Page 157 - ICSE Math 8

4 2 3 5
2 3 4 2
3 4 2
Example 1: Find the HCF of the monomials 4x y z w , 12x y z w and 18x y z .
Solution: 4 = 2 × 2 12 = 2 × 2 × 3 18 = 2 × 3 × 3
HCF of 4, 12, 18 = 2
The common variables in the terms are x, y and z.
4
3
2
Lowest power of x, out of x , x , x = 2
4
3
2
Lowest power of y, out of y , y , y = 2
4
2
3
Lowest power of z, out of z , z , z = 2
2 2 2
HCF of the variables = x y z
4 2 3 5
2 2 2
2 3 4 2
3 4 2
So, the HCF of 4x y z w , 12x y z w and 18x y z = 2x y z
Factorization By Taking Out Common Monomials
Step 1: In the algebraic expression, find the highest common factor (HCF) of its terms.
Step 2: Express each term of the algebraic expression as the product of the HCF and the quotient, when it is
divided by the HCF.
Step 3: Using distributive property of multiplication over addition, express the given algebraic expression as
the product of its HCF and the quotient obtained in step 2.
3 2
2
4 3
2 3
3 2
2
Example 2: Factorize: (a) 4x + 12 (b) 2x y – 5x y (c) 15x y + 5x y – 20x y
2 3
3 2
2
Solution: (a) 4x + 12 (b) 2x y – 5x y
2
3 2
4x = 2 × 2 × x × x 2x y = 2 × x × x × x × y × y
2 3
12 = 2 × 2 × 3 5x y = 5 × x × x × y × y × y
2 2
2 3
3 2
2
HCF of 4x and 12 = 4 HCF of 2x y and 5x y = x y
2 2
So, we take 4 common from both the So, we take x y common from both
terms. the terms.
2 2
3 2
2
2 3
2
\ 4x + 12 = 4(x + 3) \ 2x y – 5x y = x y (2x – 5y)
2
3 2
4 3
(c) 15x y + 5x y – 20x y
3 2
15x y = 3 × 5 × x × x × x × y × y Maths Info
2
5x y = 5 × x × x × y If a polynomial is divisible by 1
4 3
20x y = 2 × 2 × 5 × x × x × x × x × y × y × y and itself only, it is said to be a
prime polynomial.
2
3 2
2
4 3
HCF of 15x y , 5x y and 20x y = 5x y For example, 3x + 4, 2x + 3y ,
2
2
2
Taking 5x y common from all the terms, we get etc.
3 2
2 2
2
4 3
2
15x y + 5x y – 20x y = 5x y(3xy + 1 – 4x y )
Factorization By Taking Out Common Binomials
Step 1: Locate the common binomial.
Step 2: Write the given algebraic expression as a product of this common binomial and the quotient obtained
on dividing the given algebraic expression by the binomial.
Example 3: Factorize the following:
2
(a) 8(5x – 9y) – 12(5x – 9y) Try This
2
(b) (2x – 3y) – 6x + 9y Factorize:
2
Solution: (a) 8(5x – 9y) – 12(5x – 9y) (a) 5(3x – 4) + 7(3x – 4)
2
(b) 9(x – y) + 6(y – x)
= 4 × 2(5x – 9y) × (5x – 9y) – 4 × 3(5x – 9y)
145