NCERT Solutions Class 8 Maths Chapter 14

Q.1 Factorise.

(i) x² + x y + 8x + 8y

(ii) 15xy – 6x + 5y – 2

(iii) ax + bx – ay – by

(iv) 15 pq + 15 + 9q + 25p

(v) z – 7 + 7xy – xyz

Ans

(i) x² + x y + 8x + 8y

Rearranging the terms of the given expression as follows:

= x × x + x × y + 8 × x + 8 × y

By taking ‘x’ as common from the first two terms and ‘8’ common from the last two terms, we get

= x (x + y) + 8 (x + y)

= (x + y) (x + 8)

(ii) 15xy − 6x + 5y – 2

Rearranging the terms of the given expression as follows:

= 3 × 5 × x × y − 3 × 2 × x + 5 × y − 2

By taking ‘3x’ as common from the first two terms and ‘1’ common from the last two terms, we get

= 3x (5y − 2) + 1 (5y − 2)

= (5y − 2) (3x + 1)

(iii) ax + bx − ay − by

Rearranging the terms of the given expression as follows:

= a × x + b × x − a × y − b × y

By taking ‘x’ as common from the first two terms and ‘y’ common from the last two terms, we get

= x (a + b) − y (a + b)

= (a + b) (x − y)

(iv) 15pq + 15 + 9q + 25p = 15pq + 9q + 25p + 15

Rearranging the terms of the given expression as follows:

= 3 × 5 × p × q + 3 × 3 × q + 5 × 5 × p + 3 × 5

By taking ‘3q’ as common from the first two terms and ‘5’ common from the last two terms, we get

= 3q (5p + 3) + 5 (5p + 3)

= (5p + 3) (3q + 5)

(v) z − 7 + 7xy − xyz

Rearranging the terms of the given expression as follows:

= z − x × y × z − 7 + 7 × x × y

By taking ‘z’ as common from the first two terms and ‘−7’ common from the last two terms, we get

= z (1− xy) − 7 (1− xy)

= (1− xy) (z − 7)

Q.2

$\begin{array}{l}\text{Factorise the following expressions.}\\ {\text{(i) a}}^{\text{2}}\text{+ 8a + 16}\\ {\text{(ii) p}}^{\text{2}}-\text{10 p + 25}\\ {\text{(iii) 25m}}^{\text{2}}\text{+ 30m + 9}\\ {\text{(iv) 49y}}^{\text{2}}{\text{+ 84yz + 36z}}^{\text{2}}\\ {\text{(v) 4x}}^{\text{2}}-\text{8x + 4}\\ {\text{(vi) 121b}}^{\text{2}}-{\text{88bc + 16c}}^{\text{2}}\\ {\text{(vii) (l + m)}}^{\text{2}}-{\text{4lm (Hint: Expand ( l + m)}}^{\text{2}}\text{first)}\\ {\text{(viii) a}}^{\text{4}}{\text{+ 2a}}^{\text{2}}{\text{b}}^{\text{2}}{\text{+ b}}^{\text{4}}\end{array}$

Ans

(i) a² + 8a + 16

The given expression can be written as:

= (a)² + 2 × a × 4 + (4)²

This is of the form: x²+ y² +2xy with x = a, y = 4 and 2xy = 8a

Therefore, by using the identity: (x + y)² = x² + 2xy + y²,we get

= (a + 4)²

(ii) p² − 10p + 25

The given expression can be written as:

= (p)² − 2 × p × 5 + (5)²

This is of the form: a²+ b² −2ab with a= p, b= 5 and

2ab =10p

Therefore, by using the identity: (a − b)² = a² – 2ab + b²,we get

= (p − 5)²

(iii) 25m² + 30m + 9

The given expression can be written as:

= (5m)² + 2 × 5m × 3 + (3)²

This is of the form: a²+ b² +2ab with a = 5m, b = 3 and 2ab =30m

Therefore, by using the identity: (a + b)² = a² + 2ab + b²,we get

= (5m + 3)²

(iv) 49y² + 84yz + 36z²

The given expression can be written as:

= (7y)² + 2 × (7y) × (6z) + (6z)²

This is of the form: a² + b² +2ab with a = 7y, b= 6z and 2ab = 84yz

Therefore, by using the identity: (a + b)² = a² + 2ab + b²,we get

= (7y + 6z)²

(v) 4x² − 8x + 4

The given expression can be written as:

= (2x)² − 2 (2x) (2) + (2)²

This is of the form: a² + b² − 2ab with a = 2x, b= 2 and 2ab = 8x

Therefore, by using the identity: (a − b)² = a² − 2ab + b²,we get

= (2x − 2)²

Again, we can take out 2 common

= [(2) (x − 1)]²

= 4(x − 1)²

(vi) 121b² − 88bc + 16c²

The given expression can be written as:

= (11b)² − 2 (11b) (4c) + (4c)²

This is of the form: a² + b² − 2ab with a = 11b, b= 4c and 2ab = 88bc

Therefore, by using the identity: (a − b)² = a² − 2ab + b²,we get

= (11b − 4c)²

(vii) (l + m)² − 4lm

The given expression can be written as:

= l² + 2lm + m² − 4lm

= l² − 2lm + m²

This is of the form: a² + b² − 2ab with a = l, b= m and 2ab = 2lm

Therefore, by using the identity: (a − b)² = a² − 2ab + b²,we get

= (l − m)²

(viii) a⁴ + 2a²b² + b⁴

The given expression can be written as:

