Class 8 Factorization Exercise 7.3

\begin{array}{l} \text {Q1. Factorize } \quad 6 x(2 x-y)+7 y(2 x-y) \\ \\\text {Sol. We have,} \\ 6 x(2 x-y)+7 y(2 x-y) \\ = (6 x+7 y)(2 x-y) \quad [\text {By taking (2 x-y) as common}] \\ \\\text {Q2. Factorize } \quad 2 r(y-x)+s(x-y)\\ \\\text {Sol. We have } \\ 2 r(y-x)+s(x-y) \\ = -2 r(x-y)+s(x-y) \quad [\text {By taking (-1) as common}] \\ = (x-y)(s-2 r) \quad [\text {By taking (x-y) as common}] \\ \\\text {Q3. Factorize } \quad 7 a(2 x-3)+3 b(2 x-3)\\ \\\text {Sol. We have } \\ 7 a(2 x-3)+3 b(2 x-3) \\ = (7 a+3 b)(2 x-3) \quad [\text {By taking (2 x-3) as common}] \\ \\\text {Q4. Factorize } \quad 9 a(6 a-5 b)-12 a^{2}(6 a-5 b)\\ \\\text {Sol. We have,} \\ 9 a(6 a-5 b)-12 a^{2}(6 a-5 b) \\ =\left(9 a-12 a^{2}\right)(6 a-5 b) \quad [\text {By taking (6 a-5 b) as common}] \\ =(3\times 3 a -3a \times 4a )(6 a-5 b) \\ = 3 a(3-4 a)(6 a-5 b) \quad [\text {By taking (3a) as common}] \\ \\\text {Q5. Factorize } \quad 5(x-2 y)^{2}+3(x-2 y)\\ \\\text {Sol. We have,} \\ 5(x-2 y)^{2}+3(x-2 y) \\ =5(x-2 y)(x-2 y)+3(x-2 y) \\ = (x-2 y)[5(x-2 y)+3] \quad [\text {By taking (x-2 y) as common}] \\ =(x-2 y)(5 x-10 y+3) \\ \\\text {Q6. Factorize } \quad 16(2l-3 m)^{2}-12(3 m-2l)\\ \\\text {Sol. We have,} \\16(2l-3 m)^{2}-12(3 m-2l) \\ =16(2l-3 m)^{2}+12(2l-3 m) \quad [\text {By taking (-1) as common from (3m-2l)}] \\ =4 \times 4 (2l-3 m)(2l-3 m) + 12(2l-3 m) \\ =4(2l-3 m)[4(2l-3 m)+3] \quad [\text {By taking 4(2l-3m) as common}] \\ =4(2l-3 m)(8l-12 m+3) \\ \\\text {Q7. Factorize } \quad 3 a(x-2 y)-b(x-2 y)\\ \\\text {Sol. We have } \\ 3 a(x-2 y)-b(x-2 y) \\ =(x-2 y)(3 a-b) \quad [\text {By taking (x-2 y) as common}] \\ \\\text {Q8. Factorize } \quad a^{2}(x+y)+b^{2}(x+y)+c^{2}(x+y)\\ \\\text {Sol. We have} \\ a^{2}(x+y)+b^{2}(x+y)+c^{2}(x+y) \\ = (a^{2}+b^{2}+c^{2})(x+y) \quad [\text {By taking (x+ y) as common}] \\ \\\text {Q9. Factorize } \quad (x-y)^{2}+(x-y)\\ \\\text {Sol. We have} \\ (x-y)^{2}+(x-y) \\ = (x-y)(x-y) + (x-y) \\ = (x-y)(x-y+1) \quad [\text {By taking (x-y) as common}] \\ \\\text {Q10. Factorize } \quad 6(a+2 b)-4(a+2 b)^{2}\\ \\\text {Sol. we have} \\ 6(a+2 b)-4(a+2 b)^{2} \\ = 2 \times 3 (a +2b) – 2 \times 2 (a+2b)(a+2b) \\ =2(a+2 b)[3-2(a+2b)] \quad [\text {By taking 2(a+2 b) as common}] \\ =2(a+2 b)(3-2 a-4 b) \\\\\text {Q11. Factorize } \quad a $(x-y)+2 b(y-x)+c(x-y)^{2}\\ \\\text {Sol. We have } \\ a(x-y)+2 b(y-x)+c(x-y)^{2} \\ = a(x-y) -2 b(x-y)+c(x-y)(x-y) \\ = (x-y)[a-2 b+c(x-y)] \quad [\text {By taking (x-y) as common}] \\ =(x-y)(a-2 b+c x-c y) \\ \\\text {Q12. Factorize } \quad -4(x-2 y)^{2}+8(x-2 y)\\ \\\text {Sol. We have} \\ -4(x-2 y)^{2}+8(x-2 y) \\ = 4(x-2 y)[-(x-2 y)+2] \quad [\text {By taking 4(x-2 y) as common}] \\ =4(x-2 y)(-x+2 y+2) \\ \\\text {Q13. Factorize } \quad x^{3}(a-2 b)+x^{2}(a-2 b)\\ \\\text {Sol. We have } \\ x^{3}(a-2 b)+x^{2}(a-2 b) \\ = x^{2}(a-2 b)[x +1] \quad [\text {By taking x^{2}(a-2 b) as common}] \\ \\\text {Q14. Factorize } \quad (2 x-3 y)(a+b)+(3 x-2 y)(a+b)\\ \\\text {Sol. We have} \\ (2 x-3 y)(a+b)+(3 x-2 y)(a+b) \\ = (a + b) [2 x-3 y + 3 x-2 y ] \quad [\text {By taking (a + b) as common}] \\ = (a+b)[5 x-5 y] \\ = 5(a+b)(x-y) \quad [\text {By taking 5 as common}] \\\\\text {Q15. Factorize } \quad 4(x+y)(3 a-b)+6(x+y)(2 b-3 a)\\ \\\text {Sol. We have } \\ 4(x+y)(3 a-b)+6(x+y)(2 b-3 a) \\ =2\times 2(x+y)(3 a-b) + 2\times 3(x+y)(2 b-3 a) \\ =2(x+y)[2(3a-b) + 3(2b-3a)] \quad [\text {By taking 2(x+y) as common}] \\ =2(x+y)[6a-2 b+6 b-9 a] \\ =2(x+y)(4 b-3 a)\\ \end{array}