Question

Factorise the expression and divide them as directed
(i) $(y^2+7y+10)÷(y+5)$
(ii) $(m^2-14m-32)÷(m+2)$
(iii) $(5p^2-25p+20)÷(p-1)$

(iv) $4yz(z^2+6z-16)÷2y(z+8)$

(v) $\left(5pq\right(p^2-q^2)÷2p(p+q)$

(vi) $12xy(9x^2-16y^2)÷4xy(3x+4y)$

(vii) $39y^3(50y^2-98)÷26y^2(5y+7)$

Answer 1

$\left(i\right)$ $(y^2+7y+10)÷(y+5)$$=\frac{(y^2+5y+2y+10)}{y+5}$

$=\frac{y(y+5)+2(y+5)}{y+5}=\frac{(y+5)(y+2)}{(y+5)}$

$=y+2$

$\left(ii\right)$ $(m^2-14m-32)÷(m+2)$$=\frac{(m^2-16m+2m-32)}{m+2}$

$=\frac{m(m-16)+2(m-16)}{m+2}=\frac{(m-16)(m+2)}{(m+2)}$

$=m-16$

$\left(iii\right)$ $(5p^2-25p+20)÷(p-1)$$=\frac{(5p^2-5p-20p+20)}{p-1}$

$=\frac{5p(p-1)-20(p-1)}{p-1}=\frac{(p-1)(5p-20)}{(p-1)}$

$=5p-20=5(p-4)$

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

$=\left(2z\right)\frac{z(z+8)-2(z+8)}{z+8}=\left(2z\right)\frac{(z-2)(z+8)}{(z+8)}$

$=2z(z-2)$

$\left(v\right)$ $5pq(p^2-q^2)÷2p(p+q)$

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

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

$\left(vi\right)$ $12xy(9x^2-16y^2)÷4xy(3x+4y)$$=\left(12xy\right)\frac{\left[\right(3x{)}^2-\left(4y{)}^2\right]}{\left(4xy\right)(3x+4y)}$

$=\left(3\right)\frac{(3x+4y)(3x-4y)}{(3x+4y)}$

$=3(3x-4y)$

$\left(vii\right)$ $39y^3(50y^2-98)÷26y^2(5y+7)$$=39y^3\times 2\times \frac{(25y^2-49)}{26y^2(5y+7)}$

$=\left(3y\right)\frac{\left[\right(5y{)}^2-7^2]}{5y+7}$ $=\left(3y\right)\frac{\left[\right(5y-7\left)\right(5y+7\left)\right]}{(5y+7)}$

$=3y(5y-7)$