70 Basic Engineering Mathematics
(i) What multiplies a to make ab? Answer: b
(ii) What multiplies a to make −5ac? Answer: −5c
Hence, b − 5c appears in the bracket. Thus,
ab − 5ac = a(b − 5c)
Problem 10. Factorize 2x 2 + 14x y 3
For the numbers 2 and 14, the highest common factor
(HCF) is 2 (i.e. 2 is the largest number that divides into
both 2 and 14).
For the x terms, x 2 and x, the HCF is x.
Thus, the HCF of 2x 2 and 14x y 3 is 2x.
2x is therefore taken outside of the bracket. What goes
inside the bracket?
(i) What multiplies 2x to make 2x 2 ? Answer: x
(ii) What multiplies 2x to make 14x y 3 ? Answer: 7y 3
Hence x + 7y 3 appears inside the bracket. Thus,
2x
2
+ 14xy
3
= 2x(x + 7y
3 )
Problem 11. Factorize 3x
3 y − 12x y
2
+ 15x y
For the numbers 3, 12 and 15, the highest common factor
is 3 (i.e. 3 is the largest number that divides into 3, 12
and 15).
For the x terms, x 3 , x and x, the HCF is x.
For the y terms, y, y 2 and y, the HCF is y.
Thus, the HCF of 3x 3 y and 12x y 2 and 15x y is 3x y.
3x y is therefore taken outside of the bracket. What goes
inside the bracket?
(i) What multiplies 3x y to make 3x 3 y? Answer: x 2
(ii) What multiplies 3x y to make −12x y 2 ? Answer:
−4y
(iii) What multiplies 3x y to make 15x y? Answer: 5
Hence, x 2 − 4y + 5 appears inside the bracket. Thus,
3x
3 y − 12xy
2
+ 15xy = 3xy(x
2
− 4y + 5)
Problem 12. Factorize 25a 2 b 5 − 5a 3 b 2
For the numbers 25 and 5, the highest common factor
is 5 (i.e. 5 is the largest number that divides into 25
and 5).
For the a terms, a 2 and a 3 , the HCF is a 2 .
For the b terms, b 5 and b 2 , the HCF is b 2 .
Thus, the HCF of 25a
2 b
5 and 5a
3 b
2 is 5a
2 b
2 .
5a 2 b 2 is therefore taken outside of the bracket. What
goes inside the bracket?
(i) What multiplies 5a 2 b 2 to make 25a 2 b 5 ? Answer:
5b 3
(ii) What multiplies 5a 2 b 2 to make −5a 3 b 2 ? Answer:
−a
Hence, 5b 3 − a appears in the bracket. Thus,
25a
2 b
5
− 5a
3 b
2
= 5a
2 b
2
(5b
3
− a)
Problem 13. Factorize ax − ay + bx − by
The first two terms have a common factor of a and the
last two terms a common factor of b. Thus,
ax − ay + bx − by = a(x − y) + b(x − y)
The two newly formed terms have a common factor of
(x − y). Thus,
a(x − y) + b(x − y) = (x − y)(a + b)
Problem 14. Factorize 2ax − 3ay + 2bx − 3by
a is a common factor of the first two terms and b a
common factor of the last two terms. Thus,
2ax − 3ay + 2bx − 3by = a(2x − 3y) + b(2x − 3y)
(2x − 3y) is now a common factor. Thus,
a(2x − 3y) + b(2x − 3y) = (2x − 3y)(a + b)
Alternatively, 2x is a common factor of the original
first and third terms and −3y is a common factor of
the second and fourth terms. Thus,
2ax − 3ay + 2bx − 3by = 2x(a + b) − 3y(a + b)
(a + b) is now a common factor. Thus,
2x(a + b) − 3y(a + b) = (a + b)(2x − 3y)
as before
Problem 15. Factorize x 3 + 3x 2 − x − 3
x 2 is a common factor of the first two terms. Thus,
x
3
+ 3x
2
− x − 3 = x
2
(x + 3) − x − 3
−1 is a common factor of the last two terms. Thus,
x
2
(x + 3) − x − 3 = x
2
(x + 3) − 1(x + 3)
(i) What multiplies a to make ab? Answer: b
(ii) What multiplies a to make −5ac? Answer: −5c
Hence, b − 5c appears in the bracket. Thus,
ab − 5ac = a(b − 5c)
Problem 10. Factorize 2x 2 + 14x y 3
For the numbers 2 and 14, the highest common factor
(HCF) is 2 (i.e. 2 is the largest number that divides into
both 2 and 14).
