Class 8 Factorization Exercise 7.4

\begin{array}{l} \text {Q1. Factorize the expression} \quad qr-p r+q s-p s\\ \\\text {Sol. We have, } \\ q r-p r+q s-p s \\ =q r+q s-p r-ps \quad [\text {By re-arranging the terms}] \\ =q(r+s)-p(r+s) \\ =(r+s)(q-p) \quad [\text {By Taking (r+s) as common}] \\ \\\text {Q2. Factorize the expression} \quad p^{2} q-p r^{2}-p q+r^{2}\\\\\text {Sol. We have, } \\ p^{2} q-p r^{2}-p q+r^{2} \\ =p^{2} q-p q-p r^{2}+r^{2} \quad [\text {By re-arranging the terms}] \\ =p q(p-1)-r^{2}(p-1) \\ =(p-1)(p q-r^{2}) \quad [\text {By Taking (p-1) as common}] \\\\\text {Q3. Factorize the expression} \quad 1+x+x y+x^{2} y\\\\\text {Sol. We have, } \\ 1+x+x y+x^{2} y \\ =1(1+x)+x y(1+x) \\ =(1+x)(1+x y) \quad [\text {By Taking (1+x) as common}] \\ \\\text {Q4. Factorize the expression} \quad a x+a y-b x-b y\\ \\\text {Sol. We have, } \\ a x+a y-b x-b y \\ = a(x+y)-b(x+y) \\ =(x+y)(a-b) \quad [\text {By Taking (x+y) as common}] \\ \\\text {Q5. Factorize the expression} \quad x a^{2}+x b^{2}-y a^{2}-y b^{2}\\ \\ \text {Sol. We have, } \\ x a^{2}+x b^{2}-y a^{2}-y b^{2} \\ =x(a^{2}+b^{2})-y(a^{2}+b^{2}) \\ =(a^{2}+b^{2})(x-y) \quad [\text {By Taking } (a^{2}+b^{2}) \text { as common}] \\ \\ \text {Q6. Factorize the expression} \quad x^{2}+x y+x z+y z\\ \\\text {Sol. We have, } \\ x^{2}+x y+x z+y z \\ =x(x+y)+z(x+y) \\ =(x+y)(x+z) \quad [\text {By Taking (x+y) as common}] \\ \\\text {Q7. Factorize the expression} \quad 2 a x+b x+2 a y+b y\\ \\\text {Sol. We have, } \\ 2 a x+b x+2 a y+b y \\ =2 a x+2 a y+b x+b y \quad [\text {By re-arranging the terms}] \\ =2 a(x+y)+b(x+y) \\ =(x+y)(2 a+b) \quad [\text {By Taking (x+y) as common}] \\ \\\text {Q8. Factorize the expression} \quad a b-b y-a y+y^{2} \\ \\ \text {Sol. We have, } \\ a b-b y-a y+y^{2} \\ =ab – ay – by +y^{2} \quad [\text {By re-arranging the terms}] \\ =a(b-y)-y(b-y) \\ =(b-y)(a-y) \quad [\text {By Taking (b-y) as common}] \\ \\ \text {Q9. Factorize the expression} \quad a x y+b c x y-a z-b c z\\ \\ \text {Sol. We have, } \\ axy + bcxy – az -bcz \\ =axy – az + bcxy – bcz \quad [\text {By re-arranging the terms}] \\ =a(xy-z)+bc(xy-z) \\ =(xy-z)(a+b c) \quad [\text {By Taking (x y-z) as common}] \\ \\ \text {Q10. Factorize the expression} l {m}^{2} – m {n}^{2} – lm + {n}^{2} \\ \\ \text {Sol. We have, } \\ l {m}^{2} – m {n}^{2} – lm + {n}^{2} \\ =l {m}^{2} – lm – m {n}^{2} + {n}^{2} \quad [\text {By re-arranging and grouping the terms}] \\ =lm(m-1)-n^{2}(m-l) \\ =(m-l)(lm-n^{2}) \quad [\text {By Taking (m-l) as common}] \\ \\ \text {Q11. Factorize the expression} \quad x^{3}-y^{2}+x-x^{2} y^{2} \\ \\ \text {Sol. We have, } \\ x^{3}-y^{2}+x-x^{2} y^{2} \\ =x^{3}+x-y^{2}-x^{2} y^{2} \quad [\text {By re-arranging and grouping the terms}] \\ =x(x^{2} + 1) – y^{2}(1+x^{2}) \\ =(1+x^{2})(x-y^{2}) \quad [\text {By Taking (1+x^{2}) as common}] \\ \\ \text {Q12. Factorize the expression} \quad 6xy + 6 – 9y-4x \\ \\ \text {Sol. We have, } \\ 6xy + 6 – 9y-4x \\ =6xy – 4x – 9y + 6 \quad [\text {By re-arranging and grouping the terms}] \\ =2 x(3 y-2)-3(3 y-2) \\ =(3 y-2)(2 x-3) \quad [\text {By Taking (3 y-2) as common}] \\ \\ \text {Q13. Factorize the expression} \quad x^{2} – 2ax – 2ab + bx \\ \\ \text {Sol. We have, } \\ x^{2} – 2ax – 2ab + bx \\ =x^{2}+b x-2 a x-2 a b \quad [\text {By re-arranging and grouping the terms}] \\ =x(x+b)-2 a(x+b) \\ =(x+b)(x-2 a) \quad [\text {By Taking (x+b) as common}] \\ \\ \text {Q14. Factorize the expression} \quad x^{3}-2 x^{2} y+3 x y^{2}-6 y^{3} \\ \\ \text {Sol. We have, } \\ x^{3}-2 x^{2} y+3 x y^{2}-6 y^{3} \\ =x^{3}+3 x y^{2}-2 x^{2} y-6 y^{3} \quad [\text {By re-arranging and grouping the terms}] \\ =x(x^{2}+3 y^{2})-2 y(x^{2}+3 y^{2}) \\ =(x^{2}+3 y^{2})(x-2 y) \quad [\text {By Taking } (x^{2}+3 y^{2}) \text { as common}] \\ =(x-2 y)(x^{2}+3 y^{2}) \\ \\\text {Q15. Factorize the expression} \quad abx^{2}+(ay-b)x – y \\ \\ \text {Sol. We have, } \\ abx^{2}+(ay-b)x – y \\ =a b x^{2}-a y x-b x-y \\ =a b x^{2}-b x-a y x-y \quad [\text {By re-arranging and grouping the terms}] \\ = bx(a x-1)+y(a x-1) \\ =(b x+y)(a x-1) \quad [\text {By Taking (ax-1) as common}] \\ \\ \text {Q16. Factorize the expression} \quad (ax + by)^{2}+(bx – ay)^{2} \\ \\ \text {Sol. We have, } \\ (ax + by)^{2}+(bx – ay)^{2} \\ =a^{2} x^{2}+b^{2} y^{2}+2 a x b y+b^{2} x^{2}+a^{2} y^{2}-2 a x b y \\ =a^{2} x^{2}+b^{2} y^{2}+b^{2} x^{2}+a^{2} y^{2} \\ =a^{2} x^{2}+a^{2} y^{2}+b^{2} y^{2}+b^{2} x^{2} \quad [\text {By re-arranging and grouping the terms}] \\ =a^{2}(x^{2}+y^{2})+b^{2}(x^{2}+y^{2}) \\ =(a^{2}+b^{2})(x^{2}+y^{2}) \quad [\text {By Taking } (x^{2}+y^{2}) \text { as common}] \\ \\\text {Q17. Factorize the expression} \quad 16(a-b)^{3}-24(a-b)^{2} \\ \\ \text {Sol. We have, } \\ 16(a-b)^{3}-24(a-b)^{2} \\ =8\times 2 \times (a-b)^{3}-8 \times 3 \times (a-b)^{2} \\ =8(a-b)^{2}[2(a-b)-3] \quad [\text {By Taking } 8(a-b)^{2} \text { as common}] \\ =8(a-b)^{2}(2 a-2 b-3)\\ \\ \text {Q18. Factorize the expression} \quad ab(x^{2}+1)+x(a^{2}+b^{2}) \\ \end{array}\begin{array}{l} \text {Sol. We have, } \\ ab(x^{2}+1)+x(a^{2}+b^{2}) \\ =abx^{2}+ab+xa^{2}+xb^{2} \\ =abx^{2} + xa^{2}+a b+x b^{2} \quad [\text {By re-arranging and grouping the terms}] \\ =ax(bx+a)+b(bx+a) \\ =(ax+b)(bx+a) \\ \\\text {Q19. Factorize the expression} \quad a^{2} x^{2}+(a x^{2}+1) x+a \\ \\\text {Sol. We have, } \\ a^{2} x^{2}+(a x^{2}+1) x+a \\ =a^{2} x^{2}+a x^{3}+x+a \\ =a x^{2}(a+x)+1(x+a) \\ =(x+a)(a x^{2}+1)\\ \\\text {Q20. Factorize the expression} \quad a(a – 2b – c)+2bc\\ \\ \text {Sol. We have, } \\ a(a – 2b – c)+2bc\\ =a^{2} – 2ab – ac + 2bc \\ =a(a-2 b)-c(a-2 b) \\ =(a-2b)(a-c) \\ \\ \text {Q21. Factorize the expression} \quad a(a+b-c) – bc \\ \\ \text {Sol. We have, } \\ a(a+b-c) – bc \\ =a^{2}+a b-a c-b c \\ =a(a+b)-c(a+b) \\ =(a+b)(a-c) \\ \\ \text {Q22. Factorize the expression} \quad x^{2} – 11xy – x+11 y\\ \\ \text {Sol. We have, } \\ x^{2} – 11xy – x+11 y\\ =x^{2}-x-11 x y+11 y \quad [\text {By re-arranging and grouping the terms}] \\ =x(x-1)-11 y(x-1) \\ =(x-11 y)(x-1) \\ \\ \text {Q23. Factorize the expression} \quad ab – a – b + 1\\ \\ \text {Sol. We have, } \\ ab – a – b + 1 \\ =a(b-1)-1(b-1) \\ =(b-1)(a-1) \\ \end{array}\begin{array}{l} \text {Q24. Factorize the expression} \quad x^{2} + y – xy – x \\ \\ \end{array}\begin{array}{l} \text {Sol. We have, } \\ x^{2} + y – xy – x \\ =x^{2}-x+y-x y \quad [\text {By re-arranging and grouping the terms}] \\ = x(x-1)-y(x-1) \\ =(x-1)(x-y) \\ \end{array}