factorize
i.ax^3-ax^2-x+1
ii.a^2-(p+q)a+pq
Answers
Answer:
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)
3. Factorize:
(i) x2 + xy + 8x + 8y
(ii) 15xy – 6x + 5y – 2
(iii) ax + bx – ay – by
(iv) 15pq + 15 + 9q + 25p
(v) z – 7 + 7xy – xyz
Sol. (i) (i) x2 + xy + 8x + 8y = (x2 + xy) + (8x + 8y) = x(x + y) + 8(x + y)
= (x + y)(x + 8)
[Taking (x + y) common]
(ii) 15xy – 6x + 5y – 2
= 3 × 5 × x × y – 3 × 2 × x + 5 × y – 2
= 3x(5y – 2) + 1(5y – 2)
= (5y – 2)(3x + 1)
(iii) ax + bx – ay – by = (ax + bx) – (ay + by)
[Grouping the terms]
= (a + b)x – (a + b)y
= (a + b)(x – y)
[Taking (a + b) common]
(iv) (iv) 15pq + 15 + 9q + 25p
= 15pq + 9q + 25p + 15
= 3 × 5 × p × q + 3 × 3 × q + 5 × 5 × p + 3 × 5
= 3q(5p + 3) + 5(5p + 3)
= (5p + 3)(3q + 5)
(v) z – 7 + 7xy – xyz
= z – 7 – xyz + 7xy
= 1(z – 7) – xy(z – 7)
= (z – 7)(1 – xy)
[Taking z – 7 common]
Answer:
Hi friend ! I love anime too . Did you watch the ending in season 2 in promised neverland . I love Norman Emma and Ray so much . Seeing them wants me to try harder on my works and I just want to be like them , fearless ,hardwork, detemination, friendly etc ...
Step-by-step explanation: