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Mathematics Question on Division of Algebraic Expressions Continued (Polynomial ÷ Polynomial)

Factorise the expressions and divide them as directed.

  1. (y2+7y+10)÷(y+5)(y^ 2 + 7y + 10) ÷ (y + 5)
  2. (m2–14m–32)÷(m+2)(m^2 – 14m – 32) ÷ (m + 2)
  3. (5p2–25p+20)÷(p–1)(5p^ 2 – 25p + 20) ÷ (p – 1)
  4. 4yz(z2+6z–16)÷2y(z+8)4yz(z ^2 + 6z – 16) ÷ 2y(z + 8)
  5. 5pq(p2–q2)÷2p(p+q)5pq(p^ 2 – q^ 2 ) ÷ 2p(p + q)
  6. 12xy(9x2–16y2)÷4xy(3x+4y)12xy(9x^ 2 – 16y ^2 ) ÷ 4xy(3x + 4y)
  7. 39y3(50y2–98)÷26y2(5y+7)39y^ 3 (50y ^2 – 98) ÷ 26y ^2 (5y + 7)
Answer

(i) (y2+7y+10)=y2+2y+5y+10(y^2+ 7y + 10) = y^2+ 2y + 5y + 10

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

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

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

(ii) m2−14m−32=m2+2m−16m−32m^2 - 14m - 32 = m^2 + 2m - 16m - 32

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

= (m+2)(m−16)(m + 2) (m - 16)

⇒(m2−14m−32)(m+2)=(m+2)(m−16)(m+2)=m−16\Rightarrow\frac{(m^2-14m-32)}{(m+2)}=\frac{(m+2)(m-16)}{(m+2)}=m-16

(iii) 5p2−25p+20=5(p2−5p+4)5p^ 2 - 25p + 20 = 5(p ^2 - 5p + 4)

= 5[p2−p−4p+4]5[p^ 2 - p - 4p + 4]

= 5[p(p−1)−4(p−1)]5[p(p - 1) - 4(p - 1)]

= 5(p−1)(p−4)5(p - 1) (p - 4)

⇒(5p2−25p+20)(p−1)=5(p−1)(p−4)(p−1)=5(p−4)\Rightarrow\frac{(5p^2-25p+20)}{(p-1)}=\frac{5(p-1)(p-4)}{(p-1)}=5(p-4)

(iv) 4yz(z2+6z−16)=4yz[z2−2z+8z−16]4yz(z^ 2 + 6z - 16) = 4yz [z ^2 - 2z + 8z - 16]

= 4yz[z(z−2)+8(z−2)]4yz [z(z - 2) + 8(z - 2)]

= 4yz(z−2)(z+8)4yz(z - 2) (z + 8)

4yz(z2+6z−16)2y(z+8)=4yz(z−2)(z+8)2y(z+8)=2z(z−2)\frac{4yz(z^2+6z-16)}{2y(z+8)}=\frac{4yz(z-2)(z+8)}{2y(z+8)}=2z(z-2)

(v) 5pq(p2−q2)=5pq(p−q)(p+q)5pq(p^ 2 - q^ 2 ) = 5pq (p - q) (p + q)

5pq(p2−q2)2p(p+q)=5pq(p−q)(p+q)2p(p+q)=52q(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)}

(vi) 12xy(9x2−16y2)=12xy[(3x)2−(4y)2]=12xy(3x−4y)(3x+4y)12xy(9x ^2 - 16y^ 2 ) = 12xy[(3x)^ 2 - (4y)^ 2 ] = 12xy(3x - 4y) (3x + 4y)

12xy(9x2−16y2)4xy(3x+4y)\frac{12xy(9x^2-16y^2)}{4xy(3x+4y)}

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

= 3(3x−4y)3(3x-4y)

(vii) 39y3(50y2−98)39y^ 3 (50y^ 2 - 98)

= 3×13×y×y×y×2[(25y2−49)]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) ^2 - (7)^2 ]

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

⇒26y2(5y+7)=2×13×y×y×(5y+7)\Rightarrow26y^ 2 (5y + 7) = 2 × 13 × y × y × (5y + 7)

⇒39y3(50y2−98)26y2(5y+7)\Rightarrow\frac{39y^ 3 (50y^ 2 - 98)}{26y 2 (5y + 7)}

= 39y3×2(25y2−49)26y2(5y+7)\frac{39y^3× 2(25y^2-49)}{26y^2(5y+7)}

= 3y(5y+7)(5y−7)(5y+7)\frac{3y(5y+7)(5y-7)}{(5y+7)}

= 3y(5y−7)3y(5y-7)