ML Aggarwal Solutions for Class 9 Maths Chapter 4 Factorization are available here. Thorough understanding and study of the Class 9 ML Aggarwal Solutions will help students to prepare comprehensively for their upcoming Class examinations, and in turn, ace the exams. Our topmost experts have answered these questions in such a way that students can understand clearly, without any doubts. We suggest students download ML Aggarwal Solutions for Class 9 Maths Chapter 4 offline as well.
Chapter 4 – Factorization is when you break a number down into smaller numbers, that multiplied together, give you that original number. When you split a number into its factors or divisors, it is called factorization. In ML Aggarwal Solutions for Class 9 Maths Chapter 4, students are going to learn to solve different types of problems and topics of factorization.
ML Aggarwal Solutions for Class 9 Maths Chapter 4 Factorization
Access answers to ML Aggarwal Solutions for Class 9 Maths Chapter 4 Factorization
Exercise 4.1
Factorise the following (1 to 9):
1. (i) 8xy3 + 12x2y2
Solution:-
8xy3 + 12x2y2
Take out common in both terms,
Then, 4xy2 (2y + 3x)
Therefore, HCF of 8xy3 and 12x2y2 is 4xy2.
(ii) 15 ax3 – 9ax2
Solution:-
15 ax3 – 9ax2
Take out common in both terms,
Then, 3ax2 (5x – 3)
Therefore, HCF of 15 ax3 and 9ax2 is 3ax2.
2.
(i) 21py2 – 56py
Solution:-
21py2 – 56py
Take out common in both terms,
Then, 7py (3y – 8)
Therefore, HCF of 21py2 and 56py is 7py.
(ii) 4x3 – 6x2
Solution:-
4x3 – 6x2
Take out common in both terms,
Then, 2x2 (2x – 3)
Therefore, HCF of 4x3 and 6x2 is 2x2.
3.
(i) 2πr2 – 4πr
Solution:-
2πr2 – 4πr
Take out common in both terms,
Then, 2πr (r – 2)
Therefore, HCF of 2πr2 and 4πr is 2πr.
(ii) 18m + 16n
Solution:-
18m + 16n
Take out common in both terms,
Then, 2 (9m – 8n)
Therefore, HCF of 18m and 16n is 2.
4.
(i) 25abc2 – 15a2b2c
Solution:-
25abc2 – 15a2b2c
Take out common in both terms,
Then, 5abc (5c – 3ab)
Therefore, HCF of 25abc2 and 15a2b2c is 5abc.
(ii) 28p2q2r – 42pq2r2
Solution:-
28p2q2r – 42pq2r2
Take out common in both terms,
Then, 14pq2r (2p – 3r)
Therefore, HCF of 28p2q2r and 42pq2r2 is 14pq2r.
5.
(i) 8x3 – 6x2 + 10x
Solution:-
8x3 – 6x2 + 10x
Take out common in both terms,
Then, 2x(4x2 – 3x + 5)
Therefore, HCF of 8x3, 6x2 and 10x is 2x.
(ii) 14mn + 22m – 62p
Solution:-
14mn + 22m – 62p
Take out common in both terms,
Then, 2 (7mn + 11m – 31p)
Therefore, HCF of 14mn, 22m and 62p is 2.
6.
(i) 18p2q2 – 24pq2 + 30p2q
Solution:-
18p2q2 – 24pq2 + 30p2q
Take out common in both terms,
Then, 6pq (3pq – 4q + 5p)
Therefore, HCF of 18p2q2, 24pq2 and 30p2q is 6pq.
(ii) 27a3b3 – 18a2b3 + 75a3b2
Solution:-
27a3b3 – 18a2b3 + 75a3b2
Take out common in both terms,
Then, 3a2b2 (9a – 6b + 25a)
Therefore, HCF of 27a3b3, 18a2b3 and 75a3b2 is 3a2b2.
7.
(i) 15a (2p – 3q) – 10b (2p – 3q)
Solution:-
15a (2p – 3q) – 10b (2p – 3q)
Take out common in both terms,
Then, 5(2p – 3q) [3a – 2b]
Therefore, HCF of 15a (2p – 3q) and 10b (2p – 3q) is 5(2p – 3q).
(ii) 3a(x2 + y2) + 6b (x2 + y2)
Solution:-
3a(x2 + y2) + 6b (x2 + y2)
Take out common in all terms,
Then, 3(x2 + y2) (a + 2b)
Therefore, HCF of 3a(x2 + y2) and 6b (x2 + y2) is 3(x2 + y2).
8.
(i) 6(x + 2y)3 + 8(x + 2y)2
Solution:-
6(x + 2y)3 + 8(x + 2y)2
Take out common in all terms,
Then, 2(x + 2y)2 [3(x + 2y) + 4]
Therefore, HCF of 6(x + 2y)3 and 8(x + 2y)2 is 2(x + 2y)2.
(ii) 14(a – 3b)3 – 21p(a – 3b)
Solution:-
14(a – 3b)3 – 21p(a – 3b)
Take out common in all terms,
Then, 7(a – 3b) [2(a – 3b)2 – 3p]
Therefore, HCF of 14(a – 3b)3 and 21p(a – 3b) is 7(a – 3b).
9.
(i) 10a(2p + q)3 – 15b (2p + q)2 + 35 (2p + q)
Solution:-
10a(2p + q)3 – 15b (2p + q)2 + 35 (2p + q)
Take out common in all terms,
Then, 5(2p + q) [2a (2p + q)2 – 3b (2p + q) + 7]
Therefore, HCF of 10a(2p + q)3, 15b (2p + q)2 and 35 (2p + q) is 5(2p + q).
(ii) x(x2 + y2 – z2) + y(-x2 – y2 + z2) – z (x2 + y – z2)
Solution:-
x(x2 + y2 – z2) + y(-x2 – y2 + z2) – z (x2 + y – z2)
Take out common in all terms,
Then, (x2 + y2 – z2) [x – y – z]
Therefore, HCF of x(x2 + y2 – z2), y(-x2 – y2 + z2) and z (x2 + y – z2) is (x2 + y2 – z2)
Exercise 4.2
Factorise the following (1 to 13):
1.
(i) x2 + xy – x – y
Solution:-
x2 + xy – x – y
Take out common in all terms,
x(x + y) – 1(x + y)
(x + y) (x – 1)
(ii) y2 – yz – 5y + 5z
Solution:-
y2 – yz – 5y + 5z
Take out common in all terms,
y(y – z) – 5(y – z)
(y – z) (y – 5)
2.
(i) 5xy + 7y – 5y2 – 7x
Solution:-
5xy – 7x – 5y2 + 7y
Take out common in all terms,
x(5y – 7) – y(5y – 7)
(5y – 7) (x – y)
(ii) 5p2 – 8pq – 10p + 16q
Solution:-
5p2 – 8pq – 10p + 16q
Take out common in all terms,
p(5p – 8q) – 2(5p – 8q)
(5p – 8q) (p – 2)
3.
(i) a2b – ab2 + 3a – 3b
Solution:-
a2b – ab2 + 3a – 3b
Take out common in all terms,
ab(a – b) + 3(a – b)
(a – b) (ab + 3)
(ii) x3 – 3x2 + x – 3
Solution:-
x3 – 3x2 + x – 3
Take out common in all terms,
x2 (x – 3) + 1(x – 3)
(x – 3) (x2 + 1)
4.
(i) 6xy2 – 3xy – 10y + 5
Solution:-
6xy2 – 3xy – 10y + 5
Take out common in all terms,
3xy(2y – 1) – 5(2y – 1)
(2y – 1) (3xy – 5)
(ii) 3ax – 6ay – 8by + 4bx
Solution:-
3ax – 6ay – 8by + 4bx
Take out common in all terms,
3a(x – 2y) + 4b (x – 2y)
(x – 2y) (3a + 4b)
5.
