RD Sharma solution class 8 chapter 7 Factorization Exercise 7.7

Exercise 7.7

Page-7.27

Question 1:

Factorize each of the following algebraic expression:
x² + 12x − 45

Answer 1:

$$\begin{gathered} \mathrm{To} \mathrm{factorise} {\mathrm{x}}^2+12\mathrm{x}-45, \mathrm{we} \mathrm{will} \mathrm{find} \mathrm{two} \mathrm{numbers} \mathrm{p} \mathrm{and} \mathrm{q} \mathrm{such} \mathrm{that} \mathrm{p}+\mathrm{q}=12 \mathrm{and} \mathrm{pq}=-45. \\ \mathrm{Now}, \\ 15+(-3)=12 \\ \mathrm{and} \\ 15\times (-3)=-45 \\ \mathrm{Splitting} \mathrm{the} \mathrm{middle} \mathrm{term} 12\mathrm{x} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{quadratic} \mathrm{as} -3\mathrm{x}+15\mathrm{x}, \mathrm{we} \mathrm{get}: \\ {\mathrm{x}}^2+12\mathrm{x}-45={\mathrm{x}}^2-3\mathrm{x}+15\mathrm{x}-45 \\ =({\mathrm{x}}^2-3\mathrm{x})+(15\mathrm{x}-45) \\ =\mathrm{x}(\mathrm{x}-3)+15(\mathrm{x}-3) \\ =(\mathrm{x}+15)(\mathrm{x}-3) \end{gathered}$$

Question 2:

Factorize each of the following algebraic expression:
40 + 3x − x²

Answer 2:

$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ 40+3\mathrm{x}-{\mathrm{x}}^2 \\ \Rightarrow -({\mathrm{x}}^2-3\mathrm{x}-40) \\ \mathrm{To} \mathrm{factorise} ({\mathrm{x}}^2-3\mathrm{x}-40), \mathrm{we} \mathrm{will} \mathrm{find} \mathrm{two} \mathrm{numbers} \mathrm{p} \mathrm{and} \mathrm{q} \mathrm{such} \mathrm{that} \mathrm{p}+\mathrm{q}=-3 \mathrm{and} \mathrm{pq}=-40. \\ \mathrm{Now}, \\ 5+(-8)=-3 \\ \mathrm{and} \\ 5\times (-8)=-40 \\ \mathrm{Splitting} \mathrm{the} \mathrm{middle} \mathrm{term} -3\mathrm{x} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{quadratic} \mathrm{as} 5\mathrm{x}-8\mathrm{x}, \mathrm{we} \mathrm{get}: \\ 40+3\mathrm{x}-{\mathrm{x}}^2=-({\mathrm{x}}^2-3\mathrm{x}-40) \\ =-({\mathrm{x}}^2+5\mathrm{x}-8\mathrm{x}-40) \\ =-\left[\right({\mathrm{x}}^2+5\mathrm{x})-(8\mathrm{x}+40\left)\right] \\ =-\left[\mathrm{x}\right(\mathrm{x}+5)-8(\mathrm{x}+5\left)\right] \\ =-(\mathrm{x}-8)(\mathrm{x}+5) \\ =(\mathrm{x}+5)(-\mathrm{x}+8) \end{gathered}$$

Question 3:

Factorize each of the following algebraic expression:
a² + 3a − 88

Answer 3:

$$\begin{gathered} \mathrm{To} \mathrm{factorise} {\mathrm{a}}^2+3\mathrm{a}-88, \mathrm{we} \mathrm{will} \mathrm{find} \mathrm{two} \mathrm{numbers} \mathrm{p} \mathrm{and} \mathrm{q} \mathrm{such} \mathrm{that} \mathrm{p}+\mathrm{q}=3 \mathrm{and} \mathrm{pq}=-88. \\ \mathrm{Now}, \\ 11+(-8)=3 \\ \mathrm{and} \\ 11\times (-8)=-88 \\ \mathrm{Splitting} \mathrm{the} \mathrm{middle} \mathrm{term} 3\mathrm{a} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{quadratic} \mathrm{as} 11\mathrm{a}-8\mathrm{a}, \mathrm{we} \mathrm{get}: \\ {\mathrm{a}}^2+3\mathrm{a}-88={\mathrm{a}}^2+11\mathrm{a}-8\mathrm{a}-88 \\ =({\mathrm{a}}^2+11\mathrm{a})-(8\mathrm{a}+88) \\ =\mathrm{a}(\mathrm{a}+11)-8(\mathrm{a}+11) \\ =(\mathrm{a}-8)(\mathrm{a}+11) \end{gathered}$$

