RD Sharma Solutions for Class 8 Math Chapter 7 - Factorization (Part-7) | RD Sharma Solutions for Class 8 Mathematics PDF Download

Question 1: Factorize each of the following algebraic expression:
x² + 12x − 45
Answer 1: To factorise x2+12x−45, we will find two numbers p and q such that p+q=12 and pq=−45. 
Now,

15+(−3)=12  
and

15×(−3)=−45 
Splitting the middle term 12x in the given quadratic as −3x+15x, we get: 
x2+12x−45=x2−3x+15x−45 
=(x2−3x)+(15x−45) 
=x(x−3)+15(x−3) 
=(x+15)(x−3)

Question 2: Factorize each of the following algebraic expression:
40 + 3x − x²
Answer 2: We have: 
40+3x−x2 
⇒−(x2−3x−40)  
To factorise (x2−3x−40), we will find two numbers p and q such that p+q=−3 and pq=−40. 
Now,   
5+(−8)=−3  
and 5×(−8)=−40 Splitting the middle term −3x in the given quadratic as 5x−8x, we get: 
40+3x−x²=−(x²−3x−40)
=−(x2+5x−8x−40) 
=−[(x2+5x)−(8x+40)] 
=−[x(x+5)−8(x+5)] 
=−(x−8)(x+5) 
=(x+5)(−x+8) 
Question 3: Factorize each of the following algebraic expression:
a² + 3a − 88
Answer 3: To factorise a2+3a−88, we will find two numbers p and q such that p+q=3 and pq=−88. 
Now,  11+(−8)=3 and 11×(−8)=−88Splitting the middle term 3a in the given quadratic as 11a−8a, we get:a2+3a−88=a2+11a−8a−88 =(a2+11a)−(8a+88) 
=a(a+11)−8(a+11) 
=(a−8)(a+11) 
Question 4: Factorize each of the following algebraic expression:
a² − 14a − 51 
Answer 4: To factorise a2−14a−51, we will find two numbers p and q such that p+q=−14 and pq=−51. 
Now,  
3+(−17)=−14  
and

3×(−17)=−51 
Splitting the middle term −14a in the given quadratic as 3a−17a, we get: 
a2−14a−51=a2+3a−17a−51 
=(a2+3a)−(17a+51) 
=a(a+3)−17(a+3) 
=(a−17)(a+3) 
Question 5: Factorize each of the following algebraic expression:
x² + 14x + 45
Answer 5: To factorise x2+14x+45, we will find two numbers p and q such that p+q=14 and pq=45. 
Now, 9+5=14 and 9×5=45Splitting the middle term 14x in the given quadratic as 9x+5x, we get:x2+14x+45=x2+9x+5x+45 =(x2+9x)+(5x+45) 
=x(x+9)+5(x+9) 
=(x+5)(x+9) 
Question 6: Factorize each of the following algebraic expression:
x² − 22x + 120
Answer 6: To factorise x2−22x+120, we will find two numbers p and q such that p+q=−22 and pq=120. 
Now, (−12)+(−10)=−22 and (−12)×(−10)=120Splitting the middle term −22x in the given quadratic as −12x−10x, we get:x2−22x+12=x2−12x−10x+120 =(x2−12x)+(−10x+120) 
=x(x−12)−10(x−12) 
=(x−10)(x−12) 
Question 7: Factorize each of the following algebraic expression:
x² − 11x − 42
Answer 7: To factorise x2−11x−42, we will find two numbers p and q such that p+q=−11 and pq=−42. 
Now, 
3+(−14)=−22  
and 3×(−14)=42Splitting the middle term −11x in the given quadratic as−14x+3x, we get: x2−11x−42=x2−14x+3x−42 =(x2−14x)+(3x−42)=x(x−14)+3(x−14)=(x+3)(x−14)Question 8: Factorize each of the following algebraic expression:
a² + 2a − 3
Answer 8: To factorise a2+2a−3, we will find two numbers p and q such that p+q=2 and pq=−3. 
Now, 3+(−1)=2 and  3×(−1)=−3Splitting the middle term 2a in the given quadratic as−a+3a, we get:a2+2a−3=a2−a+3a−3 =(a2−a)+(3a−3)=a(a−1)+3(a−1)=(a+3)(a−1)Question 9: Factorize each of the following algebraic expression:
a² + 14a + 48
Answer 9: To factorise a2+14a+48, we will find two numbers p and q such that p+q=14 and pq=48. 
Now, 8+6=14 and 8×6=48Splitting the middle term 14a in the given quadratic as 8a+6a, we get:a2+14a+48=a2+8a+6a+48 =(a2+8a)+(6a+48)=a(a+8)+6(a+8)=(a+6)(a+8)Question 10: Factorize each of the following algebraic expression:
x² − 4x − 21
Answer 10: To factorise x2−4x−21, we will find two numbers p and q such that p+q=−4 and pq=−21. 
Now,3+(−7)=−4 and 3×(−7)=−21Splitting the middle term −4x in the given quadratic as −7x+3x, we get:x2−4x−21=x2−7x+3x−21 =(x2−7x)+(3x−21)=x(x−7)+3(x−7)=(x+3)(x−7)Question 11: Factorize each of the following algebraic expression:
y² + 5y − 36
Answer 11: To factorise y2+5y−36, we will find two numbers p and q such that p+q=5 and pq=−36. 
Now,9+(−4)=5 and 9×(−4)=−36Splitting the middle term 5y in the given quadratic as −4y+9y, we get: y2+5y−36=y2−4y+9y−36 =(y2−4y)+(9y−36)=y(y−4)+9(y−4)=(y+9)(y−4)Question 12: Factorize each of the following algebraic expression:
(a² − 5a)² − 36
Answer 12: (a²−5a)²−36