(i) 1 – a – b + ab
Solution:-
1 – a – b + ab
Take out common in all terms,
1(1 – a) – b(1 – a)
(1 – a) (1 – b)
(ii) a(a – 2b – c) + 2bc
Solution:-
a(a – 2b – c) + 2bc
Above question can be written as,
a2 – 2ab – ac + 2bc
Take out common in all terms,
a(a – 2b) – c(a + 2b)
(a – 2b) (a – c)
6.
(i) x2 + xy (1 + y) + y3
Solution:-
x2 + xy (1 + y) + y3
Above question can be written as,
x2 + xy + xy2 + y3
Take out common in all terms,
x(x + y) + y2(x + y)
(x + y) (x + y2)
(ii) y2 – xy (1 – x) – x3
Solution:-
y2 – xy (1 – x) – x3
Above question can be written as,
y2 – xy + x2y – x3
Take out common in all terms,
y(y – x) + x2 (y – x)
(y – x) (y + x2)
7.
(i) ab2 + (a – 1)b – 1
Solution:-
ab2 + (a – 1)b – 1
Above question can be written as,
ab2 + ab – b – 1
Take out common in all terms,
ab(b + 1) – 1(b + 1)
(b + 1) (ab – 1)
(ii) 2a – 4b – xa + 2bx
Solution:-
2a – 4b – xa + 2bx
Take out common in all terms,
2(a – 2b) – x(a – 2b)
(a – 2b) (2 – x)
8.
(i) 5ph – 10qk + 2rph – 4qrk
Solution:-
5ph – 10qk + 2rph – 4qrk
Re-arranging the given question we get,
5ph + 2rph – 10qk – 4qrk
Take out common in all terms,
ph(5 + 2r) – 2qk(5 + 2r)
(5 + 2r) (ph – 2qk)
(ii) x2 – x(a + 2b) + 2ab
Solution:-
x2 – x(a + 2b) + 2ab
Above question can be written as,
x2 – xa – 2xb + 2ab
Take out common in all terms,
x(x – a) – 2b(x – a)
(x – a) (x – 2b)
9.
(i) ab(x2 + y2) – xy(a2 + b2)
Solution:-
ab(x2 + y2) – xy(a2 + b2)
Above question can be written as,
abx2 + aby2 – xya2 – xyb2
Re-arranging the above we get,
abx2 – xyb2 + aby2 – xya2
Take out common in all terms,
bx(ax – by) + ay(by – ax)
bx(ax – by) – ay (ax – by)
(ax – by) (bx – ay)
(ii) (ax + by)2 + (bx – ay)2
Solution:-
By expanding the give question, we get,
(ax)2 + (by)2 + 2axby + (bx)2 + (ay)2 – 2bxay
a2x2 + b2y2 + b2x2 + a2y2
Re-arranging the above we get,
a2x2 + a2y2 + b2y2 + b2x2
Take out common in all terms,
a2 (x2 + y2) + b2 (x2 + y2)
(x2 + y2) (a2 + b2)
10.
(i) a3 + ab(1 – 2a) – 2b2
Solution:-
a3 + ab(1 – 2a) – 2b2
Above question can be written as,
a3 + ab – 2a2b – 2b2
Re-arranging the above we get,
a3 – 2a2b + ab – 2b2
Take out common in all terms,
a2(a – 2b) + b(a – 2b)
(a – 2b) (a2 + b)
(ii) 3x2y – 3xy + 12x – 12
Solution:-
3x2y – 3xy + 12x – 12
Take out common in all terms,
3xy(x – 1) + 12(x – 1)
(x – 1) (3xy + 12)
11. a2b + ab2 –abc – b2c + axy + bxy
Solution:-
a2b + ab2 –abc – b2c + axy + bxy
Re-arranging the above we get,
a2b – abc + axy + ab2 – b2c + bxy
Take out common in all terms,
a(ab – bc + xy) + b(ab – bc + xy)
(a + b) (ab – bc + xy)
12. ax2 – bx2 + ay2 – by2 + az2 – bz2
Solution:-
ax2 – bx2 + ay2 – by2 + az2 – bz2
Re-arranging the above we get,
ax2 + ay2 + az2 – bx2 – by2 – bz2
Take out common in all terms,
a(x2 + y2 + z2) – b(x2 + y2 + z2)
(x2 + y2 + z2) (a – b)
13. x – 1 – (x – 1)2 + ax – a
Solution:-
x – 1 – (x – 1)2 + ax – a
By expanding the above we get,
X – 1 – (x2 + 1 – 2x) + ax – a
x – 1 – x2 -1 + 2x + ax – a
2x – x2 + ax – 2 + x – a
Take out common in all terms,
x(2 – x + a) – 1(2 – x + a)
(2 – x + a) (x – 1)
Exercise 4.3
Factorise the following (1 to 17):
1. 4x2 – 25y2
Solution:-
We know that, a2 – b2 = (a + b) (a – b)
So, (2x)2 – (5y)2
Then, (2x + y) (2x – 5y)
(ii) 9x2 – 1
Solution:-
We know that, a2 – b2 = (a + b) (a – b)
So, (3x)2 – 12
Then, (3x + 1) (3x – 1)
2.
(i) 150 – 6a2
Solution:-
150 – 6a2
Take out common in all terms,
6(25 – a2)
6(52 – a2)
We know that, a2 – b2 = (a + b) (a – b)
So, 6(5 + a) (5 – a)
(ii) 32x2 – 18y2
Solution:-
32x2 – 18y2
Take out common in all terms,
2(16x2 – 9y2)
2((4x)2 – (3y)2)
We know that, a2 – b2 = (a + b) (a – b)
2(4x + 3y) (4x – 3y)
3.
(ii) (x – y)2 – 9
Solution:-
(x – y)2 – 9
(x – y)2 – 32
We know that, a2 – b2 = (a + b) (a – b)
(x – y + 3) (x – y – 3)
(ii) 9(x + y)2 – x2
Solution:-
9[(x + y)2 – x2]
We know that, a2 – b2 = (a + b) (a – b)
9[(x + y + x) (x + y – x)]
So, 9(2x + y) y
9y(2x + y)
4.
(i) 20x2 – 45y2
Solution:-
20x2 – 45y2
Take out common in all terms,
5(4x2 – 9y2)
5((2x)2 – (3y)2)
We know that, a2 – b2 = (a + b) (a – b)
5(2x + 3y) (2x – 3y)
(ii) 9x2 – 4(y + 2x)2
Solution:-
9x2 – 4(y + 2x)2
Above question can be written as,
(3x)2 – [2(y + 2x)]2
(3x)2 – (2y + 4x)2
We know that, a2 – b2 = (a + b) (a – b)
(3x + 2y + 4x) (3x – 2y – 4x)
(7x + 2y) (-x – 2y)
5.
(i) 2(x – 2y)2 – 50y2
Solution:-
2(x – 2y)2 – 50y2
Take out common in all terms,
2[(x – 2y)2 – 25y2]
2[(x – 2y)2 – (5y)2]
We know that, a2 – b2 = (a + b) (a – b)
2[(x – 2y + 5y) (x – 2y – 5y)]
2[(x + 3y) (x – 7y)]
2(x + 3y) (x – 7y)
(ii) 32 – 2(x – 4)2
Solution:-
32 – 2(x – 4)2
Take out common in all terms,
2[16 – (x – 4)2]
2[42– (x – 4)2]
We know that, a2 – b2 = (a + b) (a – b)
2[(4 + x – 4) (4 – x + 4)]
2[(x) (8 – x)]
2x (8 – x)
6.
