EXERCISE 12.3 — Solutions
1. Carry out the following divisions.
(i) \(28x^{4} \div 56\)
Divide numbers separately: \( \frac {28}{56} = \tfrac{1}{2}\).
Variables: \(x^{4}\) remains as it is since there is no \(x\) in the denominator.
Combine results: \(\tfrac{1}{2} \times x^{4} = \tfrac{1}{2}x^{4}\).
Final Answer: \(\tfrac{1}{2}x^{4}\).
(ii) \(-36y^{3} \div 9y^{2}\)
Divide numbers: \(-36 \div 9 = -4\).
Expand variables:
\[
\frac{y \times y \times y}{y \times y}
\]
Cancel common \(y \times y\):
\[
\frac{\cancel{y} \times \cancel{y} \times y}{\cancel{y} \times \cancel{y}} = y
\]
Final Answer: \(-4y\).
(iii) \(66pq^{2}r^{3} \div 11qr^{2}\)
Divide numbers: \(66 \div 11 = 6\).
Expand variables:
\[
\frac{p \times q \times q \times r \times r \times r}{q \times r \times r}
\]
Cancel common factors:
\[
\frac{p \times \cancel{q} \times q \times \cancel{r} \times \cancel{r} \times r}{\cancel{q} \times \cancel{r} \times \cancel{r}} = p \times q \times r
\]
Final Answer: \(6pqr\).
(iv) \(34x^{3}y^{3}z^{3} \div 51xy^{2}z^{3}\)
Divide numbers: \( \frac {34} {51} = \tfrac{2}{3}\).
Expand variables:
\[
\frac{x \times x \times x \times y \times y \times y \times z \times z \times z}{x \times y \times y \times z \times z \times z}
\]
Cancel common factors:
\[
\frac{\cancel{x} \times x \times x \times \cancel{y} \times y \times \cancel{y} \times \cancel{z} \times \cancel{z} \times \cancel{z}}{\cancel{x} \times \cancel{y} \times \cancel{y} \times \cancel{z} \times \cancel{z} \times \cancel{z}}
= x \times x \times y
\]
Final Answer: \(\tfrac{2}{3}x^{2}y\).
(v) \(12a^{8}b^{8} \div (-6a^{6}b^{4})\)
Divide numbers: \( 12 \div -6 = -2\).
Expand variables in grouped form:
\[
\frac{a^{6} \times a^{2} \times b^{4} \times b^{4}}{a^{6} \times b^{4}}
\]
Cancel common factors:
\[
\frac{\cancel{a^{6}} \times a^{2} \times \cancel{b^{4}} \times b^{4}}{\cancel{a^{6}} \times \cancel{b^{4}}}
= a^{2}b^{4}
\]
Final Answer: \(-2a^{2}b^{4}\).
2. Divide the given polynomial by the given monomial.
(i) \((5x^{2} - 6x) \div 3x\)
Write as two terms: \(\dfrac{5x^{2}}{3x} - \dfrac{6x}{3x}\).
First term: numbers \(5 \div 3 = \tfrac{5}{3}\), variables \(\dfrac{x \times \cancel{x}}{\cancel{x}} = x\) → \(\tfrac{5}{3}x\).
Second term: numbers \(-6 \div 3 = -2\), variables \(\dfrac{\cancel{x}}{\cancel{x}} = 1\) → \(-2\).
Final Answer: \(\tfrac{5}{3}x - 2\).
(ii) \((3y^{8} - 4y^{6} + 5y^{4}) \div y^{4}\)
Split terms: \(\dfrac{3y^{8}}{y^{4}} - \dfrac{4y^{6}}{y^{4}} + \dfrac{5y^{4}}{y^{4}}\).
First term: \(\dfrac{3 \times {\cancel{y^{4}}} \times y^{4}}{\cancel{y^{4}}} = 3y^{4}\).
Second term: \(\dfrac{4 \times {\cancel{y^{4}}} \times y^{2}}{\cancel{y^{4}}} = 4y^{2}\) → \(-4y^{2}\).
Third term: \(\dfrac{5 \times {\cancel{y^{4}}}}{\cancel{y^{4}}} = 5\).
Final Answer: \(3y^{4} - 4y^{2} + 5\).