= (a²)² + 2 (a²) (b²) + (b²)²

This is of the form: x² + y² + 2xy with x = a², y= b² and 2xy = 2a²b²

Therefore, by using the identity: (x + y)² = x² + 2xy + y²,we get

= (a² + b²)²

Q.3

$\begin{array}{l}\text{Factorise.}\\ {\text{(i)\hspace{0.33em}4p}}^{\text{2}}-{\text{\hspace{0.33em}9q}}^{\text{2}}\\ {\text{(ii)\hspace{0.33em}63a}}^{\text{2}}-{\text{\hspace{0.33em}112b}}^{\text{2}}\\ {\text{(iii)\hspace{0.33em}49x}}^{\text{2}}-\text{36}\\ {\text{(iv)\hspace{0.33em}16x}}^{\text{5}}-{\text{\hspace{0.33em}144x}}^{\text{3}}\\ {\text{(v)\hspace{0.33em}(l + m)}}^{\text{2}}-{\left(\text{l}-\text{\hspace{0.33em}m}\right)}^{\text{2}}\\ {\text{(vi)\hspace{0.33em}9x}}^{\text{2}}{\text{y}}^{\text{2}}-\text{\hspace{0.33em}16}\\ {\text{(vii)\hspace{0.33em}(x}}^{\text{2}}-{\text{\hspace{0.33em}2xy + y}}^{\text{2}}\text{)}-{\text{\hspace{0.33em}z}}^{\text{2}}\\ {\text{(viii)\hspace{0.33em}25a}}^{\text{2}}-{\text{\hspace{0.33em}4b}}^{\text{2}}\text{+ 28bc}-{\text{\hspace{0.33em}49c}}^{\text{2}}\end{array}$

Ans

(i) 4p² − 9q²

The given expression can be written as:

= (2p)² − (3q)²

This is of the form: a² − b² with a = 2p and b = 3q

Therefore, by using the identity: a² − b² = (a − b) (a + b), we get

= (2p + 3q) (2p − 3q)

(ii) 63a² − 112b²

The given expression can be written as:

= 7(9a² − 16b²)

= 7[(3a)² − (4b)²]

This is of the form: x² − y² with x = 3a and y = 4b

Therefore, by using the identity: x² − y² = (x − y) (x + y), we get

= 7(3a + 4b) (3a − 4b)

(iii) 49x² − 36

The given expression can be written as:

= (7x)² − (6)²

= (7x − 6) (7x + 6)

(iv) 16x⁵ − 144x³

The given expression can be written as:

= 16x³(x² − 9)

= 16 x³ [(x)² − (3)²]

This is of the form: a² − b² with a = x and b = 3

Therefore, by using the identity: a² − b² = (a − b)(a + b), we get

= 16x³(x − 3) (x + 3)

(v) (l + m)² − (l − m)²

This is of the form: a² − b² with a = (l + m) and

b = (l − m)

Therefore, by using the identity: a² − b² = (a − b)(a + b), we get

= [(l + m) − (l − m)] [(l + m) + (l − m)]

= (l + m − l + m) (l + m + l − m)

= 2m × 2l

= 4ml

= 4lm

(vi) 9x²y² − 16

The given expression can be written as:

= (3xy)² − (4)²

This is of the form: a² − b² with a = 3xy and b = 4

Therefore, by using the identity: a² − b² = (a − b)(a + b), we get

= (3xy − 4) (3xy + 4)

(vii) (x² − 2xy + y²) − z²

By using the identity: (a − b)² = a² − 2ab + b²,we get

= (x − y)² − (z)²

This is of the form: a² − b² with a = (x – y) and b = z

Therefore, by using the identity: a² − b² = (a − b)(a + b), we get

= (x − y − z) (x − y + z)

(viii) 25a² − 4b² + 28bc − 49c²

The given expression can be written as:

= 25a² − (4b² − 28bc + 49c²)

= (5a)² − [(2b)² − 2 × 2b × 7c + (7c)²]

By using the identity: (a − b)² = a² − 2ab + b²,we get

= (5a)² − [(2b − 7c)²]

This is of the form: x² − y² with x = 5a and y = (2b – 7c)

Therefore, by using the identity: x² − y² = (x − y)(x + y), we get

= [5a + (2b − 7c)] [5a − (2b − 7c)]

= (5a + 2b − 7c) (5a − 2b + 7c)

Q.4

$\begin{array}{l}\text{Factorise the expressions.}\\ {\text{(i)\hspace{0.33em}ax}}^{\text{2}}\text{+ bx}\\ {\text{(ii)\hspace{0.33em}7p}}^{\text{2}}{\text{+ 21q}}^{\text{2}}\\ {\text{(iii)\hspace{0.33em}2x}}^{\text{3}}{\text{+ 2xy}}^{\text{2}}{\text{+ 2xz}}^{\text{2}}\\ {\text{(iv)\hspace{0.33em}am}}^{\text{2}}{\text{+ bm}}^{\text{2}}{\text{+ bn}}^{\text{2}}{\text{+ an}}^{\text{2}}\\ \text{(v)\hspace{0.33em}(lm + l) + m + 1}\\ \text{(vi)\hspace{0.33em}y (y + z) + 9 (y + z)}\\ {\text{(vii)\hspace{0.33em}5y}}^{\text{2}}-\text{\hspace{0.33em}20y\hspace{0.33em}}-\text{\hspace{0.33em}8z + 2yz}\\ \text{(viii)\hspace{0.33em}10ab + 4a + 5b + 2}\\ \text{(ix)\hspace{0.33em}6xy\hspace{0.33em}}-\text{\hspace{0.33em}4y + 6\hspace{0.33em}}-\text{\hspace{0.33em}9x}\end{array}$

Ans

(i) ax² + bx

= a × x × x + b × x

Here, the common factor is ‘x’.

Therefore,

ax² + bx = x(ax + b)

(ii) 7p² + 21q²

= 7 × p × p + 3 × 7 × q × q

Here, the common factor is ‘7’.

Therefore,

7p² + 21q² = 7(p² + 3q²)

(iii) 2x³ + 2xy² + 2xz²

Here, the common factor is ‘2x’.

Therefore,

2x³ + 2xy² + 2xz² = 2x(x² + y² + z²)

(iv) am² + bm² + bn² + an²

The given expression can be written as:

= am² + bm² + an² + bn²

The common factor in first two terms is m² and the common factor in last two terms is n².

= m²(a + b) + n²(a + b)

= (a + b) (m² + n²)

Therefore,

am² + bm² + bn² + an² = (a + b) (m² + n²)

(v) (lm + l) + m + 1

Rearranging the given expression as follows:

= lm + m + l + 1

Taking out common factors .