Question 4:

Factorize each of the following algebraic expression:
a² − 14a − 51

Answer 4:

$$\begin{gathered} \mathrm{To} \mathrm{factorise} {\mathrm{a}}^2-14\mathrm{a}-51, \mathrm{we} \mathrm{will} \mathrm{find} \mathrm{two} \mathrm{numbers} \mathrm{p} \mathrm{and} \mathrm{q} \mathrm{such} \mathrm{that} \mathrm{p}+\mathrm{q}=-14 \mathrm{and} \mathrm{pq}=-51. \\ \mathrm{Now}, \\ 3+(-17)=-14 \\ \mathrm{and} \\ 3\times (-17)=-51 \\ \mathrm{Splitting} \mathrm{the} \mathrm{middle} \mathrm{term} -14\mathrm{a} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{quadratic} \mathrm{as} 3\mathrm{a}-17\mathrm{a}, \mathrm{we} \mathrm{get}: \\ {\mathrm{a}}^2-14\mathrm{a}-51={\mathrm{a}}^2+3\mathrm{a}-17\mathrm{a}-51 \\ =({\mathrm{a}}^2+3\mathrm{a})-(17\mathrm{a}+51) \\ =\mathrm{a}(\mathrm{a}+3)-17(\mathrm{a}+3) \\ =(\mathrm{a}-17)(\mathrm{a}+3) \end{gathered}$$

Question 5:

Factorize each of the following algebraic expression:
x² + 14x + 45

Answer 5:

$$\begin{gathered} \mathrm{To} \mathrm{factorise} {\mathrm{x}}^2+14\mathrm{x}+45, \mathrm{we} \mathrm{will} \mathrm{find} \mathrm{two} \mathrm{numbers} \mathrm{p} \mathrm{and} \mathrm{q} \mathrm{such} \mathrm{that} \mathrm{p}+\mathrm{q}=14 \mathrm{and} \mathrm{pq}=45. \\ \mathrm{Now}, \\ 9+5=14 \\ \mathrm{and} \\ 9\times 5=45 \\ \mathrm{Splitting} \mathrm{the} \mathrm{middle} \mathrm{term} 14\mathrm{x} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{quadratic} \mathrm{as} 9\mathrm{x}+5\mathrm{x}, \mathrm{we} \mathrm{get}: \\ {\mathrm{x}}^2+14\mathrm{x}+45={\mathrm{x}}^2+9\mathrm{x}+5\mathrm{x}+45 \\ =({\mathrm{x}}^2+9\mathrm{x})+(5\mathrm{x}+45) \\ =\mathrm{x}(\mathrm{x}+9)+5(\mathrm{x}+9) \\ =(\mathrm{x}+5)(\mathrm{x}+9) \end{gathered}$$

Question 6:

Factorize each of the following algebraic expression:
x² − 22x + 120

Answer 6:

$$\begin{gathered} \mathrm{To} \mathrm{factorise} {\mathrm{x}}^2-22\mathrm{x}+120, \mathrm{we} \mathrm{will} \mathrm{find} \mathrm{two} \mathrm{numbers} \mathrm{p} \mathrm{and} \mathrm{q} \mathrm{such} \mathrm{that} \mathrm{p}+\mathrm{q}=-22 \mathrm{and} \mathrm{pq}=120. \\ \mathrm{Now}, \\ (-12)+(-10)=-22 \\ \mathrm{and} \\ (-12)\times (-10)=120 \\ \mathrm{Splitting} \mathrm{the} \mathrm{middle} \mathrm{term} -22\mathrm{x} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{quadratic} \mathrm{as} -12\mathrm{x}-10\mathrm{x}, \mathrm{we} \mathrm{get}: \\ {\mathrm{x}}^2-22\mathrm{x}+12={\mathrm{x}}^2-12\mathrm{x}-10\mathrm{x}+120 \\ =({\mathrm{x}}^2-12\mathrm{x})+(-10\mathrm{x}+120) \\ =\mathrm{x}(\mathrm{x}-12)-10(\mathrm{x}-12) \\ =(\mathrm{x}-10)(\mathrm{x}-12) \end{gathered}$$

Question 7:

