Factorisation - Worksheets

CBSE Worksheet-1
Class 08 - Mathematics (Factorisation)

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General Instructions: All questions are compulsory. Q.1 to Q.2 carries one mark each. Q.3 to Q.7 carries two marks each. Q.8 and Q.9 carries three marks each. Q.10 to Q.12 carries four marks each.

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1. Factorise the following expression :–16z + 20z3
2. Factorise the following expression :5x2y – 15xy2
3. State whether the following statements are True or False:
   1. Factorized form of 2x + 4 is 2(x + 2).
   2. pq2 + qr2 + rs2 is a binomial.
   3. The product of one negative and one positive term is a negative term.
   4. The product of two negative terms is a positive term.
4. Fill In The Blanks.
   1. a2 - b2 = ____.
   2. Common factor of ax2 + bx is ___.
   3. (9)2 – (3)2 = _______.
   4. Factorised form of x2 + 8x + 15 is _____.
5. Match the following:

   Column A	Column B
   a. 2x + 16	p. 4x2 + 9y2 + 12xy
   b. (2x + 3y)2	q. 2(x + 8)
   c. P4 - 81	r. p(p + 6) + 8
   d. P2 + 6p + 8	s. (p2 + 9) (p2 - 9)

6. Factorise :14pq + 35pqr
7. Factorise :49x2 – 36
8. Find and correct the errors in the following mathematical statement. Substituting x = – 3 in the given equation 3x2+13x2=1+1=2$\frac{3x^2+1}{3x^2}=1+1=2$
9. Divide as directed: 20(y + 4) (y2 + 5y + 3) ÷ 5(y + 4)
10. Factorize x4 – y4
11. Factorise the expression and divide them as directed: (m2 – 14m – 32) ÷ (m + 2)
12. Factorize x8 – y8.

CBSE Worksheet-1
Class 08 - Mathematics (Factorisation)
Solution

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1. –16z + 20z3 = 4z (– 4 + 5z2)
2. 5x2y – 15xy2 = 5xy (x – 3y)
3. 1. True
   2. False
   3. True
   4. True
4. 1. (a + b) (a - b)
   2. x
   3. (9 + 3) (9 - 3)
   4. (x + 5) (x + 3).
5. 1. →$\to$ q
   2. →$\to$ p
   3. →$\to$ s
   4. →$\to$ r
6. 14pq + 35pqr
   14pq = 2 × 7 × p × q
   35pqr = 5 × 7 × p × q × r
   The two terms have 7, p and q as common factors.
   Therefore,
   14pq + 35pqr
   = 7 × p × q × 2 + 7 × p × q × 5 × r
   = 7 × p × q × [2 + (5 × r)] . . . . [Combining the terms]
   = 7pq × (2 + 5r)
   = 7pq(2 + 5r). . . . [Required factor form]
7. 49x2 – 36
   = (7x)2 – (6)2
   = (7x – 6) (7x + 6). . . . [Using Identity III
8. 3x2+13x2=3x23x2+13x2$\frac{3x^2+1}{3x^2}=\frac{3x^2}{3x^2}+\frac{1}{3x^2}$
   =1+13x2$=1+\frac{1}{3x^2}$
   =1+13(−3)2$=1+\frac{1}{3{(-3)}^2}$​​​​​
   =2827$=\frac{28}{27}$
9. 20(y + 4) (y2 + 5y + 3) ÷ 5(y + 4)
   =20(y+4)(y2+5y+3)5(y+4)$=\frac{20(y+4)(y^2+5y+3)}{5(y+4)}$
   = 4(y2 + 5y + 3)
10. x4 – y4 = (x2)2 – (y2)2
    = (x2 – y2)(x2 + y2) Using a2 – b2 = (a + b)(a – b)
    = (x – y)(x + y)(x2 + y2) Using a2 – b2 = (a + b)(a – b)
11. (m2 – 14m – 32) ÷ (m + 2)
    =m2−14m−32m+2$=\frac{m^2-14m-32}{m+2}$
    =m2−16m+2m−32m+2$=\frac{m^2-16m+2m-32}{m+2}$. . . . [Using Identity IV
    =m(m−16)+2(m−16)m+2$=\frac{m(m-16)+2(m-16)}{m+2}$
    =m(m−16)(m+2)m+2$=\frac{m(m-16)(m+2)}{m+2}$
    = m – 16
12. x8 – y8 = {(x4)2 - (y4)2}
    = (x4 - y4) (x4 + y4)
    = {(x2)2 – (y2)2}(x4 + y4)
    = (x2 – y2)(x2 + y2)(x4 + y4)
    = (x – y)(x + y)(x2 + y2) (x4 + y4)
    = (x – y)(x + y)(x2 + y2){(x2)2 +(y2)2 + 2x2y2 – 2x2y2}
    = (x – y )( x+y)(x2 + y2){(x2 + y2)2 –( 2xy)2}
    = (x – y)( x+y)(x2 +y2)(x2 + y2 - 2xy)(x2 + y2 + 2xy)