answer:
(i) To divide \( (5x^2 - 6x) \) by \( 3x \), we divide each term of the polynomial by the monomial:
\( \frac{5x^2 - 6x}{3x} \)
\( = \frac{5x^2}{3x} - \frac{6x}{3x} \)
\( = \frac{5}{3}x - 2 \)
This means we treat each part of the polynomial separately when dividing by a single term.

(ii) To divide \( (x^3 + 2x^2 + 3x) \) by \( 2x \), we divide each term by \( 2x \):
\( \frac{x^3 + 2x^2 + 3x}{2x} \)
First, factor out the common term \( x \) from the numerator:
\( = \frac{x(x^2 + 2x + 3)}{2x} \)
Cancel out the common \( x \) from numerator and denominator:
\( = \frac{x^2 + 2x + 3}{2} \)
Now, divide each term in the numerator by 2:
\( = \frac{x^2}{2} + \frac{2x}{2} + \frac{3}{2} \)
\( = \frac{1}{2}x^2 + x + \frac{3}{2} \)
Remember to simplify each part after division to get the final form.

(iii) To divide \( (p^3q^6 - p^6q^3) \) by \( p^3q^3 \), we apply the division to each term:
\( \frac{p^3q^6 - p^6q^3}{p^3q^3} \)
Factor out the common term \( p^3q^3 \) from the numerator:
\( = \frac{p^3q^3(q^3 - p^3)}{p^3q^3} \)
Cancel the common \( p^3q^3 \) term:
\( = \frac{q^3 - p^3}{1} \)
\( = q^3 - p^3 \)
This shows that when a common factor exists, factoring it out simplifies the division greatly.

(iv) To divide \( (3x^8 - 4x^6 + 5x^4) \) by \( x^4 \), we divide each term of the polynomial by \( x^4 \):
\( \frac{3x^8 - 4x^6 + 5x^4}{x^4} \)
\( = \frac{3x^8}{x^4} - \frac{4x^6}{x^4} + \frac{5x^4}{x^4} \)
We subtract the powers of \( x \):
\( = 3x^{8-4} - 4x^{6-4} + 5x^{4-4} \)
\( = 3x^4 - 4x^2 + 5x^0 \)
Since \( x^0 = 1 \), the expression becomes:
\( = 3x^4 - 4x^2 + 5 \)
In simple words: When dividing a polynomial by a single term, divide each part of the polynomial separately by that single term. Simplify numbers and variables in each part. You can also take out common factors first.

🎯 Exam Tip: Remember that \( x^0 = 1 \) for any non-zero \( x \). When dividing by a monomial, ensure every term in the polynomial is divided, not just the first one.

Question 3. Work out the following divisions.
(i) \( 10y(6y + 21) \div 5(2y + 7) \)
(ii) \( 9x^2y^2(3z - 24) \div 27xy(z - 8) \)
(iii) \( (10y + 14) \div 2 \)
(iv) \( (6x - 15) \div (2x - 5) \)
Answer:
(i) To divide \( 10y(6y + 21) \) by \( 5(2y + 7) \):
\( \frac{10y(6y + 21)}{5(2y + 7)} \)
Factor out common numbers from the terms in the parentheses. From \( (6y + 21) \), we can take out 3:
\( = \frac{10y \cdot 3(2y + 7)}{5(2y + 7)} \)
Now, we can cancel the common terms \( (2y+7) \) and simplify the numbers:
\( = \frac{10y \cdot 3}{5} \)
\( = 2y \cdot 3 \)
\( = 6y \)
Factoring out common terms often simplifies complex divisions quickly.

(ii) To divide \( 9x^2y^2(3z - 24) \) by \( 27xy(z - 8) \):
\( \frac{9x^2y^2(3z - 24)}{27xy(z - 8)} \)
Factor out 3 from \( (3z - 24) \):
\( = \frac{9x^2y^2 \cdot 3(z - 8)}{27xy(z - 8)} \)
Cancel the common term \( (z-8) \) and simplify the numerical and variable parts:
\( = \frac{9x^2y^2 \cdot 3}{27xy} \)
\( = \frac{27x^2y^2}{27xy} \)
\( = x^{2-1}y^{2-1} \)
\( = xy \)

(iii) To divide \( (10y + 14) \) by \( 2 \):
\( \frac{10y + 14}{2} \)
Divide each term in the numerator by 2:
\( = \frac{10y}{2} + \frac{14}{2} \)
\( = 5y + 7 \)

(iv) To divide \( (6x - 15) \) by \( (2x - 5) \):
\( \frac{6x - 15}{2x - 5} \)
Factor out the common factor 3 from the numerator \( (6x - 15) \):
\( = \frac{3(2x - 5)}{2x - 5} \)
Cancel the common term \( (2x - 5) \) from the numerator and denominator:
\( = 3 \)
In simple words: For these division problems, look for common parts in the top and bottom expressions. Factor out common numbers or terms, then cancel identical parts. This makes the expression much simpler.