= m(l + 1) + 1(l + 1)

= (l + 1) (m + 1)

Therefore,

(lm + l) + m + 1=(l + 1) (m + 1)

(vi) y (y + z) + 9 (y + z)

= (y + z) (y + 9)

Therefore,

y (y + z) + 9 (y + z) = (y + z) (y + 9)

(vii) 5y² − 20y − 8z + 2yz

The given expression can be written as follows:

5y² − 20y + 2yz − 8z

Taking out common factors.

= 5y(y − 4) + 2z(y − 4)

= (y − 4) (5y + 2z)

Therefore,

5y² − 20y − 8z + 2yz =(y − 4) (5y + 2z)

(viii) 10ab + 4a + 5b + 2

Rearranging the given expression as follows:

= 10ab + 5b + 4a + 2

Taking out the common terms

= 5b(2a + 1) + 2(2a + 1)

= (2a + 1) (5b + 2)

Therefore,

10ab + 4a + 5b + 2= (2a + 1) (5b + 2)

(ix) 6xy − 4y + 6 − 9x

Rearranging the terms of the given expression and taking out common terms.

= 6xy − 9x − 4y + 6

= 3x(2y − 3) − 2(2y − 3)

= (2y − 3) (3x − 2)

Therefore,

6xy − 4y + 6 − 9x= (2y − 3) (3x − 2)

Q.5

$\begin{array}{l}\text{Factorise.}\\ {\text{(i)\hspace{0.33em}a}}^{\text{4}}-{\text{\hspace{0.33em}b}}^{\text{4}}\\ {\text{(ii)\hspace{0.33em}p}}^{\text{4}}-\text{\hspace{0.33em}81}\\ {\text{(iii)\hspace{0.33em}x}}^{\text{4}}-{\text{\hspace{0.33em}(y + z)}}^{\text{4}}\\ {\text{(iv)\hspace{0.33em}x}}^{\text{4}}-{\text{(x – z)}}^{\text{4}}\\ {\text{(v)\hspace{0.33em}a}}^{\text{4}}-{\text{2a}}^{\text{2}}{\text{b}}^{\text{2}}\text{+ b4}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{}{\mathrm{a}}^4-{\mathrm{b}}^4\\ ={\left({\mathrm{a}}^2\right)}^2-{\left({\mathrm{b}}^2\right)}^2\\ =({\mathrm{a}}^2-{\mathrm{b}}^2)({\mathrm{a}}^2+{\mathrm{b}}^2)\left[\mathrm{By}\mathrm{using}:({\mathrm{a}}^2-{\mathrm{b}}^2)=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]\\ =(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})({\mathrm{a}}^2+{\mathrm{b}}^2)\left[\mathrm{By}\mathrm{using}:({\mathrm{a}}^2-{\mathrm{b}}^2)=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]\\ \\ \left(\mathrm{ii}\right){\mathrm{p}}^4-81\\ ={\left({\mathrm{p}}^2\right)}^2-{\left(9\right)}^2\\ =({\mathrm{p}}^2-9)({\mathrm{p}}^2+9)\left[\mathrm{By}\mathrm{using}:({\mathrm{a}}^2-{\mathrm{b}}^2)=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]\\ =[{\left(\mathrm{p}\right)}^2-{\left(3\right)}^2]({\mathrm{p}}^2+9)\\ =(\mathrm{p}-3)(\mathrm{p}+3)({\mathrm{p}}^2+9)\left[\mathrm{By}\mathrm{using}:({\mathrm{a}}^2-{\mathrm{b}}^2)=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]\\ \\ \left(\mathrm{iii}\right){\mathrm{x}}^4-{(\mathrm{y}+\mathrm{z})}^4\\ ={\left({\mathrm{x}}^2\right)}^2-{\left[{(\mathrm{y}+\mathrm{z})}^2\right]}^2\\ =[{\mathrm{x}}^2-{(\mathrm{y}+\mathrm{z})}^2][{\mathrm{x}}^2+{(\mathrm{y}+\mathrm{z})}^2]\left[\mathrm{By}\mathrm{using}:({\mathrm{a}}^2-{\mathrm{b}}^2)=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]\end{array}$

$\begin{array}{l}=[\mathrm{x}-(\mathrm{y}+\mathrm{z})][\mathrm{x}+(\mathrm{y}+\mathrm{z})][{\mathrm{x}}^2+{(\mathrm{y}+\mathrm{z})}^2]\\ \left[\mathrm{By}\mathrm{using}:({\mathrm{a}}^2-{\mathrm{b}}^2)=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]\\ =(\mathrm{x}-\mathrm{y}-\mathrm{z})(\mathrm{x}+\mathrm{y}+\mathrm{z})[{\mathrm{x}}^2+{(\mathrm{y}+\mathrm{z})}^2]\\ \\ \left(\mathrm{iv}\right){\mathrm{x}}^{\text{4}}-{(\mathrm{x}-\mathrm{z})}^{\text{4}}\\ ={\left({\mathrm{x}}^{\text{2}}\right)}^{\text{2}}-{\left[{(\mathrm{x}-\mathrm{z})}^{\text{2}}\right]}^{\text{2}}\\ =[{\mathrm{x}}^{\text{2}}-{(\mathrm{x}-\mathrm{z})}^{\text{2}}][{\mathrm{x}}^{\text{2}}+{(\mathrm{x}-\mathrm{z})}^{\text{2}}]\left[\mathrm{By}\mathrm{using}:({\mathrm{a}}^2-{\mathrm{b}}^2)=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]\\ =[\mathrm{x}-(\mathrm{x}-\mathrm{z})][\mathrm{x}+(\mathrm{x}-\mathrm{z})][{\mathrm{x}}^{\text{2}}+{(\mathrm{x}-\mathrm{z})}^{\text{2}}]\\ \left[\mathrm{By}\mathrm{using}:({\mathrm{a}}^2-{\mathrm{b}}^2)=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]\\ =\mathrm{z}(\text{2}\mathrm{x}-\mathrm{z})[{\mathrm{x}}^{\text{2}}+{\mathrm{x}}^{\text{2}}-\text{2}\mathrm{xz}+{\mathrm{z}}^{\text{2}}]\\ =\mathrm{z}(\text{2}\mathrm{x}-\mathrm{z})(\text{2}{\mathrm{x}}^{\text{2}}-\text{2}\mathrm{xz}+{\mathrm{z}}^{\text{2}})\\ \\ \left(\mathrm{v}\right){\mathrm{a}}^{\text{4}}-\text{2}{\mathrm{a}}^{\text{2}}{\mathrm{b}}^{\text{2}}+{\mathrm{b}}^{\text{4}}\\ ={\left({\mathrm{a}}^{\text{2}}\right)}^{\text{2}}-\text{2}\left({\mathrm{a}}^{\text{2}}\right)\left({\mathrm{b}}^{\text{2}}\right)+{\left({\mathrm{b}}^{\text{2}}\right)}^{\text{2}}\\ ={({\mathrm{a}}^{\text{2}}-{\mathrm{b}}^{\text{2}})}^{\text{2}}\\ ={\left[(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})\right]}^{\text{2}}\left[\mathrm{By}\mathrm{using}:({\mathrm{a}}^2-{\mathrm{b}}^2)=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]\\ ={(\mathrm{a}-\mathrm{b})}^{\text{2}}{(\mathrm{a}+\mathrm{b})}^{\text{2}}\end{array}$