Factorize each of the following algebraic expression:
x² − 11x − 42

Answer 7:

$$\begin{gathered} \mathrm{To} \mathrm{factorise} {\mathrm{x}}^2-11\mathrm{x}-42, \mathrm{we} \mathrm{will} \mathrm{find} \mathrm{two} \mathrm{numbers} \mathrm{p} \mathrm{and} \mathrm{q} \mathrm{such} \mathrm{that} \mathrm{p}+\mathrm{q}=-11 \mathrm{and} \mathrm{pq}=-42. \\ \mathrm{Now}, \\ 3+(-14)=-22 \\ \mathrm{and} \\ 3\times (-14)=42 \\ \mathrm{Splitting} \mathrm{the} \mathrm{middle} \mathrm{term} -11\mathrm{x} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{quadratic} \mathrm{as}-14\mathrm{x}+3\mathrm{x}, \mathrm{we} \mathrm{get}: \\ {\mathrm{x}}^2-11\mathrm{x}-42={\mathrm{x}}^2-14\mathrm{x}+3\mathrm{x}-42 \\ =({\mathrm{x}}^2-14\mathrm{x})+(3\mathrm{x}-42) \\ =\mathrm{x}(\mathrm{x}-14)+3(\mathrm{x}-14) \\ =(\mathrm{x}+3)(\mathrm{x}-14) \end{gathered}$$

Question 8:

Factorize each of the following algebraic expression:
a² + 2a − 3

Answer 8:

$$\begin{gathered} \mathrm{To} \mathrm{factorise} {\mathrm{a}}^2+2\mathrm{a}-3, \mathrm{we} \mathrm{will} \mathrm{find} \mathrm{two} \mathrm{numbers} \mathrm{p} \mathrm{and} \mathrm{q} \mathrm{such} \mathrm{that} \mathrm{p}+\mathrm{q}=2 \mathrm{and} \mathrm{pq}=-3. \\ \mathrm{Now}, \\ 3+(-1)=2 \\ \mathrm{and} \\ 3\times (-1)=-3 \\ \mathrm{Splitting} \mathrm{the} \mathrm{middle} \mathrm{term} 2\mathrm{a} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{quadratic} \mathrm{as}-\mathrm{a}+3\mathrm{a}, \mathrm{we} \mathrm{get}: \\ {\mathrm{a}}^2+2\mathrm{a}-3={\mathrm{a}}^2-\mathrm{a}+3\mathrm{a}-3 \\ =({\mathrm{a}}^2-\mathrm{a})+(3\mathrm{a}-3) \\ =\mathrm{a}(\mathrm{a}-1)+3(\mathrm{a}-1) \\ =(\mathrm{a}+3)(\mathrm{a}-1) \end{gathered}$$

Question 9:

Factorize each of the following algebraic expression:
a² + 14a + 48

Answer 9:

$$\begin{gathered} \mathrm{To} \mathrm{factorise} {\mathrm{a}}^2+14\mathrm{a}+48, \mathrm{we} \mathrm{will} \mathrm{find} \mathrm{two} \mathrm{numbers} \mathrm{p} \mathrm{and} \mathrm{q} \mathrm{such} \mathrm{that} \mathrm{p}+\mathrm{q}=14 \mathrm{and} \mathrm{pq}=48. \\ \mathrm{Now}, \\ 8+6=14 \\ \mathrm{and} \\ 8\times 6=48 \\ \mathrm{Splitting} \mathrm{the} \mathrm{middle} \mathrm{term} 14\mathrm{a} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{quadratic} \mathrm{as} 8\mathrm{a}+6\mathrm{a}, \mathrm{we} \mathrm{get}: \\ {\mathrm{a}}^2+14\mathrm{a}+48={\mathrm{a}}^2+8\mathrm{a}+6\mathrm{a}+48 \\ =({\mathrm{a}}^2+8\mathrm{a})+(6\mathrm{a}+48) \\ =\mathrm{a}(\mathrm{a}+8)+6(\mathrm{a}+8) \\ =(\mathrm{a}+6)(\mathrm{a}+8) \end{gathered}$$

Question 10:

