**GSEB Solutions for Class 8 Mathematics – Revision 2 (Sem – II) (English Medium)**

GSEB SolutionsMathsScience

**Revision 2**

**Solution 1:**

**Solution 2A:**

**Solution 2B:**

- ab
^{3}+ ab= ab(b^{2 }+ 1) - x
^{3}– 5x^{2}+ 2x – 10= x^{2}(x – 5) + 2(x – 5)= (x – 5)(x^{2 }+ 2) - 4a
^{2}b^{2}+ 4ab + 1= (2ab)^{2 }+ 2(2ab)(1) + (1)^{2}= (2ab + 1)^{2} - 4x
^{2}+ 9y^{2}+ 25z^{2}+ 12xy + 30yz + 20zx= (2x)^{2}+ (3y)^{2}+ (5z)^{2}+ 2(2x)(3y) + 2(3y)(5z) + 2(5z)(2x)= (2x + 3y + 5z)^{2} - m
^{4}– 8m^{2}+ 16= (m^{2})^{2}– 2(m^{2})(4) + (4)^{2}= (m^{2}– 4)^{2} - 64m
^{3}– mn^{2 }= m(64m^{2}– n^{2}) = m[(8m)^{2}– (n)^{2}] = m(8m + n)(8m – n) - 4x
^{2}+ 4x + 1 – y^{2}+ 10y – 25= (4x^{2}+ 4x +1 ) – (y^{2}– 10y + 25)= (2x + 1)^{2}– (y – 5)^{2 }= m^{2 }– n^{2}(Taking 2x + 1 = m and y – 5 = n)= (m + n)(m – n)

= [(2x + 1) + (y – 5)][(2x + 1) – (y – 5)] (Putting back m = 2x + 1 and n = y – 5)

= (2x + 1 + y – 5)(2x + 1 – y + 5)

= (2x + y – 4)(2x – y + 6) - 4x
^{4}– 12x^{2}+ 9= (2x^{2})^{2 }– 2(2x^{2})(3) + (3)^{2}= (2x^{2 }– 3)^{2}

**Solution 3A(1):**

**Solution 3A(2):**

**Solution 3B(1):**

**Solution 3B(2):**

**Solution 3B(3):**

**Solution 3B(4):**

**solution 4(1):**

**Solution 4(2):**

**Solution 4(3):**

**Solution 5(1):**

**Solution 5(2):**

**Solution 5(3):**

**Solution 5(4):**