Q.6

$\begin{array}{l}\text{Factorise the following expressions.}\\ {\text{(i) p}}^{\text{2}}\text{+ 6p + 8}\\ {\text{(ii) q}}^{\text{2}}-\text{10q + 21}\\ {\text{(iii) p}}^{\text{2}}\text{+ 6p}-\text{16}\end{array}$

Ans

(i) p² + 6p + 8

Here, 8 = 4 × 2 and

6 = 4 + 2

Therefore, p² + 6p + 8 can be written as:

p² + 2p + 4p + 8

= p(p + 2) + 4(p + 2)

= (p + 2) (p + 4)

(ii) q² − 10q + 21

Here, 21 = (−7) × (−3) and

−10 = (−7) + (−3)

Therefore, q² − 10q + 21 can be written as:

q² − 7q − 3q + 21

= q(q − 7) − 3(q − 7)

= (q − 7) (q − 3)

(iii) p² + 6p − 16

Here, 16 = (−2) × 8 and

6 = 8 + (−2)

Therefore, p² + 6p – 16 can be written as:
= p² + 8p − 2p − 16

= p(p + 8) − 2(p + 8)

= (p + 8) (p – 2)

Q.7

$\begin{array}{l}\text{Carry out the following divisions.}\\ {\text{(i) \hspace{0.33em}28x}}^{\text{4}}\text{÷ 56x}\\ \text{(ii)\hspace{0.33em}}-{\text{36y}}^{\text{3}}{\text{÷ 9y}}^{\text{2}}\\ {\text{(iii) 66pq}}^{\text{2}}{\text{r}}^{\text{3}}{\text{÷ 11qr}}^{\text{2}}\\ {\text{(iv)\hspace{0.33em}34x}}^{\text{3}}{\text{y}}^{\text{3}}{\text{z}}^{\text{3}}{\text{÷ 51xy}}^{\text{2}}{\text{z}}^{\text{3}}\\ {\text{(v) 12a}}^{\text{8}}{\text{b}}^{\text{8}}\text{÷ (}-{\text{6a}}^{\text{6}}{\text{b}}^{\text{4}}\text{)}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)28{\mathrm{x}}^4÷56\mathrm{x}\\ 28{\mathrm{x}}^4=2\times 2\times 7\times \mathrm{x}\times \mathrm{x}\times \mathrm{x}\times \mathrm{x}\\ 56\mathrm{x}=2\times 2\times 2\times 7\times \mathrm{x}\\ \\ \therefore \frac{28{\mathrm{x}}^4}{56\mathrm{x}}\\ =\frac{\bcancel{2}\times \bcancel{2}\times \bcancel{7}\times \mathrm{x}\times \mathrm{x}\times \mathrm{x}\times \bcancel{\mathrm{x}}}{\bcancel{2}\times \bcancel{2}\times 2\times \bcancel{7}\times \bcancel{\mathrm{x}}}\\ =\frac{{\mathrm{x}}^3}{2}\\ \\ \left(\mathrm{ii}\right)-36{\mathrm{y}}^3÷9{\mathrm{y}}^2\\ 36{\mathrm{y}}^3=2\times 2\times 3\times 3\times \mathrm{y}\times \mathrm{y}\times \mathrm{y}\\ 9{\mathrm{y}}^2=3\times 3\times \mathrm{y}\times \mathrm{y}\\ \\ \therefore \frac{-36{\mathrm{y}}^3}{9{\mathrm{y}}^2}\\ =\frac{-\bcancel{3}\times \bcancel{3}\times 2\times 2\times \mathrm{y}\times \bcancel{\mathrm{y}}\times \bcancel{\mathrm{y}}}{\bcancel{3}\times \bcancel{3}\times \bcancel{\mathrm{y}}\times \bcancel{\mathrm{y}}}\\ =-4\mathrm{y}\\ \\ \left(\mathrm{iii}\right)66{\mathrm{pq}}^2{\mathrm{r}}^3÷11{\mathrm{qr}}^2\\ 66{\mathrm{pq}}^2{\mathrm{r}}^3=2\times 3\times 11\times \mathrm{p}\times \mathrm{q}\times \mathrm{q}\times \mathrm{r}\times \mathrm{r}\times \mathrm{r}\\ 11{\mathrm{qr}}^2=11\times \mathrm{q}\times \mathrm{r}\times \mathrm{r}\\ \\ \therefore \frac{66{\mathrm{pq}}^2{\mathrm{r}}^3}{11{\mathrm{qr}}^2}\\ =\frac{\bcancel{11}\times 3\times 2\times \mathrm{p}\times \bcancel{\mathrm{q}}\times \mathrm{q}\times \bcancel{\mathrm{r}\times \mathrm{r}}\times \mathrm{r}}{\bcancel{11}\times \bcancel{\mathrm{q}}\times \bcancel{\mathrm{r}\times \mathrm{r}}}\\ =6\mathrm{pqr}\end{array}$

