factorise the following
(a) 1+a+b+c+ab+bc+ca+abc
(b) x^3-x^2-ax+x-a-1
(c) x^2+x+1/4
(d) (1-x^2)(1-y^2)+4xy
(e)a^6-7a^3-8
Answers
Answer:
Factorise the following expressions
(i) 7x – 42
(ii) 6p – 12q
(iii) 7a2 + 14a
(iv) –16z + 20z3
(v) 20l2m + 30alm
(vi) 5x2y – 15xy2
(vii) 10a2 – 15b2 + 20c2
(viii) – 4a2 + 4ab – 4ca
(x) ax2y + bxy2 + cxyz
Sol. (i) Whe have 7x = 7 × x and 42 = 2 × 3 × 7
The two terms have 7 as a common factor
7x – 42 = (7 × x) – 2 × 3 × 7
= 7 × (x – 2 × 3) = 7(x – 6)
(ii) 6p = 2 × 3 × p
12q = 2 × 2 × 3 × q
The common factors are 2 and 3.
∴6p – 12q = (2 × 3 × p) – (2 × 2 × 3 × q)
= 2 × 3[p – (2 × q)]
= 6(p – 2q)
(iii) We have, 7a2 = 7 × a × a
and , 14a = 2 × 7 × a
The two terms have 7 and a as common factors
7a2 + 14a = (7 × a × a) + (2 × 7 × a)
= 7 × a × (a + 2) = 7a(a + 2)
(iv) 16z = 2 × 2 × 2 × 2 × z
20z3 = 2 × 2 × 5 × z × z × z
The common factors are 2, 2, and z.
∴ –16z + 20z3 = –(2 × 2 × 2 × 2 × z)
+ (2 × 2 × 5 × z × z × z
= (2 × 2 × z)[–(2 × 2) + (5 × z × z)]
= 4z(– 4 + 5z2)
(v) We have, 20l2m = 2 × 2 × 5 × l × l × m
and, 30alm = 3 × 2 × 5 × a × l × m
The two terms have 2, 5, l and m as common factors.
∴ 20 l2m+ 30alm = (2 × 2 × 5 × l × l × m)
+ (3 × 2 × 5 × a × l × m)
= 2 × 5 × l × m × (2 × l + 3 × a)
= 10lm(2l + 3a)
(vi) 5x2y = 5 × x × x × y
15xy2 = 3 × 5 × x × y × y
The common factors are 5, x, and y.
∴5x2y – 15xy2 = (5 × x × x × y)
– (3 × 5 × x × y × y)
= 5 × x × y[x – (3 × y)]
= 5xy(x – 3y)
(vii) We have, 10a2 = 2 × 5 × a × a,
15b2 = 3 × 5 × b × b
and 20c2 = 2 × 2 × 5 × c × c
The three terms have 5 as a common factor
10a2 – 15b2 + 20c2 = (2 × 5 × a × a)
– (3 × 5 × b × b) + (2 × 2 × 5 × c × c)
= 5 × (2 × a × a – 3 × b × b + 4 × c × c)
= 5(2a2 – 3b2 + 4c2)
(viii) We have, 4a2 = 2 × 2 × a × a,
4ab = 2 × 2 × a × b
and, 4ca = 2 × 2 × c × a
The three terms have 2, 2 and a as common factors
∴–4a2 + 4ab – 4ca = – (2 × 2 × a × a)
+ (2 × 2 × a × b) – (2 × 2 × c × a)
= 2 × 2 × a × (–a + b – c)
= 4a(–a + b – c)
(ix) x2yz = x × x × y × z
xy2z = x × y × y × z
xyz2 = x × y × z × z
The common factors are x, y, and z.
∴ x2yz + xy2z + xyz2 = (x × x × y × z) + (x × y × y × z) + (x × y × z × z)
= xyz(x + y + z)
(x) We have, ax2y = a × x × x × y
bxy2 = b × x × y × y
and, cxyz = c × x × y × z
The three terms have x andy as common factors.
ax2y + bxy2 + cxyz = (a × x × x × y)
+ (b × x × y × y) + (c × x × y × z)
= x × y × (a × x + b × y + c × z)
= xy(ax + by + cz)