Q. 34.2( 140 Votes )

# Factorise.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Answer :

(i) ax^{2} + bx

Here common factor is x

ax^{2} + bx = x(ax + b)

(ii) 7p^{2} + 21q^{2}

Here common factor is 7

7p^{2} + 21q^{2} = 7(p^{2} + 3q^{2})

(iii) 2x^{3} + 2xy^{2} + 2xz^{2}

Here common factor is 2x

2x^{3} + 2xy^{2} + 2xz^{2} = 2x(x^{2} + y^{2} + z^{2})

(iv) am^{2} + bm^{2} + bn^{2} + an^{2}

Here common factor is m^{2} in first two terms and n^{2} in the last two terms

am^{2} + bm^{2} + bn^{2} + an^{2} = m^{2}(a + b) + n^{2}(a + b)

am^{2} + bm^{2} + bn^{2} + an^{2} = (m^{2} + n^{2}) + (a + b)

(v) (lm + l) + m + 1

On opening the bracket, we get

lm + l + m + 1

Here common factor is l in first two terms and 1 in the last two terms

lm + l + m + 1 = l(m + 1) + 1(m + 1)

lm + l + m + 1 = (l + 1) + (m + 1)

(vi) y(y + z) + 9(y + z)

In the brackets y + z is common,

(y + z) (9 + y) = (y + z)(y + 9)

(vii) 5y^{2} - 20y – 8z + 2yz

Taking 5y common in first two pairs and 2z in the last two terms

5y(y - 4)-2z(y - 4) = (y - 4)(5y – 2z)

(viii) 10ab + 4a + 5b + 2

Taking 2a common in first two pairs and 1 in the last two terms

10ab + 4a + 5b + 2 = 2a(5b + 2) + 1(5b + 2)

10ab + 4a + 5b + 2 = (5b + 2) + (2a + 1)

(ix) 6xy – 4y + 6 – 9x

Taking 2y common in first two pairs and -3 in the last two terms

6xy – 4y + 6 – 9x = 2y(3x - 2) - 3(3x - 2)

6xy – 4y + 6 – 9x = 2y(3x - 2) + 3(2 - 3x)For taking (3x - 2) common, we will change the sign of second expression, we get,

6xy – 4y + 6 – 9x = 2y(3x - 2) - 3(3x - 2)

6xy – 4y + 6 – 9x = (3x - 2) + (2y - 3)

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