$\begin{array}{l}\left(\mathrm{iv}\right)34{\mathrm{x}}^3{\mathrm{y}}^3{\mathrm{z}}^3÷51{\mathrm{xy}}^2{\mathrm{z}}^3\\ 34{\mathrm{x}}^3{\mathrm{y}}^3{\mathrm{z}}^3=2\times 17\times \mathrm{x}\times \mathrm{x}\times \mathrm{x}\times \mathrm{y}\times \mathrm{y}\times \mathrm{y}\times \mathrm{z}\times \mathrm{z}\times \mathrm{z}\\ 51{\mathrm{xy}}^2{\mathrm{z}}^3=3\times 17\times \mathrm{x}\times \mathrm{y}\times \mathrm{y}\times \mathrm{z}\times \mathrm{z}\times \mathrm{z}\\ \therefore \frac{34{\mathrm{x}}^3{\mathrm{y}}^3{\mathrm{z}}^3}{51{\mathrm{xy}}^2{\mathrm{z}}^3}\\ =\frac{2\times \bcancel{17}\times \mathrm{x}\times \bcancel{\mathrm{x}}\times \mathrm{x}\times \bcancel{\mathrm{y}}\times \bcancel{\mathrm{y}}\times \mathrm{y}\times \bcancel{\mathrm{z}}\times \bcancel{\mathrm{z}}\times \bcancel{\mathrm{z}}}{3\times \bcancel{17}\times \bcancel{\mathrm{x}}\times \bcancel{\mathrm{y}}\times \bcancel{\mathrm{y}}\times \bcancel{\mathrm{z}}\times \bcancel{\mathrm{z}}\times \bcancel{\mathrm{z}}}\\ =\frac{2{\mathrm{x}}^2\mathrm{y}}{3}\\ \\ \left(\mathrm{v}\right)12{\mathrm{a}}^8{\mathrm{b}}^8÷(-6{\mathrm{a}}^6{\mathrm{b}}^4)\\ 12{\mathrm{a}}^8{\mathrm{b}}^8=2\times 2\times 3\times \mathrm{a}\times \mathrm{a}\times \mathrm{a}\times \mathrm{a}\times \mathrm{a}\times \mathrm{a}\times \mathrm{a}\times \mathrm{a}\times \mathrm{b}\times \mathrm{b}\times \mathrm{b}\times \mathrm{b}\times \mathrm{b}\times \mathrm{b}\times \mathrm{b}\times \mathrm{b}\\ 6{\mathrm{a}}^6{\mathrm{b}}^4=3\times 2\times \mathrm{a}\times \mathrm{a}\times \mathrm{a}\times \mathrm{a}\times \mathrm{a}\times \mathrm{a}\times \mathrm{b}\times \mathrm{b}\times \mathrm{b}\times \mathrm{b}\\ \therefore \frac{12{\mathrm{a}}^8{\mathrm{b}}^8}{(-6{\mathrm{a}}^6{\mathrm{b}}^4)}\\ =\frac{2\times \bcancel{2\times 3}\times \bcancel{\mathrm{a}}\times \bcancel{\mathrm{a}}\times \bcancel{\mathrm{a}}\times \bcancel{\mathrm{a}}\times \bcancel{\mathrm{a}}\times \bcancel{\mathrm{a}}\times \mathrm{a}\times \mathrm{a}\times \bcancel{\mathrm{b}}\times \bcancel{\mathrm{b}}\times \bcancel{\mathrm{b}}\times \bcancel{\mathrm{b}}\times \mathrm{b}\times \mathrm{b}\times \mathrm{b}\times \mathrm{b}}{-\bcancel{3\times 2}\times \bcancel{\mathrm{a}}\times \bcancel{\mathrm{a}}\times \bcancel{\mathrm{a}}\times \bcancel{\mathrm{a}}\times \bcancel{\mathrm{a}}\times \bcancel{\mathrm{a}}\times \bcancel{\mathrm{b}}\times \bcancel{\mathrm{b}}\times \bcancel{\mathrm{b}}\times \bcancel{\mathrm{b}}}\\ =-2{\mathrm{a}}^2{\mathrm{b}}^4\end{array}$

Q.8

$\begin{array}{l}\text{Divide the given polynomial by the given monomial.}\\ {\text{(i) (5x}}^{\text{2}}-\text{\hspace{0.33em}6x) ÷ 3x}\\ \left(\mathrm{ii}\right)(3{\mathrm{y}}^{\text{8}}-{\text{\hspace{0.33em}4y}}^{\text{6}}{\text{+ 5y}}^{\text{4}}{\text{) ÷ y}}^{\text{4}}\\ {\text{(iii)\hspace{0.33em}8(x}}^{\text{3}}{\text{y}}^{\text{2}}{\text{z}}^{\text{2}}{\text{+ x}}^{\text{2}}{\text{y}}^{\text{3}}{\text{z}}^{\text{2}}{\text{+ x}}^{\text{2}}{\text{y}}^{\text{2}}{\text{z}}^{\text{3}}{\text{) ÷ 4x}}^{\text{2}}{\text{y}}^{\text{2}}{\text{z}}^{\text{2}}\\ {\text{(iv)\hspace{0.33em}(x}}^{\text{3}}{\text{+ 2x}}^{\text{2}}\text{+ 3x) ÷ 2x}\\ {\text{(v)\hspace{0.33em}(p}}^{\text{3}}{\text{q}}^{\text{6}}-{\text{\hspace{0.33em}p}}^{\text{6}}{\text{q}}^{\text{3}}{\text{) ÷ p}}^{\text{3}}{\text{q}}^{\text{3}}\end{array}$