Factorize each of the following algebraic expression:
x² − 4x − 21

Answer 10:

$$\begin{gathered} \mathrm{To} \mathrm{factorise} {\mathrm{x}}^2-4\mathrm{x}-21, \mathrm{we} \mathrm{will} \mathrm{find} \mathrm{two} \mathrm{numbers} \mathrm{p} \mathrm{and} \mathrm{q} \mathrm{such} \mathrm{that} \mathrm{p}+\mathrm{q}=-4 \mathrm{and} \mathrm{pq}=-21. \\ \mathrm{Now}, \\ 3+(-7)=-4 \\ \mathrm{and} \\ 3\times (-7)=-21 \\ \mathrm{Splitting} \mathrm{the} \mathrm{middle} \mathrm{term} -4\mathrm{x} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{quadratic} \mathrm{as} -7\mathrm{x}+3\mathrm{x}, \mathrm{we} \mathrm{get}: \\ {\mathrm{x}}^2-4\mathrm{x}-21={\mathrm{x}}^2-7\mathrm{x}+3\mathrm{x}-21 \\ =({\mathrm{x}}^2-7\mathrm{x})+(3\mathrm{x}-21) \\ =\mathrm{x}(\mathrm{x}-7)+3(\mathrm{x}-7) \\ =(\mathrm{x}+3)(\mathrm{x}-7) \end{gathered}$$

Question 11:

Factorize each of the following algebraic expression:
y² + 5y − 36

Answer 11:

$$\begin{gathered} \mathrm{To} \mathrm{factorise} {\mathrm{y}}^2+5\mathrm{y}-36, \mathrm{we} \mathrm{will} \mathrm{find} \mathrm{two} \mathrm{numbers} \mathrm{p} \mathrm{and} \mathrm{q} \mathrm{such} \mathrm{that} \mathrm{p}+\mathrm{q}=5 \mathrm{and} \mathrm{pq}=-36. \\ \mathrm{Now}, \\ 9+(-4)=5 \\ \mathrm{and} \\ 9\times (-4)=-36 \\ \mathrm{Splitting} \mathrm{the} \mathrm{middle} \mathrm{term} 5\mathrm{y} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{quadratic} \mathrm{as} -4\mathrm{y}+9\mathrm{y}, \mathrm{we} \mathrm{get}: \\ {\mathrm{y}}^2+5\mathrm{y}-36={\mathrm{y}}^2-4\mathrm{y}+9\mathrm{y}-36 \\ =({\mathrm{y}}^2-4\mathrm{y})+(9\mathrm{y}-36) \\ =\mathrm{y}(\mathrm{y}-4)+9(\mathrm{y}-4) \\ =(\mathrm{y}+9)(\mathrm{y}-4) \end{gathered}$$

Question 12:

Factorize each of the following algebraic expression:
(a² − 5a)² − 36

Answer 12:

$$\begin{gathered} (a^2-5a{)}^2-36 \\ =(a^2-5a{)}^2-6^2 \\ =\left[\right(a^2-5a)-6]\left[\right(a^2-5a)+6] \\ =(a^2-5a-6)(a^2-5a+6) \\ In order to factorise a^2-5a-6, we will find two numbers p and q such that p+q=-5 and pq=-6 \\ Now, \\ (-6)+1=-5 \\ and \\ (-6)\times 1=-6 \\ Splitting the middle term -5 in the given quadratic as -6a+a, we get: \\ {a}^2-5a-6=a^2-6a+a-6 \\ =(a^2-6a)+(a-6) \\ =a(a-6)+(a-6) \\ =(a+1)(a-6) \\ Now, \\ In order to factorise a^2-5a+6, we will find two numbers p and q such that p+q=-5 and pq=6 \\ Clearly, \\ (-2)+(-3)=-5 \\ and \\ (-2)\times (-3)=6 \\ Splitting the middle term -5 in the given quadratic as -2a-3a, we get: \\ {a}^2-5a+6=a^2-2a-3a+6 \\ =(a^2-2a)-(3a-6) \\ =a(a-2)-3(a-2) \\ =(a-3)(a-2) \\ \therefore (a^2-5a-6)(a^2-5a+6)=(a-6)(a+1)(a-3)(a-2) \\ =(a+1)(a-2)(a-3)(a-6) \end{gathered}$$

Question 13:

Factorize each of the following algebraic expression:
(a + 7)(a − 10) + 16

Answer 13:

$$\begin{gathered} (a+7)(a-10)+16 \\ =a^2-10a+7a-70+16 \\ =a^2-3a-54 \\ To factorise a^2-3a-54 , we will find two numbers p and q such that p+q=-3 and pq=-54. \\ Now, \\ 6+(-9)=-3 \\ and \\ 6\times (-9)=-54 \\ Splitting the middle term -3a in the given quadratic as -9a+6a, we get: \\ a^2-3a-54=a^2-9a+6a-54 \\ =(a^2-9a)+(6a-54) \\ =a(a-9)+6(a-9) \\ =(a+6)(a-9) \end{gathered}$$