(i) 108a2 – 3(b – c)2
Solution:-
108a2 – 3(b – c)2
Take out common in all terms,
3[36a2 – (b – c)2]
3[(6a)2 – (b – c)2]
We know that, a2 – b2 = (a + b) (a – b)
3[(6a + b – c) (6a – b + c)]
(ii) πa5 – π3ab2
Solution:-
πa5 – π3ab2
Take out common in all terms,
πa(a4 – π2b2)
πa((a2)2 – (πb)2)
We know that, a2 – b2 = (a + b) (a – b)
πa(a2 + πb) (a2 – πb)
7.
(i) 50x2 – 2(x – 2)2
Solution:-
50x2 – 2(x – 2)2
Take out common in all terms,
2[25x2 – (x – 2)2]
2[(5x)2 – (x – 2)2]
We know that, a2 – b2 = (a + b) (a – b)
2[(5x + x – 2) (5x – x + 2)]
2[(6x – 2) (4x + 2)]
2(6x – 2) (4x + 2)
(ii) (x – 2)(x + 2) + 3
Solution:-
We know that, a2 – b2 = (a + b) (a – b)
(x2 – 22) + 3
X2 – 4 + 3
X2 – 1
Then,
(x + 1) (x – 1)
8.
(i) x – 2y – x2 + 4y2
Solution:-
x – 2y – x2 + 4y2
x – 2y – (x2 + (2y)2)
We know that, a2 – b2 = (a + b) (a – b)
x – 2y – [(x + 2y) (x – 2y)]
Take out common in all terms,
(x – 2y) (1 – (x + 2y))
(x – 2y) (1 – x – 2y)
(ii) 4a2 – b2 + 2a + b
Solution:-
4a2 – b2 + 2a + b
(2a)2 – b2 + 2a + b
We know that, a2 – b2 = (a + b) (a – b)
((2a + b) (2a – b)) + 1(2a + b)
Take out common in all terms,
(2a + b) (2a – b + 1)
9.
(i) a(a – 2) – b(b – 2)
Solution:-
a(a – 2) – b(b – 2)
Above question can be written as,
a2 – 2a – b2 – 2b
Rearranging the above terms, we get,
a2 – b2 – 2a – 2b
We know that, a2 – b2 = (a + b) (a – b)
[(a + b)(a – b)] – 2(a – b)Take out common in all terms,
(a – b) (a + b – 2)
(ii) a(a – 1) – b(b – 1)
Solution:-
a(a – 1) – b(b – 1)
Above question can be written as,
a2 – a – b2 + b
Rearranging the above terms, we get,
a2 – b2 – a + b
We know that, a2 – b2 = (a + b) (a – b)
[(a + b) (a – b)] – 1 (a – b)Take out common in all terms,
(a – b) (a + b – 1)
10.
(i) 9 – x2 + 2xy – y2
Solution:-
9 – x2 + 2xy – y2
9 – x2 + 2xy – y2
Above terms can be written as,
9 – x2 + xy + xy – y2
Now,
9 – x2 + xy + 3x – 3x + 3y – 3y + xy – y2
Rearranging the above terms, we get,
9 – 3x + 3y + 3x – x2 + xy + xy – 3y – y2
Take out common in all terms,
3(3 – x + y) + x(3 – x + y) + y (-3 – y + x)
3(3 – x + y) + x(3 – x + y) – y(3 – x + y)
(3 – x + y) (3 + x – y)
(ii) 9x4 – (x2 + 2x + 1)
Solution:-
9x4 – (x2 + 2x + 1)
Above terms can be written as,
(3x2)2 – (x + 1)2 … [because (a + b)2 = a2 + 2ab + b2]
We know that, a2 – b2 = (a + b) (a – b)
So, (3x2 + x + 1) (3x2 – x – 1)
11.
(i) 9x4 – x2 – 12x – 36
Solution:-
9x4 – x2 – 12x – 36
Above terms can be written as,
9x4 – (x2 + 12x + 36)
We know that, (a + b)2 = a2 + 2ab + b2
(3x2)2 – (x2 + (2 × 6 × x) + 62)
So, (3x2)2 – (x + 6)2
We know that, a2 – b2 = (a + b) (a – b)
(3x2 + x + 6) (3x2 – x – 6)
(ii) x3 – 5x2 – x + 5
Solution:-
x3 – 5x2 – x + 5
Take out common in all terms,
x2(x – 5) – 1(x – 5)
(x – 5) (x2 – 1)
(x – 5) (x2 – 12)
We know that, a2 – b2 = (a + b) (a – b)
(x – 5) (x + 1) (x – 1)
12.
(i) a4 – b4 + 2b2 – 1
Solution:-
a4 – b4 + 2b2 – 1
Above terms can be written as,
a4 – (b4 – 2b2 + 1)
We know that, (a – b)2 = a2 – 2ab + b2
a4 – ((b2)2) – (2 × b2 × 1) + 12)
(a2)2 – (b2 – 1)2
We know that, a2 – b2 = (a + b) (a – b)
(a2 + b2 – 1) (a2 – b2 + 1)
(ii) x3 – 25x
Solution:-
x3 – 25x
Take out common in all terms,
x(x2 – 25)
Above terms can be written as,
x(x2 – 52)
We know that, a2 – b2 = (a + b) (a – b)
x(x + 5) (x – 5)
13.
(i) 2x4 – 32
Solution:-
2x4 – 32
Take out common in all terms,
2(x4 – 16)
Above terms can be written as,
2((x2)2 – 42)
We know that, a2 – b2 = (a + b) (a – b)
2(x2 + 4) (x2 – 4)
2(x2 + 4) (x2 – 22)
2(x2 + 4) (x + 2) (x – 2)
(ii) a2(b + c) – (b + c)3
Solution:-
a2(b + c) – (b + c)3
Take out common in all terms,
(b + c) (a2 – (b + c)2)
We know that, a2 – b2 = (a + b) (a – b)
(b + c) (a + b + c) (a – b – c)
14.
(i) (a + b)3 – a – b
Solution:-
(a + b)3 – a – b
Above terms can be written as,
(a + b)3 – (a + b)
Take out common in all terms,
(a + b) [(a + b)2 – 1]
(a + b) [(a + b)2 – 12]
We know that, a2 – b2 = (a + b) (a – b)
(a + b) (a + b + 1) (a + b – 1)
(ii) x2 – 2xy + y2 – a2 – 2ab – b2
Solution:-
x2 – 2xy + y2 – a2 – 2ab – b2
Above terms can be written as,
(x2 – 2xy + y2) – (a2 + 2ab + b2)
We know that, (a + b)2 = a2 + 2ab + b2 and (a – b)2 = a2 – 2ab + b2
(x2 – (2 × x × y) + y2) – (a2 + (2 × a × b) + b2)
(x – y)2 – (a + b)2
We know that, a2 – b2 = (a + b) (a – b)
[(x – y) + (a + b)] [(x – y) – (a + b)](x – y + a + b) (x – y – a – b)
15.
(i) (a2 – b2) (c2 – d2) – 4abcd
Solution:-
(a2 – b2) (c2 – d2) – 4abcd
a2(c2 – d2) – b2 (c2 – d2) – 4abcd
a2c2 – a2d2 – b2c2 + b2d2 – 4abcd
a2c2 + b2d2 – a2d2 – b2c2 – 2abcd – 2abcd
Rearranging the above terms, we get,
a2c2 + b2d2 – 2abcd – a2d2 – b2c2 – 2abcd
We know that, (a + b)2 = a2 + 2ab + b2 and (a – b)2 = a2 – 2ab + b2
(ac – bd)2 – (ad – bc)2
(ac – bd + ad – bc) (ac – bd – ad + bc)
(ii) 4x2 – y2 – 3xy + 2x – 2y
Solution:-
4x2 – y2 – 3xy + 2x – 2y
Above terms can be written as,
x2 + 3x2 – y2 – 3xy + 2x – 2y
Rearranging the above terms, we get,
(x2 – y2) + (3x2 – 3xy) + (2x – 2y)
We know that, a2 – b2 = (a + b) (a – b) and take out common terms,
(x + y) (x – y) + 3x(x – y) + 2(x – y)
(x – y) [(x + y) + 3x + 2]
(x – y) (x + y + 3x + 2)
(x – y) (4x + y + 2)
16.