=(a²−5a)²−6²=[(a2−5a)−6][(a2−5a)+6]=(a2−5a−6)(a2−5a+6) In order to factorise a2−5a−6, we will find two numbers p and q such that p+q=−5 and pq=−6 
Now, (−6)+1=−5 and (−6)×1=−6Splitting the middle term −5 in the given quadratic as −6a+a, we get:a2−5a−6=a2−6a+a−6                   =(a2−6a)+(a−6)                   =a(a−6)+(a−6)                   =(a+1)(a−6) Now, 
In order to factorise a2−5a+6, we will find two numbers p and q such that p+q=−5 and pq=6 
Clearly,(−2)+(−3)=−5 and (−2)×(−3)=6Splitting the middle term −5 in the given quadratic as −2a−3a, we get:a2−5a+6=a2−2a−3a+6=(a2−2a)−(3a−6)=a(a−2)−3(a−2)=(a−3)(a−2)∴ (a2−5a−6)(a2−5a+6)=(a−6)(a+1)(a−3)(a−2)=(a+1)(a−2)(a−3)(a−6)Question 13: Factorize each of the following algebraic expression:
(a + 7)(a − 10) + 16
Answer 13: (a+7)(a−10)+16=a2−10a+7a−70+16=a2−3a−54 To factorise a2−3a−54 , we will find two numbers p and q such that p+q=−3 and pq=−54. 
Now, 6+(−9)=−3 and 6×(−9)=−54Splitting the middle term −3a in the given quadratic as −9a+6a, we get: a2−3a−54=a2−9a+6a−54 =(a2−9a)+(6a−54)=a(a−9)+6(a−9)=(a+6)(a−9)

Question 14: Factorize each of the following quadratic polynomials by using the method of  completing the square:
p² + 6p + 8
Answer 14: p²+6p+8
=p2+6p+(6/2)2−(6/2)2+8    [Adding and subtracting (6/2)2, that is, 32]
=p2+6p+32−32+8=p2+2×p×3+32−9+8=p2+2×p×3+32−1=(p+3)2−12                              [Completing the square]=[(p+3)−1][(p+3)+1]=(p+3−1)(p+3+1)=(p+2)(p+4)Question 15: Factorize each of the following quadratic polynomials by using the method of completing the square:
q2 − 10q + 21
Answer 15: q2−10q+21 
=q2−10q+(10/2)2−(10/2)2+21   [Adding and subtracting (10/2)2, that is, 52] 
=q2−2×q×5+52−52+21=(q−5)2−4                                        [Completing the square]=(q−5)2−22 =[(q−5)−2][(q−5)+2]
=(q−5−2)(q−5+2)
=(q−7)(q−3)
Question 16: Factorize each of the following quadratic polynomials by using the method of completing the square:
4y² + 12y + 5
Answer 16: 4y2+12y+5 
=4(y²+3y+5/4)                                    [Making the coefficient of y²=1]

=4[y2+3y+(3/2)2−3/2)2+5/4]        [Adding and subtracting (3/2)2] 
=4[(y+3/2)²−9/4+5/4]
=4[(y+3/2)2−12]                                  [Completing the square]
 =4[(y+3/2)−1][(y+3/2)+1]
=4(y+3/2 −1)(y+3/2+1)
=4(y+1/2)(y+5/2)
=(2y+1)(2y+5)
Question 17: Factorize each of the following quadratic polynomials by using the method of completing the square:
p² + 6p − 16
Answer 17: p2+6p−16 
=p²+6p+(6/2)2−(6/2)²−16    [Adding and subtracting (6/2)², that is, 32]
=p²+6p+3²−9−16
=(p+3)²−25                                [Completing the square]
=(p+3)²−52
=[(p+3)−5][(p+3)+5]
=(p+3−5)(p+3+5)
=(p−2)(p+8) 
Question 18: Factorize each of the following quadratic polynomials by using the method of completing the square:
x² + 12x + 20
Answer 18: x2+12x+20 
=x²+12x+(12/2)²−(12/2)²+20       [Adding and subtracting (12/2)², that is, 6²]
=x2+12x+62−62+20=(x+6)2−16                                          [Completing the square]
=(x+6)2−42 
=[(x+6)−4][(x+6)+4] 
=(x+6−4)(x+6+4) 
=(x+2)(x+10)