Grade 8 Factorisation


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1 ID : ae8factorisation [1] Grade 8 Factorisation For more such worksheets visit Answer the questions (1) Find factors of following polynomial A) y 22xy + 3y  6x B) 3y 212xy  2y + 8x (2) Factorize (49z 2100). (3) Write the following polynomials in factored form. A) 8xy + 6x 2 y 2 z 3 B) 16a 3 c + 72bc C) 32a 3 b a 2 bc D) 20x 2 yz yz 3 (4) Write the following polynomials in factored form: A) 30p 2 q p 2 q 4 B) 28p 3 q + 8p 2 q 3 C) 44xy xy 3 D) 45x 3 y x 2 y 2 (5) Find factors of polynomial (a 24ab + 4b 24c 2 ). (6) Factorize (4x + 7) 2112x. (7) Find factors of following polynomial A) 45xy + 35y + 63x + 49 B) 3xy  21y + 5x  35 (8) Factorize (16y 240y + 25). (9) Find factors of following polynomial A) p 2 + 7p + 10 B) 5b 218b  8 (10) Factorize (625x 4200x ) Edugain ( All Rights Reserved Many more such worksheets can be generated at
2 Answers ID : ae8factorisation [2] (1) A) (y + 3)(y  2x) Reorder some of the terms as following, y 22xy + 3y  6x = y 2 + 3y  2xy  6x Now we can see that y is common in first two terms, and 2x is common in last two terms. Lets rewrite the expression as following, = y(y + 3) 2x(y + 3) = (y + 3)(y  2x) Thus, the factors of the polynomial, y 22xy + 3y  6x are (y  2x)( y + 3). B) (3y  2)(y  4x) Reorder some of the terms as following, 3y 212xy  2y + 8x = 3y 22y  12xy + 8x Now we can see that y is common in first two terms, and 4x is common in last two terms. Lets rewrite the expression as following, = y(3y  2) 4x(3y  2) = (3y  2)(y  4x) Thus, the factors of the polynomial, 3y 212xy  2y + 8x are (y  4x)( 3y  2). (2) (7z + 10) (7z  10) We know that, a 2  b 2 = (a + b)(a  b). Let's write (49z 2100) as: (7z) 2  (10) 2 = (7z + 10)(7z  10)
3 ID : ae8factorisation [3] (3) A) 2xy(4 + 3xyz 3 ) If we look at the given polynomial carefully, we observe that 2xy is common to both the terms. By taking out the common factor 2xy from the polynomial 8xy + 6x 2 y 2 z 3, we get: 2xy(4 + 3xyz 3 ) Thus, the polynomial 8xy + 6x 2 y 2 z 3 can be written in the factored form as 2xy(4 + 3xyz 3 ). B) 8c(2a 3 + 9b) If we look at the given polynomial carefully, we observe that 8c is common to both the terms. By taking out the common factor 8c from the polynomial 16a 3 c + 72bc, we get: 8c(2a 3 + 9b) Thus, the polynomial 16a 3 c + 72bc can be written in the factored form as 8c(2a 3 + 9b). C) 16a 2 b(2ab + 3c) If we look at the given polynomial carefully, we observe that 16a 2 b is common to both the terms. By taking out the common factor 16a 2 b from the polynomial 32a 3 b a 2 bc, we get: 16a 2 b(2ab + 3c) Thus, the polynomial 32a 3 b a 2 bc can be written in the factored form as 16a 2 b(2ab + 3c).
4 ID : ae8factorisation [4] D) 20yz 3 (x 2 + 1) If we look at the given polynomial carefully, we observe that 20yz 3 is common to both the terms. By taking out the common factor 20yz 3 from the polynomial 20x 2 yz yz 3, we get: 20yz 3 (x 2 + 1) Thus, the polynomial 20x 2 yz yz 3 can be written in the factored form as 20yz 3 (x 2 + 1). (4) A) 30p 2 q 3 (1 + q) If we look at the given polynomial carefully, we observe that the factor 30p 2 q 3 is common to both terms of the polynomial 30p 2 q p 2 q 4. By taking out the common factor 30p 2 q 3 from the polynomial 30p 2 q p 2 q 4, we get the following factorization: 30p 2 q p 2 q 4 = 30p 2 q 3 (1 + q) Thus, the polynomial 30p 2 q p 2 q 4 can be written in factored form as 30p 2 q 3 (1 + q). B) 4p 2 q(7p + 2q 2 ) If we look at the given polynomial carefully, we observe that the factor 4p 2 q is common to both terms of the polynomial 28p 3 q + 8p 2 q 3. By taking out the common factor 4p 2 q from the polynomial 28p 3 q + 8p 2 q 3, we get the following factorization: 28p 3 q + 8p 2 q 3 = 4p 2 q(7p + 2q 2 ) Thus, the polynomial 28p 3 q + 8p 2 q 3 can be written in factored form as 4p 2 q(7p + 2q 2 ).
