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5.Factorise the expressions 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) \(5pq (p^2 – q^2) ÷2p (p + q)\)

(vi) \(12xy (9x^2 – 16y^2) ÷ 4xy (3x + 4y)\)

(vii) \(39y^3 (50y^2 – 98)÷ 26y^2 (5y + 7)\)

Answer :

(i)We have, \(y^2 +7y +10=y^2 +2y +5y +10\)

\(=y(y+2)+5(y+2)\)

\(=(y+2)(y+5)\)....(1)

∴\(\frac{y^2 +7y +10}{y + 5}=\frac{(y+2)(y+5)}{(y + 5)}\quad\)[Substituting (1)]

\(=y+5\)

(ii)We have, \(m^2 – 14m – 32=m^2 +2m – 16m – 32\)

\(=m(m+2)-16(m+2)\)

\(=(m+2)(m-16)\)....(1)

∴\(\frac{m^2 – 14m – 32}{m + 2}=\frac{(m+2)(m-16)}{(m + 2)}\quad\)[Substituting (1)]

\(=m-16\)

(iii)We have, \(5p^2 – 25p + 20=5(p^2 – 5p +4)\)

\(=5(p^2 -p – 4p +4)\)

\(=5(p(p-1)-4(p-1))\)

\(=5(p-1)(p-4)\)....(1)

∴\(\frac{5p^2 – 25p + 20}{p-1}=\frac{5(p-1)(p-4)}{(p – 1))}\quad\)[Substituting (1)]

\(=5(p-4)\)

(iv)We have, \(4yz (z^2 + 6z – 16)=4yz(z^2 + 8z-2z – 16)\)

\(=4yz(z^2 + 8z-2z – 16)\)

\(=4yz(z(z+8)-2(z+8))\)

\(=4yz(z+8)(z-2)\)....(1)

∴\(\frac{4yz (z^2 + 6z – 16)}{2y (z + 8)}=\frac{4yz(z+8)(z-2)}{2y (z + 8)}\quad\)[Substituting (1)]

\(=\frac{2z(z-2)}1=2z(z-2)\)

(v)We have, \(5pq (p^2 – q^2)=5pq(p+q)(p-q)\)....(1)

∴\(\frac{5pq (p^2 – q^2)}{2p (p + q)}=\frac{5\times p \times q \times (p+q) \times (p-q)}{2p \times (p + q)}\quad\)[Substituting (1)]

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

(vi)We have, \(12xy (9x^2 – 16y^2)=12xy((3x)^2 – (4y)^2)\)

\(=12xy(3x-4y)(3x+4y)\)....(1)

∴\(\frac{12xy (9x^2 – 16y^2)}{4xy (3x + 4y)}=\frac{12xy(3x-4y)(3x+4y)}{4xy (3x + 4y)}\quad\)[Substituting (1)]

\(=3(3x-4y)\)

(vii)We have, \(39y^3 (50y^2 – 98)=39y^3 \times 2(25y^2 – 49\)

\(=78y^2(5y^2-7^2)\)

\(=78y^2(5y+7)(5y-7)\)....(1)

∴\(\frac{39y^3 (50y^2 – 98)}{26y^2 (5y + 7)}=\frac{78y^2(5y+7)(5y-7)}{26y^2 (5y + 7)}\quad\)[Substituting (1)]

\(=3(5y-7)\)