Ans

$\begin{array}{l}\\ \left(\mathrm{i}\right)(5{\mathrm{x}}^2-6\mathrm{x})÷3\mathrm{x}\\ \frac{(5{\mathrm{x}}^2-6\mathrm{x})}{3\mathrm{x}}\\ =\frac{\bcancel{\mathrm{x}}(5\mathrm{x}-6)}{3\bcancel{\mathrm{x}}}\\ =\frac{(5\mathrm{x}-6)}{3}\\ \\ \left(\mathrm{ii}\right)(3{\mathrm{y}}^8-4{\mathrm{y}}^6+5{\mathrm{y}}^4)÷{\mathrm{y}}^4\\ \frac{(3{\mathrm{y}}^8-4{\mathrm{y}}^6+5{\mathrm{y}}^4)}{{\mathrm{y}}^4}\\ =\frac{\bcancel{{\mathrm{y}}^4}(3{\mathrm{y}}^4-4{\mathrm{y}}^2+5)}{\bcancel{{\mathrm{y}}^4}}\\ =3{\mathrm{y}}^4-4{\mathrm{y}}^2+5\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right)8({\mathrm{x}}^3{\mathrm{y}}^2{\mathrm{z}}^2+{\mathrm{x}}^2{\mathrm{y}}^3{\mathrm{z}}^2+{\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{z}}^3)÷4{\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{z}}^2\\ \\ \frac{8({\mathrm{x}}^3{\mathrm{y}}^2{\mathrm{z}}^2+{\mathrm{x}}^2{\mathrm{y}}^3{\mathrm{z}}^2+{\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{z}}^3)}{4{\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{z}}^2}\\ =\frac{8\bcancel{{\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{z}}^2}(\mathrm{x}+\mathrm{y}+\mathrm{z})}{4\bcancel{{\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{z}}^2}}\\ =2(\mathrm{x}+\mathrm{y}+\mathrm{z})\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\left(\mathrm{iv}\right)({\mathrm{x}}^3+2{\mathrm{x}}^2+3\mathrm{x})÷2\mathrm{x}\\ \\ \frac{({\mathrm{x}}^3+2{\mathrm{x}}^2+3\mathrm{x})}{2\mathrm{x}}\\ =\frac{\bcancel{\mathrm{x}}({\mathrm{x}}^2+2\mathrm{x}+3)}{2\bcancel{\mathrm{x}}}\\ =\frac{({\mathrm{x}}^2+2\mathrm{x}+3)}{2}\\ \\ \left(\mathrm{v}\right)({\mathrm{p}}^3{\mathrm{q}}^6-{\mathrm{p}}^6{\mathrm{q}}^3)÷{\mathrm{p}}^3{\mathrm{q}}^3\\ \\ \frac{{\mathrm{p}}^3{\mathrm{q}}^6-{\mathrm{p}}^6{\mathrm{q}}^3}{{\mathrm{p}}^3{\mathrm{q}}^3}\\ =\frac{\bcancel{{\mathrm{p}}^3{\mathrm{q}}^3}({\mathrm{q}}^3-{\mathrm{p}}^3)}{\bcancel{{\mathrm{p}}^3{\mathrm{q}}^3}}\\ ={\mathrm{q}}^3-{\mathrm{p}}^3\end{array}$

Q.9 Work out the following divisions.

(i) (10x – 25) ÷ 5

(ii) (10x – 25) ÷ (2x – 5)

(iii) 10y(6y + 21) ÷ 5(2y + 7)

(iv) 9x²y²(3z – 24) ÷ 27xy(z – 8)

(v) 96abc(3a – 12) (5b – 30) ÷ 144(a – 4) (b – 6)

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{}(10\mathrm{x}–25)÷5\\ \frac{\mathrm{}(10\mathrm{x}–25)}{5}\\ =\frac{\mathrm{}5(2\mathrm{x}–5)}{5}\\ =2\mathrm{x}–5\\ \\ \left(\mathrm{ii}\right)(10\mathrm{x}–25)÷(2\mathrm{x}–5)\\ \frac{(10\mathrm{x}–25)}{(2\mathrm{x}–5)}\\ =\frac{5(2\mathrm{x}–5)}{(2\mathrm{x}–5)}\\ =5\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right)\mathrm{}10\mathrm{y}(6\mathrm{y}+21)÷5(2\mathrm{y}+7)\\ \frac{10\mathrm{y}(6\mathrm{y}+21)}{5(2\mathrm{y}+7)}\\ =\frac{10\mathrm{y}\times 3\bcancel{(2\mathrm{y}+7)}}{5\bcancel{(2\mathrm{y}+7)}}\\ =6\mathrm{y}\\ \\ \left(\mathrm{iv}\right)\mathrm{}9{\mathrm{x}}^2{\mathrm{y}}^2(3\mathrm{z}–24)÷27\mathrm{xy}(\mathrm{z}–8)\\ \frac{9{\mathrm{x}}^2{\mathrm{y}}^2(3\mathrm{z}–24)}{27\mathrm{xy}(\mathrm{z}–8)}\\ =\frac{9{\mathrm{x}}^2{\mathrm{y}}^2\times 3\bcancel{(\mathrm{z}–8)}}{27\mathrm{xy}\bcancel{(\mathrm{z}–8)}}\\ =\mathrm{xy}\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\left(\mathrm{v}\right)96\mathrm{abc}(3\mathrm{a}–12)(5\mathrm{b}–30)÷144(\mathrm{a}–4)(\mathrm{b}–6)\\ \frac{96\mathrm{abc}(3\mathrm{a}–12)(5\mathrm{b}–30)}{144(\mathrm{a}–4)(\mathrm{b}–6)}\\ =\frac{96\mathrm{abc}\times 3\bcancel{(\mathrm{a}–4)}\times 5\bcancel{(\mathrm{b}–\text{6})}}{144\bcancel{(\mathrm{a}–4)}\bcancel{(\mathrm{b}–6)}}\\ =10\mathrm{abc}\end{array}$