5 ID : ae8factorisation [5] C) 22xy 2 (2 + y) If we look at the given polynomial carefully, we observe that the factor 22xy 2 is common to both terms of the polynomial 44xy xy 3. By taking out the common factor 22xy 2 from the polynomial 44xy xy 3, we get the following factorization: 44xy xy 3 = 22xy 2 (2 + y) Thus, the polynomial 44xy xy 3 can be written in factored form as 22xy 2 (2 + y). D) 5x 2 y 2 (9x + 4) If we look at the given polynomial carefully, we observe that the factor 5x 2 y 2 is common to both terms of the polynomial 45x 3 y x 2 y 2. By taking out the common factor 5x 2 y 2 from the polynomial 45x 3 y x 2 y 2, we get the following factorization: 45x 3 y x 2 y 2 = 5x 2 y 2 (9x + 4) Thus, the polynomial 45x 3 y x 2 y 2 can be written in factored form as 5x 2 y 2 (9x + 4). (5) (a  2b + 2c) (a  2b  2c) We know that, (a  b) 2 = a 22ab + b 2, a 2  b 2 = (a + b)(a  b) The factors of the polynomial (a 24ab + 4b 24c 2 ) can be found using above identities as following, a 24ab + 4b 24c 2 = {(a) 24ab + (2b) 2 }  (2c) 2 = (a  2b) 2  (2c) 2 = (a  2b + 2c) (a  2b  2c)
6 (6) (4x  7) (4x  7) ID : ae8factorisation [6] Using algebraic identity (a + b) 2 = a 2 + 2ab + b 2 (4x + 7) 2112x = (4x) x + (7) 2112x (4x + 7) 2112x = (4x) 256x + (7) 2 (4x + 7) 2112x = (4x) 22(7)(4x) + (7) 2 Now using algebraic identity (a  b) 2 = a 22ab + b 2 (4x + 7) 2112x = (4x  7) 2 (4x + 7) 2112x = (4x  7) (4x  7) (7) A) (9x + 7)(5y + 7) The factors of the polynomial, 45xy + 35y + 63x + 49 can be found as, 45xy + 35y + 63x + 49 = 5y(9x + 7) + 7(9x + 7) = (9x + 7)(5y + 7) Thus, the factors of the polynomial, 45xy + 35y + 63x + 49 are ( 9x + 7)(5y + 7). B) (x  7)(3y + 5) The factors of the polynomial, 3xy  21y + 5x  35 can be found as, 3xy  21y + 5x  35 = 3y(x  7) + 5(x  7) = (x  7)(3y + 5) Thus, the factors of the polynomial, 3xy  21y + 5x  35 are ( x  7)(3y + 5). (8) (4y  5) 2 We know that, (a  b) 2 = a 22ab + b 2. (16y 240y + 25) can be factorized as, 16y 240y + 25 = (4y) 22(5)(4y) + (5) 2 = (4y  5) 2
7 (9) A) (p + 5)(p + 2) ID : ae8factorisation [7] In order to find factors of polynomial p 2 + 7p + 10, we need to find two numbers whose sum is 7 and product is (10 1 = 10) We notice that 5 and 2 are such numbers, since (5) (2) = 10, and (5) + (2) = 7 Now we can rewrite polynomial as following, p 2 + 7p + 10 = p 2 + 5p + 2p + 10 = p(p + 5) + 2(p + 5) = (p + 5)(p + 2) Step 4 Thus, the factors of the polynomial p 2 + 7p + 10 are (p + 5) and (p + 2). B) (5b + 2)(b  4) In order to find factors of polynomial 5b 218b  8, we need to find two numbers whose sum is 18 and product is (8 5 = 40) We notice that 2 and 20 are such numbers, since (2) (20) = 40, and (2) + (20) =  18 Now we can rewrite polynomial as following, 5b 218b  8 = 5b 2 + 2b  20b  8 = b(5b + 2) 4(5b + 2) = (5b + 2)(b  4) Step 4 Thus, the factors of the polynomial 5b 218b  8 are (5b + 2) and (b  4).
8 (10) (5x + 2) (5x + 2) (5x  2) (5x  2) ID : ae8factorisation [8] Let's factorize 625x 4200x = 625x 4200x = (25x 2 ) 2  (2 4 25) Since we know that (a  b) 2 = a 22ab + b 2 ] = (25x 24) 2 = {(5x) } 2 = {(5x + 2)(5x  2)} 2...[Since we know that a 2  b 2 = (a + b)(a  b)] = (5x + 2)(5x + 2)(5x  2)(5x  2) Thus, the factors of (625x 4200x ) are (5x + 2), (5x + 2), (5x  2) and (5x  2).
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