(i) x2 + 1/x2 – 11
Solution:-
x2 + 1/x2 – 11
Above terms can be written as,
x2 + (1/x2) – 2 – 9
Then, (x2 + (1/x2) – 2) – 32
We know that, (a – b)2 = a2 – 2ab + b2,
(x2 – (2 × x2 × (1/x2)) + (1/x)2)
(x – 1/x)2 – 32
We know that, a2 – b2 = (a + b) (a – b)
(x – 1/x + 3) (x – 1/x – 3)
(ii) x4 + 5x2 + 9
Solution:-
x4 + 5x2 + 9
x4 + 6x2 – x2 + 9
(x4 + 6x2 + 9) – x2
((x2)2 + (2 × x2 × 3) + 32)
We know that, (a + b)2 = a2 + 2ab + b2,
((x2)2 + (2 × x2 × 3) + 32)
So, (x2 + 3)2 – x2
We know that, a2 – b2 = (a + b) (a – b)
(x2 + 3 + x) (x2 + 3 – x)
17.
(i) a4 + b4 – 7a2b2
Solution:-
a4 + b4 – 7a2b2
Above terms can be written as,
a4 + b4 + 2a2b2 – 9a2b2
We know that, (a + b)2 = a2 + 2ab + b2,
[(a2)2 + (b2)2 + (2 × a2 × b2)] – (3ab)2(a2 + b2)2 – (3ab)2
We know that, a2 – b2 = (a + b) (a – b)
(a2 + b2 + 3ab) (a2 + b2 – 3ab)
(ii) x4 – 14x2 + 1
Solution:-
x4 – 14x2 + 1
Above terms can be written as,
x4 + 2x2 + 1 – 16x2
We know that, (a + b)2 = a2 + 2ab + b2,
So, [(x2)2 + (2 × x2 × 1) + 12] – 16x2
(x2 + 1)2 – (4x)2
We know that, a2 – b2 = (a + b) (a – b)
(x2 + 1 + 4x) (x2 + 1 – 4x)
18. Express each of the following as the difference of two squares:
(i) (x2 – 5x + 7) (x2 + 5x + 7)
Solution:-
(x2 – 5x + 7) (x2 + 5x + 7)
Rearranging the above terms, we get,
((x2 + 7) – 5x) ((x2 + 7) + 5x)
As, we know that, a2 – b2 = (a + b) (a – b)
So, (x2 + 7)2 – (5x)2
(x2 + 7)2 -25x2
(ii) (x2 – 5x + 7) (x2 – 5x – 7)
Solution:-
(x2 – 5x + 7) (x2 – 5x – 7)
[(x2 – 5x) + 7) ((x2 – 5x) – 7)As, we know that, a2 – b2 = (a + b) (a – b)
(x2 – 5x)2 – 72
(x2 – 5x)2 – 49
(iii) (x2 + 5x – 7) (x2 – 5x + 7)
Solution:-
(x2 + 5x – 7) (x2 – 5x + 7)
[x2 + (5x – 7)] [x2 – (5x – 7)]As, we know that, a2 – b2 = (a + b) (a – b)
x2 – (5x – 7)2
We know that, (a – b)2 = a2 – 2ab + b2,
X2 – [(5x)2 – (2 × 5x × 7) + 72]
X2 – (25x2 – 70x + 49)
X2 – 25x2 + 70x – 49
-24x2 + 70x – 49
19. Evaluate the following by using factors:
(i) (979)2 – (21)2
(ii) (99.9)2 – (0.1)2
Solution:
(i) (979)2 – (21)2
We know that
= (979 + 21) (979 – 21)
So we get
= 1000 ×958
= 958000
(ii) (99.9)2 – (0.1)2
We know that
= (99.9 + 0.1) (99.9 – 0.1)
So we get
= 100 × 99.8
= 9980
Exercise 4.4
Factorise the following (1 to 18):
1.
(i) x2 + 5x + 6
Solution:-
x2 + 5x + 6
x2 + 3x + 2x + 6
Take out common in all terms we get,
x(x + 3) + 2 (x + 3)
(x + 3) (x + 2)
(ii) x2 – 8x + 7
Solution:-
x2 – 8x + 7
x2 – 7x – x + 7
Take out common in all terms we get,
x(x – 7) – 1(x – 7)
(x – 7) (x – 1)
2.
(i) x2 + 6x – 7
Solution:-
x2 + 6x – 7
x2 + 7x – x – 7
Take out common in all terms we get,
x(x + 7) – 1(x + 7)
(x + 7) (x – 1)
(ii) y2 + 7y – 18
Solution:-
y2 + 7y – 18
y2 + 9y – 2y – 18
Take out common in all terms we get,
y(y + 9) – 2(y + 9)
(y + 9) (y – 2)
3.
(i) y2 – 7y – 18
Solution:-
y2 – 7y – 18
y2 + 2y – 9y – 18
Take out common in all terms we get,
y(y + 2) – 9(y + 2)
(y + 2) (y – 9)
(ii) a2 – 3a – 54
Solution:-
a2 – 3a – 54
a2 + 6a – 9a – 54
Take out common in all terms we get,
a(a + 6) – 9(a + 6)
So, (a + 6) (a – 9)
4.
(i) 2x2 – 7x + 6
Solution:-
2x2 – 7x + 6
2x2 – 4x – 3x + 6
Take out common in all terms we get,
2x(x – 2) – 3(x – 2)
(x – 2) (2x – 3)
(ii) 6x2 + 13x – 5
Solution:-
6x2 + 13x – 5
6x2 + 15x – 2x – 5
Take out common in all terms we get,
3x(2x + 5) – 1(2x + 5)
(2x + 5) (3x – 1)
5.
(i) 6x2 + 11x – 10
Solution:-
6x2 + 11x – 10
6x2 + 15x – 4x – 10
Take out common in all terms we get,
3x(2x + 5) – 2(2x + 5)
(2x + 5) (3x – 2)
(ii) 6x2 – 7x – 3
Solution:-
6x2 – 7x – 3
6x2 – 9x + 2x – 3
Take out common in all terms we get,
3x(2x – 3) + 1(2x – 3)
(2x – 3) (3x + 1)
6.
(i) 2x2 – x – 6
Solution:-
2x2 – x – 6
2x2 – 4x + 3x – 6
Take out common in all terms we get,
2x(x – 2) + 3(x – 2)
(x – 2) (2x + 3)
(ii) 1 – 18y – 63y2
Solution:-
1 – 18y – 63y2
1 – 21y + 3y – 63y2
Take out common in all terms we get,
1(1 – 21y) + 3y(1 – 21y)
(1 – 21y) (1 + 3y)
7.
(i) 2y2 + y – 45
Solution:-
2y2 + y – 45
2y2 + 10y – 9y – 45
Take out common in all terms we get,
2y (y + 5) – 9(y + 5)
(y + 5) (2y – 9)
(ii) 5 – 4x – 12x2
Solution:-
5 – 4x – 12x2
5 – 10x + 6x – 12x2
Take out common in all terms we get,
5(1 – 2x) + 6x(1 – 2x)
(1 – 2x) (5 + 6x)
8.