For the x terms, x 2 and x, the HCF is x.
Thus, the HCF of 2x 2 and 14x y 3 is 2x.
2x is therefore taken outside of the bracket. What goes
inside the bracket?
(i) What multiplies 2x to make 2x 2 ? Answer: x
(ii) What multiplies 2x to make 14x y 3 ? Answer: 7y 3
Hence x + 7y 3 appears inside the bracket. Thus,
2x
2
+ 14xy
3
= 2x(x + 7y
3 )
Problem 11. Factorize 3x
3 y − 12x y
2
+ 15x y
For the numbers 3, 12 and 15, the highest common factor
is 3 (i.e. 3 is the largest number that divides into 3, 12
and 15).
For the x terms, x 3 , x and x, the HCF is x.
For the y terms, y, y 2 and y, the HCF is y.
Thus, the HCF of 3x 3 y and 12x y 2 and 15x y is 3x y.
3x y is therefore taken outside of the bracket. What goes
inside the bracket?
(i) What multiplies 3x y to make 3x 3 y? Answer: x 2
(ii) What multiplies 3x y to make −12x y 2 ? Answer:
−4y
(iii) What multiplies 3x y to make 15x y? Answer: 5
Hence, x 2 − 4y + 5 appears inside the bracket. Thus,
3x
3 y − 12xy
2
+ 15xy = 3xy(x
2
− 4y + 5)
Problem 12. Factorize 25a 2 b 5 − 5a 3 b 2
For the numbers 25 and 5, the highest common factor
is 5 (i.e. 5 is the largest number that divides into 25
and 5).
For the a terms, a 2 and a 3 , the HCF is a 2 .
For the b terms, b 5 and b 2 , the HCF is b 2 .
Thus, the HCF of 25a
2 b
5 and 5a
3 b
2 is 5a
2 b
2 .
5a 2 b 2 is therefore taken outside of the bracket. What
goes inside the bracket?
(i) What multiplies 5a 2 b 2 to make 25a 2 b 5 ? Answer:
5b 3
(ii) What multiplies 5a 2 b 2 to make −5a 3 b 2 ? Answer:
−a
Hence, 5b 3 − a appears in the bracket. Thus,
25a
2 b
5
− 5a
3 b
2
= 5a
2 b
2
(5b
3
− a)
Problem 13. Factorize ax − ay + bx − by
The first two terms have a common factor of a and the
last two terms a common factor of b. Thus,
ax − ay + bx − by = a(x − y) + b(x − y)
The two newly formed terms have a common factor of
(x − y). Thus,
a(x − y) + b(x − y) = (x − y)(a + b)
Problem 14. Factorize 2ax − 3ay + 2bx − 3by
a is a common factor of the first two terms and b a
common factor of the last two terms. Thus,
2ax − 3ay + 2bx − 3by = a(2x − 3y) + b(2x − 3y)
(2x − 3y) is now a common factor. Thus,
a(2x − 3y) + b(2x − 3y) = (2x − 3y)(a + b)
Alternatively, 2x is a common factor of the original
first and third terms and −3y is a common factor of
the second and fourth terms. Thus,
2ax − 3ay + 2bx − 3by = 2x(a + b) − 3y(a + b)
(a + b) is now a common factor. Thus,
2x(a + b) − 3y(a + b) = (a + b)(2x − 3y)
as before
Problem 15. Factorize x 3 + 3x 2 − x − 3
x 2 is a common factor of the first two terms. Thus,
x
3
+ 3x
2
− x − 3 = x
2
(x + 3) − x − 3
−1 is a common factor of the last two terms. Thus,
x
2
(x + 3) − x − 3 = x
2
(x + 3) − 1(x + 3)