Q.10

$\begin{array}{l}\text{Divide as directed.}\\ \text{(i)\hspace{0.33em}5(2x + 1) (3x + 5) ÷ (2x + 1)}\\ \text{(ii)\hspace{0.33em}26xy(x + 5) (y}-\text{\hspace{0.33em}4) ÷ 13x(y – 4)}\\ \text{(iii)\hspace{0.33em}52pqr (p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p)}\\ {\text{(iv)\hspace{0.33em}20(y + 4) (y}}^{\text{2}}\text{+ 5y + 3) ÷ 5(y + 4)}\\ \text{(v)\hspace{0.33em}x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)5(2\mathrm{x}+1)(3\mathrm{x}+5)÷(2\mathrm{x}+1)\\ \frac{5(2\mathrm{x}+1)(3\mathrm{x}+5)}{(2\mathrm{x}+1)}\\ =\frac{5\bcancel{(2\mathrm{x}+1)}(3\mathrm{x}+5)}{\bcancel{(2\mathrm{x}+1)}}\\ =5(3\mathrm{x}+5)\\ \\ \left(\mathrm{ii}\right)26\mathrm{xy}(\mathrm{x}+5)(\mathrm{y}-4)÷13\mathrm{x}(\mathrm{y}-4)\\ \frac{26\mathrm{xy}(\mathrm{x}+5)(\mathrm{y}-4)}{13\mathrm{x}(\mathrm{y}-4)}\\ =\frac{26\bcancel{\mathrm{x}}\mathrm{y}(\mathrm{x}+5)\bcancel{(\mathrm{y}-4)}}{13\bcancel{\mathrm{x}}\bcancel{(\mathrm{y}-4)}}\\ =2\mathrm{y}(\mathrm{x}+5)\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right)52\mathrm{pqr}(\mathrm{p}+\mathrm{q})(\mathrm{q}+\mathrm{r})(\mathrm{r}+\mathrm{p})÷104\mathrm{pq}(\mathrm{q}+\mathrm{r})(\mathrm{r}+\mathrm{p})\\ \\ \frac{52\mathrm{pqr}(\mathrm{p}+\mathrm{q})(\mathrm{q}+\mathrm{r})(\mathrm{r}+\mathrm{p})}{104\mathrm{pq}(\mathrm{q}+\mathrm{r})(\mathrm{r}+\mathrm{p})}\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}=\frac{\bcancel{2\times 2}\times \bcancel{13}\times \bcancel{\mathrm{pq}}\mathrm{r}(\mathrm{p}+\mathrm{q})\bcancel{(\mathrm{q}+\mathrm{r})}\bcancel{(\mathrm{r}+\mathrm{p})}}{\bcancel{2\times 2}\times 2\times \bcancel{13}\times \bcancel{\mathrm{pq}}\bcancel{(\mathrm{q}+\mathrm{r})}\bcancel{(\mathrm{r}+\mathrm{p})}}\\ =\frac{\mathrm{r}(\mathrm{p}+\mathrm{q})}{2}\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\left(\mathrm{iv}\right)20(\mathrm{y}+4)({\mathrm{y}}^2+5\mathrm{y}+3)÷5(\mathrm{y}+4)\\ \\ \frac{20(\mathrm{y}+4)({\mathrm{y}}^2+5\mathrm{y}+3)}{5(\mathrm{y}+4)}\\ =\frac{\bcancel{5}\times 4\bcancel{(\mathrm{y}+4)}({\mathrm{y}}^2+5\mathrm{y}+3)}{\bcancel{5}\bcancel{(\mathrm{y}+4)}}\\ =4({\mathrm{y}}^2+5\mathrm{y}+3)\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\left(\mathrm{v}\right)\mathrm{x}(\mathrm{x}+1)(\mathrm{x}+2)(\mathrm{x}+3)÷\mathrm{x}(\mathrm{x}+1)\\ \\ \frac{\mathrm{x}(\mathrm{x}+1)(\mathrm{x}+2)(\mathrm{x}+3)}{\mathrm{x}(\mathrm{x}+1)}\\ =\frac{\bcancel{\mathrm{x}(\mathrm{x}+1)}(\mathrm{x}+2)(\mathrm{x}+3)}{\bcancel{\mathrm{x}(\mathrm{x}+1)}}\\ =(\mathrm{x}+2)(\mathrm{x}+3)\end{array}$