(i) x(12x + 7) – 10
Solution:-
x(12x + 7) – 10
Above terms can be written as,
12x2 + 7x – 10
12x2 + 15x – 8x – 10
Take out common in all terms we get,
3x(4x + 5) – 2(4x + 5)
(4x + 5) (3x – 2)
(ii) (4 – x)2 – 2x
Solution:-
(4 – x)2 – 2x
We know that, (a – b)2 = a2 – 2ab + b2
So, (42 – (2 × 4 × x) + x2) – 2x
16 – 8x + x2 – 2x
x2 – 10x + 16
x2 – 8x – 2x + 16
Take out common in all terms we get,
x(x – 8) – 2(x – 8)
(x – 8) (x -2)
9.
(i) 60x2 – 70x – 30
Solution:-
60x2 – 70x – 30
Take out common in all terms we get,
10(6x2 – 7x – 3)
10(6x2 – 9x + 2x – 3)
Again, take out common in all terms we get,
10(3x(2x – 3) + 1(2x – 3))
10(2x – 3) (3x + 1)
(ii) x2 – 6xy – 7y2
Solution:-
x2 – 6xy – 7y2
x2 – 7xy + xy – 7y2
Take out common in all terms we get,
x(x – 7y) + y(x – 7y)
(x – 7y) (x + y)
10.
(i) 2x2 + 13xy – 24y2
Solution:-
2x2 + 13xy – 24y2
2x2 + 16xy – 3xy – 24y2
Take out common in all terms we get,
2x(x + 8y) – 3y(x + 8y)
(x + 8y) (2x – 3y)
(ii) 6x2 – 5xy – 6y2
Solution:-
6x2 – 5xy – 6y2
6x2 – 9xy + 4xy – 6y2
Take out common in all terms we get,
3x(2x – 3y) + 2y (2x – 3y)
(2x – 3y) (3x + 2y)
11.
(i) 5x2 + 17xy – 12y2
Solution:-
5x2 + 17xy – 12y2
5x2 + 20xy – 3xy – 12y2
Take out common in all terms we get,
5x(x + 4y) – 3y(x + 4y)
(x + 4y) (5x – 3y)
(ii) x2y2 – 8xy – 48
Solution:-
x2y2 – 8xy – 48
x2y2 – 12xy + 4xy – 48
Take out common in all terms we get,
xy(xy – 12) + 4(xy – 12)
(xy – 12) (xy + 4)
12.
(i) 2a2b2 – 7ab – 30
Solution:-
2a2b2 – 7ab – 30
2a2b2 – 12ab + 5ab – 30
Take out common in all terms we get,
2ab(ab – 6) + 5 (ab – 6)
(ab – 6) (2ab + 5)
(ii) a(2a – b) – b2
Solution:-
a(2a – b) – b2
Above terms can be written as,
2a2 – ab – b2
2a2 – 2ab + ab – b2
Take out common in all terms we get,
2a(a – b) + b(a – b)
(a – b) (2a + b)
13.
(i) (x – y)2 – 6(x – y) + 5
Solution:-
(x – y)2 – 6(x – y) + 5
Above terms can be written as,
(x – y)2 – 5(x – y) – (x – y) + 5
(x – y) (x – y – 5) – 1(x – y – 5)
Then,
(x – y – 5) (x – y – 1)
(ii) (2x – y)2 – 11(2x – y) + 28
Solution:-
(2x – y)2 – 11(2x – y) + 28
Above terms can be written as,
(2x – y)2 – 7(2x – y) – 4(2x – y) + 28
(2x – y) (2x – y – 7) – 4(2x – y – 7)
(2x – y – 7) (2x – y – 4)
14.
(i) 4(a – 1)2 – 4(a – 1) – 3
Solution:-
4(a – 1)2 – 4(a – 1) – 3
Above terms can be written as,
4(a – 1)2 – 6(a – 1) + 2(a – 1) – 3
Take out common in all terms we get,
2(a – 1) [2(a – 1) – 3] + 1[2(a – 1) – 3]
(2(a – 1) – 3) (2(a – 1) + 1)
(2a – 2 – 3) (2a – 2 + 1)
(2a – 5) (2a – 1)
(ii) 1 – 2a – 2b – 3(a + b)2
Solution:-
1 – 2a – 2b – 3(a + b)2
Above terms can be written as,
1 – 2(a + b) – 3(a + b)2
1 – 3(a + b) + (a + b) – 3(a + b)2
Take out common in all terms we get,
1(1 – 3(a + b)) + (a + b) (1 – (a + b))
(1 – 3(a + b)) (1 + (a + b))
(1 – 3a + 3b) (1 + a + b)
15.
(i) 3 – 5a – 5b – 12(a + b)2
Solution:-
3 – 5a – 5b – 12(a + b)2
Above terms can be written as,
3 – 5(a + b) – 12(a + b)2
3 – 9(a + b) + 4(a + b) – 12(a + b)2
Take out common in all terms we get,
3(1 – 3(a + b)) + 4(a + b) (1 – 3(a + b))
(1 – 3(a + b)) (3 + 4(a + b))
(1 – 3a – 3b) (3 + 4a + 4b)
(ii) a4 – 11a2 + 10
Solution:-
a4 – 11a2 + 10
Above terms can be written as,
a4 – 10a2 – a2 + 10
Take out common in all terms we get,
a2 (a2 – 10) – 1(a2 – 10)
(a2 – 10) (a2 – 1)
16.
(i) (x + 4)2 – 5xy -20y – 6y2
Solution:-
(x + 4)2 – 5xy -20y – 6y2
Above terms can be written as,
(x + 4)2 – 5y(x + 4) – 6y2
(x + 4)2 – 6y(x + 4) + y(x + 4) – 6y2
Take out common in all terms we get,
(x + 4) (x + 4 – 6y) + y(x + 4 – 6y)
(x – 6y + 4) (x + 4 + y)
(ii) (x2 – 2x2) – 23(x2 – 2x) + 120
Solution:-
(x2 – 2x2) – 23(x2 – 2x) + 120
Above terms can be written as,
(x2 – 2x)2 – 15(x2 – 2x) – 8(x2 – 2x) + 120
Take out common in all terms we get,
(x2 – 2x) (x2 – 2x – 15) – 8(x2 – 2x – 15)
(x2 – 2x – 15) (x2 – 2x – 8)
17. 4(2a – 3)2 – 3(2a – 3) (a – 1) – 7 (a – 1)2
Solution:-
4(2a – 3)2 – 3(2a – 3) (a – 1) – 7 (a – 1)2
Let us assume, 2a – 3 = p and a – 1 = q
So, 4p2 – 3pq – 7q2
Then, 4p2 – 7pq + 4pq – 7q2
Take out common in all terms we get,
P(4p – 7q) + q(4p – 7q)
(4p – 7q) (p + q)
Now, substitute the value of p and q we get,
(4(2a – 3) – 7(a – 1)) (2a – 3 + a – 1)
(8a – 12 – 7a + 7) (3a – 4)
(a – 5) (3a – 4)
18. (2x2 + 5x) (2x2 + 5x – 19) + 84
Solution:-
(2x2 + 5x) (2x2 + 5x – 19) + 84
Let us assume, 2x2 + 5x = p
So, (p) (p – 19) + 84
p2 – 19p + 84
p2 – 12p – 7p + 84
p(p – 12) – 7(p – 12)
(p – 12) (p – 7)
Now, substitute the value of p we get,
(2x2 + 5x – 12) (2x2 + 5x – 7)
Exercise 4.5
Factorise the following (1 to 13):
1.
