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

(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)]

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