Math, asked by omaimaansari15, 1 month ago

factorize
i.ax^3-ax^2-x+1
ii.a^2-(p+q)a+pq

Answers

Answered by thanvi13
2

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]

Answered by Preetha2006
2

Answer:

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