(iii) \(8(x^{3}y^{2}z^{2} + x^{2}y^{3}z^{2} + x^{2}y^{2}z^{3}) \div 4x^{2}y^{2}z^{2}\)
Coefficient: \(8 \div 4 = 2\).
Split terms: \(\dfrac{x^{3}y^{2}z^{2}}{x^{2}y^{2}z^{2}} + \dfrac{x^{2}y^{3}z^{2}}{x^{2}y^{2}z^{2}} + \dfrac{x^{2}y^{2}z^{3}}{x^{2}y^{2}z^{2}}\).
First term: \(\dfrac{x \times \cancel{x^{2}} \times \cancel{y^{2}} \times \cancel{z^{2}}}{\cancel{x^{2}} \times \cancel{y^{2}} \times \cancel{z^{2}}} = x\).
Second term: \(\dfrac{\cancel{x^{2}} \times y \times \cancel{y^{2}} \times \cancel{z^{2}}}{\cancel{x^{2}} \times \cancel{y^{2}} \times \cancel{z^{2}}} = y\).
Third term: \(\dfrac{\cancel{x^{2}} \times \cancel{y^{2}} \times z \times \cancel{z^{2}}}{\cancel{x^{2}} \times \cancel{y^{2}} \times \cancel{z^{2}}} = z\).
Final Answer: \(2(x + y + z)\).
(iv) \((x^{3} + 2x^{2} + 3x) \div 2x\)
Write as three terms: \(\dfrac{x^{3}}{2x} + \dfrac{2x^{2}}{2x} + \dfrac{3x}{2x}\).
Split and cancel each term:
First: \(\dfrac{x^{3}}{2x}=\dfrac{1\times x^{3}}{2\times x}
= \dfrac{1\times \cancel{x}\times x^{2}}{2\times \cancel{x}}=\tfrac{1}{2}x^{2}.\)
Second: \(\dfrac{2x^{2}}{2x}=\dfrac{2\times x^{2}}{2\times x}
= \dfrac{2\times \cancel{x}\times x}{2\times \cancel{x}}=x.\)
Third: \(\dfrac{3x}{2x}=\dfrac{3\times \cancel{x}}{2\times \cancel{x}}=\tfrac{3}{2}.\)
Final Answer: \(\tfrac{1}{2}x^{2} + x + \tfrac{3}{2}\).
(v) \((p^{3}q^{6} - p^{6}q^{3}) \div p^{3}q^{3}\)
Split terms: \(\dfrac{p^{3}q^{6}}{p^{3}q^{3}} - \dfrac{p^{6}q^{3}}{p^{3}q^{3}}\).
First: \(\dfrac{p^{3}q^{6}}{p^{3}q^{3}}=\dfrac{\cancel{p^{3}}\times q^{3}\times q^{3}}
{\cancel{p^{3}}\times \cancel{q^{3}}}=q^{3}.\)
Second: \(\dfrac{p^{6}q^{3}}{p^{3}q^{3}}=\dfrac{p^{3}\times \cancel{p^{3}}\times \cancel{q^{3}}}
{\cancel{p^{3}}\times \cancel{q^{3}}}=p^{3}.\)
Final Answer: \(q^{3} - p^{3}\).
3. Work out the following divisions.
(i) \((10x - 25) \div 5\)
Split terms: \(\dfrac{10x}{5} - \dfrac{25}{5}\).
First: \(\dfrac{10x}{5} = \dfrac{10}{5} \times x = 2x.\)
Second: \(\dfrac{25}{5} = 5.\)
Final Answer: \(2x - 5\).
(ii) \((10x - 25) \div (2x - 5)\)
Takeout common factor in numerator: \(10x - 25 = 5(2x - 5).\)
Now divide: \( \dfrac{5\; \cancel{(2x - 5)}} { \cancel{ 2x - 5}} \) = 5.
Final Answer: \(5\).
(iii) \(10y(6y + 21) \div 5(2y + 7)\)
Factor numerator so that it matches denominator: \(10y(6y + 21) = 10y \times 3(2y + 7) = 30y(2y+7).\)
Denominator: \(5(2y+7).\)
Cancel \((2y+7): \dfrac{30y \;\cancel{(2y+7)}}{5 \;\cancel{(2y+7)}} = \dfrac{30y}{5} = 6y.\)
Final Answer: \(6y\).
