Question

Factorise the expressions and divide them as directed. 

1. $(y^ 2 + 7y + 10) ÷ (y + 5)$ 
2.  $(m^2 – 14m – 32) ÷ (m + 2)$ 
3.  $(5p^ 2 – 25p + 20) ÷ (p – 1)$ 
4.  $4yz(z ^2 + 6z – 16) ÷ 2y(z + 8)$
5.  $5pq(p^ 2 – q^ 2 ) ÷ 2p(p + q)$ 
6.  $12xy(9x^ 2 – 16y ^2 ) ÷ 4xy(3x + 4y)$ 
7.  $39y^ 3 (50y ^2 – 98) ÷ 26y ^2 (5y + 7)$

Answer 1

(i) $(y^2+ 7y + 10) = y^2+ 2y + 5y + 10$

= $y (y + 2) + 5 (y + 2)$

= $(y + 2) (y + 5)$

$\Rightarrow\frac{(y2+7y+10)}{(y+5)}=\frac{(y+5)(y+2)}{(y+5)}=y+2$

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(ii) $m^2 - 14m - 32 = m^2 + 2m - 16m - 32$ 

= $m (m + 2) - 16 (m + 2)$

= $(m + 2) (m - 16)$

$\Rightarrow\frac{(m^2-14m-32)}{(m+2)}=\frac{(m+2)(m-16)}{(m+2)}=m-16$

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(iii) $5p^ 2 - 25p + 20 = 5(p ^2 - 5p + 4)$ 

= $5[p^ 2 - p - 4p + 4]$ 

= $5[p(p - 1) - 4(p - 1)]$ 

= $5(p - 1) (p - 4)$

$\Rightarrow\frac{(5p^2-25p+20)}{(p-1)}=\frac{5(p-1)(p-4)}{(p-1)}=5(p-4)$

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(iv) $4yz(z^ 2 + 6z - 16) = 4yz [z ^2 - 2z + 8z - 16]$ 

= $4yz [z(z - 2) + 8(z - 2)]$ 

= $4yz(z - 2) (z + 8)$

$\frac{4yz(z^2+6z-16)}{2y(z+8)}=\frac{4yz(z-2)(z+8)}{2y(z+8)}=2z(z-2)$

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(v) $5pq(p^ 2 - q^ 2 ) = 5pq (p - q) (p + q)$ 

$\frac{5pq(p^2-q^2)}{2p(p+q)}=\frac{5pq(p-q)(p+q)}{2p(p+q)}=\frac{5}{2q(p-q)}$

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(vi) $12xy(9x ^2 - 16y^ 2 ) = 12xy[(3x)^ 2 - (4y)^ 2 ] = 12xy(3x - 4y) (3x + 4y)$ 

$\frac{12xy(9x^2-16y^2)}{4xy(3x+4y)}$

=$\frac{2 ×2×3×x×y×(3x-4y)×(3x+4y)}{2×2×x×y×(3x+4y)}$

= $3(3x-4y)$

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(vii) $39y^ 3 (50y^ 2 - 98)$ 

= $3 × 13 × y × y × y × 2[(25y ^2 - 49)]$ 

= $3 × 13 × 2 × y × y × y × [(5y) ^2 - (7)^2 ]$ 

= $3 × 13 × 2 × y × y × y (5y - 7) (5y + 7)$ 

$\Rightarrow26y^ 2 (5y + 7) = 2 × 13 × y × y × (5y + 7)$ 

$\Rightarrow\frac{39y^ 3 (50y^ 2 - 98)}{26y 2 (5y + 7)}$ 

= $\frac{39y^3× 2(25y^2-49)}{26y^2(5y+7)}$

= $\frac{3y(5y+7)(5y-7)}{(5y+7)}$

= $3y(5y-7)$