🎯 Exam Tip: Always look for common factors within parentheses first. Factoring them out can often reveal a common term that can be cancelled directly, greatly simplifying the division.

Question 4. Factorize the expressions and divide them as directed
(i) \( (y^2 + 7y + 10) \div (y + 5) \)
(ii) \( (5x^2 - 25x + 20) \div (x - 1) \)
(iii) \( 12xy(9x^2 - 16y^2) \div 4xy(3x + 4y) \)
(iv) \( 4yz (z^2 + 6z - 16) \div 2y (z + 8) \)
Answer:
(i) To divide \( (y^2 + 7y + 10) \) by \( (y + 5) \):
First, factorize the quadratic expression \( y^2 + 7y + 10 \). We need two numbers that multiply to 10 and add to 7 (which are 2 and 5):
\( y^2 + 7y + 10 = y^2 + 2y + 5y + 10 \)
\( = y(y + 2) + 5(y + 2) \)
\( = (y + 2)(y + 5) \)
Now, perform the division:
\( \frac{(y + 2)(y + 5)}{y + 5} \)
Cancel the common term \( (y + 5) \):
\( = y + 2 \)
Factoring is a key step in simplifying algebraic fractions.

(ii) To divide \( (5x^2 - 25x + 20) \) by \( (x - 1) \):
First, factorize the numerator \( 5x^2 - 25x + 20 \). Take out the common factor 5:
\( 5x^2 - 25x + 20 = 5(x^2 - 5x + 4) \)
Now, factorize the quadratic expression \( x^2 - 5x + 4 \). We need two numbers that multiply to 4 and add to -5 (which are -1 and -4):
\( 5(x^2 - 5x + 4) = 5(x - 1)(x - 4) \)
Now, perform the division:
\( \frac{5(x - 1)(x - 4)}{x - 1} \)
Cancel the common term \( (x - 1) \):
\( = 5(x - 4) \)

(iii) To divide \( 12xy(9x^2 - 16y^2) \) by \( 4xy(3x + 4y) \):
\( \frac{12xy(9x^2 - 16y^2)}{4xy(3x + 4y)} \)
Factorize the term \( (9x^2 - 16y^2) \) using the difference of squares identity \( (a^2 - b^2) = (a - b)(a + b) \). Here, \( a = 3x \) and \( b = 4y \):
\( 9x^2 - 16y^2 = (3x)^2 - (4y)^2 = (3x - 4y)(3x + 4y) \)
Substitute this back into the expression:
\( = \frac{12xy(3x - 4y)(3x + 4y)}{4xy(3x + 4y)} \)
Cancel the common terms \( 4xy \) and \( (3x + 4y) \):
\( = \frac{12(3x - 4y)}{4} \)
\( = 3(3x - 4y) \)

(iv) To divide \( 4yz (z^2 + 6z - 16) \) by \( 2y (z + 8) \):
\( \frac{4yz (z^2 + 6z - 16)}{2y (z + 8)} \)
First, factorize the quadratic expression \( z^2 + 6z - 16 \). We need two numbers that multiply to -16 and add to 6 (which are 8 and -2):
\( z^2 + 6z - 16 = (z + 8)(z - 2) \)
Substitute this back into the division expression:
\( = \frac{4yz(z + 8)(z - 2)}{2y(z + 8)} \)
Cancel the common terms \( 2y \) and \( (z + 8) \):
\( = \frac{4z(z - 2)}{2} \)
\( = 2z(z - 2) \)
In simple words: To solve these, first break down the expressions into their simpler parts by factoring. Look for common factors like \( (a+b) \) or use formulas like the difference of squares. Once factors are found, cancel out any identical parts from the top and bottom of the division.

🎯 Exam Tip: Mastering factorization (common factors, quadratic trinomials, and difference of squares) is essential for solving division problems involving polynomials. Always check if you can factorize before attempting to cancel terms.