(i) 8x3 + y3
Solution:-
8x3 + y3
Above terms can be written as,
(2x)3 + y3
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = 2x, b = y
Then, (2x)3 + y3 = (2x + y) ((2x)2 – (2x × y) + y2)
= (2x + y) (4x2 – 2xy + y2)
(ii) 64x3 – 125y3
Solution:-
64x3 – 125y3
Above terms can be written as,
(4x)3 – (5y)3
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
Where, a = 4x, b = 5y
Then, (4x)3 – (5y)3 = (4x – 5y) ((4x)2 + (4x × 5y) + 5y2)
= (4x – 5y) (16x2 + 2oxy + 25y2)
2.
(i) 64x3 + 1
Solution:-
64x3 + 1
Above terms can be written as,
(4x)3 + 13
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = 4x, b = 1
Then, (4x)3 + 13 = (4x + 1) ((4x)2 – (4x × 1) + 12)
= (4x + 1) (16x2 – 4x + 1)
(ii) 7a3 + 56b3
Solution:-
7a3 + 56b3
Take out common in all terms we get,
7(a3 + 8b3)
Above terms can be written as,
7(a3 + (2b)3)
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = a, b = 2b
Then, 7[(a)3 + (2b)3] = 7[(a + 2b) ((a)2 – (a × 2b) + (2b)2)]
= 7(a + 2b) (a2 – 2ab + 4b2)
3.
(i) (x6/343) + (343/x6)
Solution:-
(x6/343) + (343/x6)
Above terms can be written as,
(x2/7)3 + (7/x2)3
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = (x2/7), b = (7/x2)
Then, (x2/7)3 + (7/x2)3 = [(x2/7) + (7/x2)] [(x2/7)2 – ((x2/7) × (7/x2)) + (7/x2)2]
= [(x2/7) + (7/x2)] [(x4/49) – 1 + (49/x4)]
(ii) 8x3 – 1/27y3
Solution:-
8x3 – 1/27y3
Above terms can be written as,
(2x)3 – (1/3y)3
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
Where, a = 2x, b = (1/3y)
Then, (2x)3 – (1/3y)3 = (2x – (1/3y)) ((2x)2 + (2x × (1/3y)) + (3y)2)
= (2x – (1/3y)) (4x2 + (2x/3y) + 9y2)
4.
(i) x2 + x5
Solution:-
x2 + x5
Take out common in all terms we get,
x2(1 + x3)
x2(13 + x3)
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = 1, b = x
= x2 [(1 + x) (12 – (1 × x) + x2)]
= x2 (1 + x) (1 – x + x2)
(ii) 32x4 – 500x
Solution:-
32x4 – 500x
Take out common in all terms we get,
4x(8x3 – 125)
Above terms can be written as,
4x((2x)3 – 53)
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
Where, a = 2x, b = 5
= 4x(2x – 5) ((2x)2 + (2x × 5) + 52)
= 4x(2x – 5) (4x2 + 10x + 25)
5.
(i) 27x3y3 – 8
Solution:-
27x3y3 – 8
Above terms can be written as,
(3xy)3 – 23
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
Where, a = 3xy, b = 2
= (3xy – 2) ((3xy)2 + (3xy × 2) + 22)
= (3xy – 2) (9x2y2 + 6xy + 4)
(ii) 27(x + y)3 + 8(2x – y)3
Solution:-
27(x + y)3 + 8(2x – y)3
Above terms can be written as,
33(x + y)3 + 23(2x – y)3
(3(x + y))3 + (2(x – y))3
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = 3(x + y), b = 2(x – y)
= [3(x + y) + 2(2x – y)] [(3(x + y))3 – (3(x + y) × 2(2x – y)) + (2(2x – y))2]
= [3x + 3y + 4x – 2y] [9(x + y)2 – 6(x + y)(2x – y) + 4(2x – y)2]
= (7x – y) [9(x2 + y2 + 2xy) – 6(2x2 – xy + 2xy – y2) + 4(4x2 + y2 – 4xy)]
= (7x – y) [9x2 + 9y2 + 18xy – 12x2 – 6xy – 6y2 + 16x2 + 4y2 – 16xy]
= (7x – y) [13x2 – 4xy + 19y2]
6.
(i) a3 + b3 + a + b
Solution:-
a3 + b3 + a + b
(a3 + b3) + (a + b)
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
[(a + b) (a2 – ab + b2)] + (a + b)(a + b) (a2 – ab + b2 + 1)
(ii) a3 – b3 – a + b
Solution:-
a3 – b3 – a + b
(a3 – b3) – (a – b)
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
[(a – b) (a2 + ab + b2)] – (a – b)(a – b) (a2 + ab + b2 – 1)
7.
(i) x3 + x + 2
Solution:-
x3 + x + 2
Above terms can be written as,
x3 + x + 1 + 1
Rearranging the above terms, we get
(x3 + 1) (x + 1)
(x3 + 13) (x + 1)
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
[(x + 1) (x2 – x + 1)] + (x + 1)(x + 1) (x2 – x + 1 + 1)
(x + 1) (x2 – x + 2)
(ii) a3 – a – 120
Solution:-
a3 – a – 120
Above terms can be written as,
a3 – a – 125 + 5
Rearranging the above terms, we get
a3 – 125 – a + 5
(a3 – 125) – (a – 5)
(a3 – 53) – (a – 5)
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
[(a – 5) (a2 + 5a + 52)] – (a – 5)(a – 5) (a2 + 5a + 25) – (a – 5)
(a – 5) (a2 + 5a + 25 – 1)
(a – 5) (a2 + 5a + 24)
8.
(i) x3 + 6x2 + 12x + 16
Solution:-
x3 + 6x2 + 12x + 16
x3 + 6x2 + 12x + 8 + 8
Above terms can be written as,
(x3 + (3 × 2 × x2) + (3 × 22 × x) + 23) + 8
We know that, (a + b)3 = a3 + b3 + 3a2b + 3ab2
Now a = x and b = 2
So, (x + 2)3 + 23
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
(x + 2 + 2) ((x + 2)2 – (2 × (x + 2)) + 22)
(x + 4) (x2 + 4 + 4x – 2x – 4 + 4)
(x + 4) (x2 + 2x + 4)
(ii) a3 – 3a2b + 3ab2 – 2b3
Solution:-
a3 – 3a2b + 3ab2 – 2b3
Above terms can be written as,
a3 – 3a2b + 3ab2 – b3 – b3
We know that, (a – b)3 = a3 – b3 – 3a2b + 3ab2
So, (a – b)3 + b3
We also know that, a3 – b3 = (a – b) (a2 + ab + b2)
Where, a = a – b, b = b
(a – b – b) ((a – b)2 + (a – b)b + b2)
(a – 2b) (a2 + b2 – 2ab + ab – b2 + b2)
(a – 2b) (a2 + b2 – ab)
9.
(i) 2a3 + 16b3 – 5a – 10b
Solution:-
2a3 + 16b3 – 5a – 10b
Above terms can be written as,
2(a3 + 8b3) – 5(a + 2b)
2(a3 + (2b)3) – 5(a + 2b)
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
2[(a + 2b) (a2 – 2ab + 4b2)] – 5(a + 2b)
(a + 2b) (2a2 – 4ab + 8b2 – 5)
(ii) a3 – (1/a3) – 2a + 2/a
Solution:-
a3 – (1/a3) – 2a + 2/a
(a3 – (1/a)3) – 2a + 2/a
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
[(a – 1/a) – (a2 + (a × 1/a) + (1/a)2] – 2(a – 1/a)(a – 1/a) (a2 + 1 + 1/a2) – 2(a – 1/a)
(a – 1/a) (a2 + 1 + 1/a2 – 2)
(a – 1/a) (a2 + (1/a2) – 1)
10.
