Math, asked by ashish123458633, 9 months ago

Please answer my question ​

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Answered by ashy12
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Answer:

1.

(i) 12x, 36

= (2 × 6 × x), (6 × 6)

The common factor is 6

(ii) 2y, 22xy

= (2 × y), (2 × 11 × x × y)

The common factors are 2, y

(iii) 14pq, 28p²

= (2 × 7 × p × q), (2 × 2 × 7 × p × p × q × q)

The common factors are 2, 7, p, q

(iv) 2x, 3x², 4

= (2 × x), (3 × x × x), (2 × 2)

There are no common factors

(v) 6abc, 24abc², 12a²b

= (6 × a × b × c), (4 × 6 × a × b × c × c), (2 × 6 × a × a × b)

The common factor are 6, a, b

(vi) 16x³, -4x², 32x

= (4 × 4 × x × x × x), (-1 × 4 × x × x), (4 × 8 × x)

The common factors are 4, x

(vii) 10pq, 20qr, 30rp

= (10 × p × q), (2 × 10 × q × r), (3 × 10 × r × p)

The common factor is 10

(viii) 3x², 10x³, 6x²z

= (3 × x × x × y × y × y), (2 × 5 × x × x × x × y × y), (2 × 3 × x × x × y × y × z)

The common factors are x²,

2.

(i) 7x - 42

= 7(x - 6)

(ii) 6p - 12q

= 6(p - 2q)

(iii) 7a² + 14a

= 7a(a + 2)

(iv) -16z + 20z³

= 4z(-4 + 5z²)

(v) 20l²m + 30alm

= 10lm(2l + 3a)

(vi) 5x²y - 15xy²

= 5xy(x - 3y)

(vii) 10a² - 15b² + 20c²

= 5(2a² - 3b² + 4c²)

(viii) -4a² + 4ab - 4ca

= 4a(-a + b - c)

(ix) yz + xy²z + xyz²

= xyz(x + y + z)

(x) ax²y + bxy² + cxyz

= xy(ax + by + cz)

3.

(i) + xy + 8x + 8y

= x(x + y) + 8(x + y)

= (x + y)(x + 8)

(ii) 15xy - 6x + 5y - 2

= 15xy + 5y - 6x - 2

= 5y(3x + 1) - 2(3x + 1)

= (3x + 1)(5y - 2)

I HOPE IT HELPS YOU

MARK AS BRAINLIEST :)

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