(iv) \(9x^{2}y^{2}(3z - 24) \div 27xy(z - 8)\)
Factor numerator to match denominator: \(9x^{2}y^{2}(3z - 24) = 9x^{2}y^{2}\times 3(z-8) = 27x^{2}y^{2}(z-8).\)
Denominator: \(27xy(z-8).\)
Divide: \(\dfrac{27x^{2}y^{2}(z-8)}{27xy(z-8)}\)
Now cancel common factors:
Cancel \(27\): \(\dfrac{27}{27}=1.\)
Cancel \(x\): \(\dfrac{x^{2}}{x} = x.\)
Cancel \(y\): \(\dfrac{y^{2}}{y} = y.\)
cancel \( (z-8) \): \dfrac{(z-8)}{(z-8)} = 1.\)
Cancel \(z-8: \dfrac{\cancel{(z-8)}}{\cancel{(z-8)}} = 1.\)
Final Answer: \(xy\).
(v) \(96abc(3a - 12)(5b - 30) \div 144(a - 4)(b - 6)\)
Factor inside brackets:
\(3a - 12 = 3(a-4), \quad 5b - 30 = 5(b-6).\)
So numerator = \(96abc \times 3(a-4)\times 5(b-6) = 1440abc(a-4)(b-6).\)
Denominator = \(144(a-4)(b-6).\)
Now divide: \(\dfrac{1440abc (a-4)(b-6)}{144(a-4)(b-6)}\)
Cancel common factors:
Cancel \((a-4)\) and \((b-6)\).
Cancel coefficient: \(1440 \div 144 = 10.\)
Final Answer: \(10abc\).
4. Divide as directed.
(i) \(5(2x + 1)(3x + 5) \div (2x + 1)\)
Write numerator and denominator and cancel the common factor \((2x+1)\):
\[
\frac{5\,( \cancel{2x+1} )\, (3x+5)}{\cancel{2x+1}} = 5(3x+5).
\]
Final Answer: \(5(3x+5)\).
(ii) \(26xy(x+5)(y-4) \div 13x(y-4)\)
Show coefficients, variables and factors, then cancel common parts:
\[
\frac{\cancel {26} ^{13 \times 2} \times \cancel x \times y \times (x+5) \times \cancel{(y-4)}}
{\cancel {13}^{13 \times 1} \times \cancel x \times \cancel{(y-4)}} \]
\[
= 2 \times y \times (x+5)
\]
Final Answer: \(2y(x+5)\).
(iii) \(52pqr(p+q)(q+r)(r+p) \div 104pq(q+r)(r+p)\)
Show coefficients, variables and factors, then cancel common parts:
\[
\frac{\cancel{52}^{52 \times 1} \times \cancel{p} \times \cancel{q} \times r \times (p+q) \times \cancel{(q+r)} \times \cancel{(r+p)}}
{\cancel{104}^{52 \times 2} \times \cancel{p} \times \cancel{q} \times \cancel{(q+r)} \times \cancel{(r+p)}} \]
\[
= \tfrac{1}{2} \times r \times (p+q)
\]
Final Answer: \(\tfrac{1}{2}r(p+q)\).
(iv) \(20(y+4)(y^{2}+5y+3) \div 5(y+4)\)
Show coefficients and cancel the common factor and coefficient:
\[
\frac{\cancel{20}^{5\times4}\times \cancel{(y+4)}\times (y^{2}+5y+3)}
{\cancel{5}^{5\times1}\times \cancel{(y+4)}} \]
\[
= 4 \times (y^{2}+5y+3)
\]
Final Answer: \(4(y^{2}+5y+3)\).
(v) \(x(x+1)(x+2)(x+3) \div x(x+1)\)
Show variables and cancel common factors \(x\) and \((x+1)\):
\[
\frac{\cancel{x}\times \cancel{(x+1)}\times (x+2)\times (x+3)}
{\cancel{x}\times \cancel{(x+1)}}
= (x+2)(x+3)
\]
Final Answer: \((x+2)(x+3)\).
5. Factorise the expressions and divide them as directed.
(i) \((y^{2}+7y+10) \div (y+5)\)
In Numerator, split middle term: \(y^2+2y+5y+10\).
Group Terms: \((y^2+2y)+(5y+10)\).
Take out common terms: \(y(y+2)+5(y+2)=(y+5)(y+2)\).
Now divide and cancel:
\[
\frac{(y+5)(y+2)}{\cancel{(y+5)}}=(y+2)
\]
Final Answer: \(y+2\).
(ii) \((m^{2}-14m-32) \div (m+2)\)
Split middle term: \(m^2-16m+2m-32\).
Group terms: \((m^2-16m)+(2m-32)\).
Take out common terms: \(m(m-16)+2(m-16)=(m-16)(m+2)\).
Now divide and cancel:
\[
\frac{(m-16)(m+2)}{\cancel{(m+2)}}=(m-16)
\]
Final Answer: \(m-16\).
(iii) \((5p^{2}-25p+20) \div (p-1)\)
Take out 5: \(5(p^2-5p+4)\).
Split middle term inside: \(p^2-4p-p+4\).
Group: \((p^2-4p)-(p-4)\).
Factor: \(p(p-4)-1(p-4)=(p-1)(p-4)\).
So now numerator is : \(5(p-1)(p-4)\).
Now divide and cancel:
\[
\frac{5(p-1)(p-4)}{\cancel{(p-1)}}=5(p-4)
\]
Final Answer: \(5(p-4)\).
(iv) \(4yz(z^{2}+6z-16) \div 2y(z+8)\)
Split middle term: \(z^2+8z-2z-16\).
Group: \((z^2+8z)-(2z+16)\).
Factor: \(z(z+8)-2(z+8)=(z-2)(z+8)\).
So now numerator is: \(4yz(z-2)(z+8)\).
Now divide and cancel:
\[
\frac{4yz(z-2)\cancel{(z+8)}}{2y\cancel{(z+8)}}=2z(z-2)
\]
Final Answer: \(2z(z-2)\).
(v) \(5pq(p^{2}-q^{2}) \div 2p(p+q)\)
For (\(p^2-q^2\)): Use identity \( (a^2 -b^2)= (a+b)(a-b) \)
So : \((p^2-q^2)=(p+q)(p-q)\).
So now numerator is : \(5pq(p+q)(p-q)\).
Now divide and cancel:
\[
\frac{5pq\cancel{(p+q)}(p-q)}{2p\cancel{(p+q)}}=\tfrac{5}{2}q(p-q)
\]
Final Answer: \(\tfrac{5}{2}q(p-q)\).
(vi) \(12xy(9x^{2}-16y^{2}) \div 4xy(3x+4y)\)
Rewrite Numerator as : \(12xy[(3x)^{2}-(4y)^{2}]\)
Now for: \([(3x)^{2}-(4y)^{2}]\): Use identity \( (a^2 -b^2)= (a+b)(a-b) \)
So: \((3x)^{2}-(4y)^{2}\) = \((3x-4y)(3x+4y)\).
So now numerator becomes: \(12xy(3x-4y)(3x+4y)\).
Now divide and cancel:
\[
\frac{12xy(3x-4y)\cancel{(3x+4y)}}{4xy\cancel{(3x+4y)}}=3(3x-4y)
\]
Final Answer: \(3(3x-4y)\).
(vii) \(39y^{3}(50y^{2}-98) \div 26y^{2}(5y+7)\)
Factor inside: \( 50y^2-98=2(25y^2-49)=2[(5y)^2-7^2] \).
Now for: \([(5y)^2-7^2] \): Use identity \( (a^2 -b^2)= (a+b)(a-b) \)
So numerator becomes: \(39y^{3}\times 2(5y-7)(5y+7)\).
Now divide and cancel:
\[
\frac{39 \times { \cancel y^{2} \times y} \times \cancel 2 \times (5y-7)\cancel{(5y+7)}}{13 \times \cancel 2 \times \cancel y^{2}\times \cancel{(5y+7)}} \]
\[
= \frac {\cancel {39}^{(13 \times 3)}} {\cancel {13}^{(13 \times 1)}} \times y \times (5y-7)
= {3} \times y \times(5y-7)
\]
Final Answer: \({3} y(5y-7)\).
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