(i) a6 – b6
Solution:-
a6 – b6
Above terms can be written as,
(a2)3 – (b2)3
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
So, a = a2, b = b2
(a2 – b2) ((a2)2) + a2b2 + (b2)2)
(a2 – b2) (a4 + a2b2 + b4)
(ii) x6 – 1
Solution:-
x6 – 1
Above terms can be written as,
(x2)3 – 13
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
So, a = x2, b = 1
(x2 – 1) ((x2)2 + (x2 × 1) + 12)
(x2 – 1) (x4 + x2 + 1)
11.
(i) 64x6 – 729y6
Solution:-
64x6 – 729y6
Above terms can be written as,
(2x)6 – (3y)6
[(2x)2]3 – [(3y)2]3We know that, a3 – b3 = (a – b) (a2 + ab + b2)
So, a = (2x)2, b = (3y)2
[(2x)2 – (3y)2] [((2x)2)2 + ((2x)2× (3y)2) + ((3y)2)2](4x2 – 9y2) [16x4 + (4x2 × 9y2) + (9y2)2]
(4x2 – 9y2) [16x4 + 36x2y2 + 81y4] [(2x)2 – (3y)2] [16x4 + 36x2y2 + 81y4]
(2x + 3y) (2x – 3y) (16x4 + 36x2y2 + 81y4)
(ii) x3 – (8/x)
Solution:-
x3 – (8/x)
Above terms can be written as,
(1/x) (x3 – 8)
(1/x) [(x)3 – (2)3]
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
So, a = x, b = 2
(1/x) (x – 2) (x2 + 2x + 4)
12.
(i) 250 (a – b)3 + 2
Solution:-
250 (a – b)3 + 2
Take out common in all terms we get,
2(125(a – b)3 + 1)
2[(5(a – b))3 + 13]
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
= 2[(5a – 5b + 1) ((5a – 5b)2 – (5a – 5b)1 + 12)]
= 2(5a – 5b + 1) (25a2 + 25b2 – 50ab – 5a + 5b + 1)
(ii) 32a2x3 – 8b2x3 – 4a2y3 + b2y3
Solution:-
32a2x3 – 8b2x3 – 4a2y3 + b2y3
Take out common in all terms we get,
8x3(4a2 – b2) – y3(4a2 – b2)
(4a2 – b2) (8x3 – y3)
Above terms can be written as,
((2a)2 – b2) ((2x)3 – y3)
We know that, a3 – b3 = (a – b) (a2 + ab + b2) and (a2 – b2) = (a + b) (a – b)
(2a + b) (2a – b) [(2x – y) ((2x)2 + 2xy + y2)]
(2a + b) (2a – b) (2x – y) (4x2 + 2xy + y2)
13.
(i) x9 + y9
Solution:-
x9 + y9
Above terms can be written as,
(x3)3 + (y3)3
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = x3, b = y3
(x3 + y3) ((x3)2 – x3y3 + (y3)2)
(x3 + y3) (x6 – x3y3 + y6)
Then, (x3 + y3) in the form of (a3 + b3)
(x + y)(x2 – xy + y2) (x6 – x3y3 + y6)
(ii) x6 – 7x3 – 8
Solution:-
X6 – 7x3 – 8
Above terms can be written as,
(x2)3 – 7x3 – x3 + x3 – 8
(x2)3 – 8x3 + x3 – 23
(((x2)3) – (2x)3) + (x3 – 23)
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
(x2 – 2x) ((x2)2 + (x2 × 2x) + (2x)2) + (x – 2) (x2 + 2x + 22)
(x2 – 2x) (x4 + 2x3 + 4x2) + (x – 2) (x2 + 2x + 4)
x(x – 2) x2(x2 + 2x + 4) + (x – 2) (x2 + 2x + 4)
Take out common in all terms we get,
(x – 2) (x2 + 2x + 4) ((x × x2) + 1)
(x – 2) (x2 + 2x + 4) (x3 + 1)
So, above terms are in the form of a3 + b3
Therefore, (x – 2) (x2 + 2x + 4) (x + 1) (x2 – x + 1)
Chapter test
Factorise the following (1 to 12):
1.
(i) 15(2x – 3)3 – 10(2x – 3)
Solution:-
15(2x – 3)3 – 10(2x – 3)
Take out common in both terms,
Then, 5(2x – 3) [3(2x – 3)2 – 2]
(ii) a(b – c) (b + c) – d(c – b)
Solution:-
a(b – c) (b + c) – d(c – b)
Above terms can be written as,
a(b – c) (b + c) + d(b – c)
Take out common in both terms,
(b – c) [a(b + c) + d]
(b – c) (ab + ac + d)
2.
(i) 2a2x – bx + 2a2 – b
Solution:-
2a2x – bx + 2a2 – b
Rearrange the above terms we get,
2a2x + 2a – bx – b
Take out common in both terms,
2a2(x + 1) – b(x + 1)
(x + 1) (2a2 – b)
(ii) p2 – (a + 2b)p + 2ab
Solution:-
p2 – (a + 2b)p + 2ab
Above terms can be written as,
p2 – ap – 2bp + 2ab
Take out common in both terms,
p(p – a) – 2b(p – a)
(p – a) (p – 2b)
3.
(i) (x2 – y2)z + (y2 – z2)x
Solution:-
(x2 – y2)z + (y2 – z2)x
Above terms can be written as,
zx2 – zy2 + xy2 – xz2
Rearrange the above terms we get,
zx2 – xz2 + xy2 – zy2
Take out common in both terms,
zx(x – z) + y2(x – z)
(x – z) (zx + y2)
(ii) 5a4 – 5a3 + 30a2 – 30a
Solution:-
5a4 – 5a3 + 30a2 – 30a
Take out common in both terms,
5a(a3 – a2 + 6a – 6)
5a[a2(a – 1) + 6(a – 1)]
5a(a – 1) (a2 + 6)
4.
(i) b(c -d)2 + a(d – c) + 3c – 3d
Solution:-
b(c -d)2 + a(d – c) + 3c – 3d
Above terms can be written as,
b(c – d)2 – a(c – d) + 3c – 3d
b(c – d)2 – a(c – d) + 3(c – d)
Take out common in both terms,
(c – d) [b(c – d) – a + 3]
(c – d) (bc – bd – a + 3)
(ii) x3 – x2 – xy + x + y – 1
Solution:-
x3 – x2 – xy + x + y – 1
Rearrange the above terms we get,
x3 – x2 – xy + y + x – 1
Take out common in both terms,
x2(x – 1) – y(x – 1) + 1(x – 1)
(x – 1) (x2 – y + 1)
5.
(i) x(x + z) – y (y + z)
Solution:-
x(x + z) – y (y + z)
x2 + xz – y2 – yz
Rearrange the above terms we get,
x2 – y2 + xz – yz
We know that, (a2 – b2) = (a + b) (a – b)
So, (x + y) (x – y) + z(x – y)
(x – y) (x + y + z)
(ii) a12x4 – a4x12
Solution:-
a12x4 – a4x12
Take out common in both terms,
a4x4 (a8 – x8)
a4x4((a4)2 – (x4)2)
We know that, (a2 – b2) = (a + b) (a – b)
a4x4 (a4 + x4) (a4 – x4)
a4x4 (a4 + x4) ((a2)2 – (x2)2)
a4x4(a4 + x4) (a2 + x2) (a2 – x2)
a4x4 (a4 + x4) (a2 + x2) (a + x) (a – x)
6.
(i) 9x2 + 12x + 4 – 16y2
Solution:-
9x2 + 12x + 4 – 16y2
Above terms can be written as,
(3x)2 + (2 × 3x × 2) + 22 – 16y2
Then, (3x + 2)2 + (4y)2
(3x + 2 + 4y) (3x + 2 – 4y)
(ii) x4 + 3x2 + 4
Solution:-
x4 + 3x2 + 4
Above terms can be written as,
(x2)2 + 3(x2) + 4
(x2)2 + (2)2 + 4x2 – x2
(x2 + 2)2 – (x2)
We know that, (a2 – b2) = (a + b) (a – b)
(x2 + 2 + x) (x2 + 2 – x)
(x2 + x + 2) (x2 – x + 2)
7.
(i) 21x2 – 59xy + 40y2
Solution:-
21x2 – 59xy + 40y2
By multiplying the first and last term we get, 21 × 40 = 840
Then, (-35) × (-24) = 840
So, 21x2 – 35xy – 24xy + 40y2
7x(3x – 5y) – 8y(3x – 5y)
(3x – 5y) (7x – 8y)
(ii) 4x3y – 44x2y + 112xy
Solution:-
4x3y – 44x2y + 112xy
Take out common in all terms,
4xy(x2 – 11x + 28)
Then, 4xy (x2 – 7x – 4x + 28)
4xy [x(x – 7) – 4(x + 7)]
4xy (x – 7) (x – 4)
8.
(i) x2y2 – xy – 72
Solution:-
x2y2 – xy – 72
x2y2 – 9xy + 8xy – 72
Take out common in all terms,
xy(xy – 9) + 8(xy – 9)
(xy – 9) (xy + 8)
(ii) 9x3y + 41x2y2 + 20xy3
Solution:-
9x3y + 41x2y2 + 20xy3
Take out common in all terms,
xy(9x2 + 41xy + y2)
Above terms can be written as,
xy (9x2 + 36xy + 5xy + 20y2)
xy [9x(x + 4y) + 5y(x + 4y)]
xy (x + 4y) (9x + 5y)
9.
(i) (3a – 2b)2 + 3(3a – 2b) – 10
Solution:-
(3a – 2b)2 + 3(3a – 2b) – 10
Let us assume, (3a – 2b) = p
p2 + 3p – 10
p2 + 5p – 2p – 10
Take out common in all terms,
p(p + 5) – 2(p + 5)
(p + 5) (p – 2)
Now, substitute the value of p
(3a – 2b + 5) (3a – 2b – 2)
(ii) (x2 – 3x) (x2 – 3x + 7) + 10
Solution:-
(x2 – 3x) (x2 – 3x + 7) + 10
Let us assume, (x2 – 3x) = q
q (q + 7) + 10
q2 + 7q + 10
q2 + 5q + 2q + 10
q(q + 5) + 2(q + 5)
(q + 5) (q + 2)
Now, substitute the value of q
(x2 – 3x + 5) (x2 – 3x + 2)
10.
(i) (x2 – x) (4x2 – 4x – 5) – 6
Solution:-
(x2 – x) (4x2 – 4x – 5) – 6
(x2 – x) [(4x2 – 4x) – 5] – 6
(x2 – x) [4(x2 – x) – 5] – 6
Let us assume x2 – x = q
So, q[4q – 5] – 6
4q2 – 5q – 6
4q2 – 8q + 3q – 6
4q(q – 2) + 3(q – 2)
(q – 2) (4q + 3)
Now, substitute the value of q
(x2 – x – 2) (4(x2 – x) + 3)
(x2 – x – 2) (4x2 – 4x + 3)
(x2 – 2x + x – 2) (4x2 – 4x + 3)
[x(x – 2) + 1(x – 2)] (4x2 – 4x + 3)(x – 2) (x + 1) (4x2 – 4x + 3)
(ii) x4 + 9x2y2 + 81y4
Solution:-
x4 + 9x2y2 + 81y4
Above terms can be written as,
x4 + 18x2y2 + 81y4 – 9x2y2
((x2)2 + (2 × x2 × 9y2) + (9y2)2) – 9x2y2
We know that, (a + b)2 = a2 + 2ab + b2
(x2 + 9y2)2 – (3xy)2
(x2 + 9y2 + 3xy) (x2 + 9y2 – 3xy)
11.
(i) (8/27)x3 – (1/8)y3
Solution:-
(8/27)x3 – (1/8)y3
Above terms can be written as,
((2/3)x)3 – (½y)3
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
((2/3)x – ½y) [(2/3)x + (2/3)x (1/2)y + ((1/2)y)2]
((2/3)x – (1/2)y) [(4/9)x2 + (xy/3) + (y2/4)]
(ii) x6 + 63x3 – 64
Solution:-
x6 + 63x3 – 64
Above terms can be written as,
x6 + 64x3 – x3 – 64
Take out common in all terms,
x3 (x3 + 64) – 1(x3 + 64)
(x3 + 64) (x3 – 1)
(x3 + 43) (x3 – 13)
We know that, a3 – b3 = (a – b) (a2 + ab + b2) and a3 + b3 = (a + b) (a2 – ab + b2)
So, (x + 4) [x2 – 4x + 42] (x – 1) [x2 + x + 12]
(x + 4) (x2 – 4x + 16) (x – 1) (x2 + x + 1)
12.
(i) x3 + x2 – (1/x2) + (1/x3)
Solution:-
x3 + x2 – (1/x2) + (1/x3)
Rearranging the above terms, we get,
x3 + (1/x3) + x2 – (1/x2)
We know that, a3 – b3 = (a – b) (a2 + ab + b2) and (a2 – b2) = (a + b) (a – b)
(x + 1/x) (x2 – 1 + 1/x2) + (x + 1/x) (x – 1/x)
(x + 1/x) [x2 – 1 + 1/x2 + x – 1/x]
(ii) (x + 1)6 – (x – 1)6
Solution:-
(x + 1)6 – (x – 1)6
Above terms can be written as,
((x + 1)3)2 – ((x – 1)3)2
We know that, (a2 – b2) = (a + b) (a – b)
[(x + 1)3 + (x – 1)3] [(x + 1)3 – (x – 1)3] [(x + 1) + (x – 1)][(x + 1)2 – (x – 1) (x + 1) + (x – 1)2] [(x + 1) – (x – 1)][(x + 1)2 + (x – 1) (x + 1) + (x – 1)2](x + 1 + x – 1) [x2 + 2x + 1 – x2 + 1 + x2 + 1 – 2x(x + 1) – x + 1] [x2 + 2x + 1 + x2 – 1 + x2 – 2x + 1]
By simplifying we get,
2x(x2 + 3) 2(3x2 + 1)
4x(x2 + 3) (3x2 + 1)
13. Show that (97)3 + (14)3 is divisible by 111
Solution:-
From the question,
(97)3 + (14)3
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
So, (97 + 14) [(97)2 – (97 × 14) + (14)2]
111 [(97)2 – (97 × 14) + (14)2]
Therefore, it is clear that the given expression is divisible by 111.
14. If a + b = 8 and ab = 15, find the value of a4 + a2b2 + b4.
Solution:-
a4 + a2b2 + b4
Above terms can be written as,
a4 + 2a2b2 + b4 – a2b2
(a2)2 + 2a2b2 + (b2)2 – (ab)2
(a2 + b2)2 – (ab)2
(a2 + b2 + ab) (a2 + b – ab)
a + b = 8, ab = 15
So, (a + b)2 = 82
a2 + 2ab + b2 = 64
a2 + 2(15) + b2 = 64
a2 + b2 + 30 = 64
By transposing,
a2 + b2 = 64 – 30
a2 + b2 = 34
Then, a4 + a2b2 + b4
= (a2 + b2 + ab) (a2 + b2 – ab)
= (34 + 15) (34 – 15)
= 49 × 19
= 931