Q.11

$\begin{array}{l}\text{Factorise the expressions and divide them as directed.}\\ {\text{(i) (y}}^{\text{2}}\text{+ 7y + 10) ÷ (y + 5)}\\ {\text{(ii) (m}}^{\text{2}}-\text{\hspace{0.33em}14m\hspace{0.33em}}-\text{\hspace{0.33em}32) ÷ (m + 2)}\\ {\text{(iii) (5p}}^{\text{2}}-\text{\hspace{0.33em}25p + 20) ÷ (p\hspace{0.33em}}-\text{\hspace{0.33em}1)}\\ {\text{(iv) 4yz(z}}^{\text{2}}\text{+ 6z\hspace{0.33em}}-\text{\hspace{0.33em}16) ÷ 2y(z + 8)}\\ {\text{(v) 5pq(p}}^{\text{2}}-{\text{\hspace{0.33em}q}}^{\text{2}}\text{) ÷ 2p(p + q)}\\ {\text{(vi) 12xy(9x}}^{\text{2}}-{\text{\hspace{0.33em}16y}}^{\text{2}}\text{) ÷ 4xy(3x + 4y)}\\ {\text{(vii) 39y}}^{\text{3}}{\text{(50y}}^{\text{2}}-{\text{\hspace{0.33em}98) ÷ 26y}}^{\text{2}}\text{(5y + 7)}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)({\mathrm{y}}^2+7\mathrm{y}+10)÷(\mathrm{y}+5)\\ \text{First we will factorise}({\mathrm{y}}^2+7\mathrm{y}+10)\\ ={\mathrm{y}}^2+2\mathrm{y}+5\mathrm{y}+10\\ =\mathrm{y}(\mathrm{y}+2)+5(\mathrm{y}+2)\\ =(\mathrm{y}+2)(\mathrm{y}+5)\\ \\ \therefore \frac{({\mathrm{y}}^2+7\mathrm{y}+10)}{(\mathrm{y}+5)}\\ =\frac{(\mathrm{y}+2)\bcancel{(\mathrm{y}+5)}}{\bcancel{(\mathrm{y}+5)}}\\ =(\mathrm{y}+2)\end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)({\mathrm{m}}^2-14\mathrm{m}-32)÷(\mathrm{m}+2)\\ \mathrm{First}\mathrm{we}\mathrm{will}\mathrm{factorise}({\mathrm{m}}^2-14\mathrm{m}-32)\\ ={\mathrm{m}}^2-16\mathrm{m}+2\mathrm{m}-32\\ =\mathrm{m}(\mathrm{m}-16)+2(\mathrm{m}-16)\\ =(\mathrm{m}-16)(\mathrm{m}+2)\\ \\ \therefore \frac{({\mathrm{m}}^2-14\mathrm{m}-32)}{(\mathrm{m}+2)}\\ =\frac{(\mathrm{m}-16)\bcancel{(\mathrm{m}+2)}}{\bcancel{(\mathrm{m}+2)}}\\ =(\mathrm{m}-16)\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\left(\mathrm{iii}\right)(5{\mathrm{p}}^2-25\mathrm{p}+20)÷(\mathrm{p}-1)\\ \mathrm{First}\mathrm{we}\mathrm{will}\mathrm{factorise}(5{\mathrm{p}}^2-25\mathrm{p}+20)\\ =5{\mathrm{p}}^2-20\mathrm{p}-5\mathrm{p}+20\\ =5\mathrm{p}(\mathrm{p}-4)-5(\mathrm{p}-4)\\ =5(\mathrm{p}-1)(\mathrm{p}-4)\\ \\ \therefore \frac{(5{\mathrm{p}}^2-25\mathrm{p}+20)}{(\mathrm{p}-1)}\\ =\frac{5(\mathrm{p}-4)\bcancel{(\mathrm{p}-1)}}{\bcancel{(\mathrm{p}-1)}}\\ =5(\mathrm{p}-4)\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\left(\mathrm{iv}\right)4\mathrm{yz}({\mathrm{z}}^2+6\mathrm{z}-16)÷2\mathrm{y}(\mathrm{z}+8)\\ \mathrm{First}\mathrm{we}\mathrm{will}\mathrm{factorise}4\mathrm{yz}({\mathrm{z}}^2+6\mathrm{z}-16)\\ =4\mathrm{yz}({\mathrm{z}}^2+8\mathrm{z}-2\mathrm{z}-16)\\ =4\mathrm{yz}\left[\mathrm{z}(\mathrm{z}+8)-2(\mathrm{z}+8)\right]\\ =4\mathrm{yz}\left[(\mathrm{z}+8)(\mathrm{z}-2)\right]\\ \\ \therefore \frac{4\mathrm{yz}({\mathrm{z}}^2+6\mathrm{z}-16)}{2\mathrm{y}(\mathrm{z}+8)}\\ =\frac{4\bcancel{\mathrm{y}}\mathrm{z}\left[\bcancel{(\mathrm{z}+8)}(\mathrm{z}-2)\right]}{2\bcancel{\mathrm{y}}\bcancel{(\mathrm{z}+8)}}\\ =2\mathrm{z}(\mathrm{z}-2)\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\left(\mathrm{v}\right)5\mathrm{pq}({\mathrm{p}}^2-{\mathrm{q}}^2)÷2\mathrm{p}(\mathrm{p}+\mathrm{q})\\ \mathrm{First}\mathrm{we}\mathrm{will}\mathrm{factorise}5\mathrm{pq}({\mathrm{p}}^2-{\mathrm{q}}^2)\\ =5\mathrm{pq}(\mathrm{p}-\mathrm{q})(\mathrm{p}+\mathrm{q})\left[\mathrm{Byusing}:{\mathrm{a}}^2-{\mathrm{b}}^2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]\\ \\ \therefore \frac{5\mathrm{pq}({\mathrm{p}}^2-{\mathrm{q}}^2)}{2\mathrm{p}(\mathrm{p}+\mathrm{q})}\\ =\frac{5\bcancel{\mathrm{p}}\mathrm{q}(\mathrm{p}-\mathrm{q})\bcancel{(\mathrm{p}+\mathrm{q})}}{2\bcancel{\mathrm{p}}\bcancel{(\mathrm{p}+\mathrm{q})}}\\ =\frac{5\mathrm{q}(\mathrm{p}-\mathrm{q})}{2}\\ \end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\left(\mathrm{vi}\right)12\mathrm{xy}(9{\mathrm{x}}^2-16{\mathrm{y}}^2)÷4\mathrm{xy}(3\mathrm{x}+4\mathrm{y})\\ \mathrm{First}\mathrm{we}\mathrm{will}\mathrm{factorise}12\mathrm{xy}(9{\mathrm{x}}^2-16{\mathrm{y}}^2)\\ =12\mathrm{xy}[{\left(3\mathrm{x}\right)}^2-{\left(4\mathrm{y}\right)}^2]\\ =12\mathrm{xy}\left[(3\mathrm{x}-4\mathrm{y})(3\mathrm{x}+4\mathrm{y})\right]\left[\mathrm{Byusing}:{\mathrm{a}}^2-{\mathrm{b}}^2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]\\ \\ \therefore \frac{12\bcancel{\mathrm{xy}}\left[\bcancel{(3\mathrm{x}+4\mathrm{y})}(3\mathrm{x}-4\mathrm{y})\right]}{4\bcancel{\mathrm{xy}}\bcancel{(3\mathrm{x}+4\mathrm{y})}}\\ =3(3\mathrm{x}-4\mathrm{y})\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\left(\mathrm{vii}\right)39{\mathrm{y}}^3(50{\mathrm{y}}^2-98)÷26{\mathrm{y}}^2(5\mathrm{y}+7)\\ \mathrm{First}\mathrm{we}\mathrm{will}\mathrm{factorise}39{\mathrm{y}}^3(50{\mathrm{y}}^2-98)\\ =13\times 3\times {\mathrm{y}}^3\times 2(25{\mathrm{y}}^2-49)\\ =13\times 3\times {\mathrm{y}}^3\times 2[{\left(5\mathrm{y}\right)}^2-7^2]\\ =13\times 3\times {\mathrm{y}}^3\times 2\left[(5\mathrm{y}-7)(5\mathrm{y}+7)\right]\\ \left[\mathrm{Byusing}:{\mathrm{a}}^2-{\mathrm{b}}^2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]\\ \\ \therefore \frac{39{\mathrm{y}}^3(50{\mathrm{y}}^2-98)}{26{\mathrm{y}}^2(5\mathrm{y}+7)}\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}=\frac{\bcancel{13}\times 3\times {\mathrm{y}}^3\times \bcancel{2}\left[(5\mathrm{y}-7)\bcancel{(5\mathrm{y}+7)}\right]}{\bcancel{2}\times \bcancel{13}\times {\mathrm{y}}^2\bcancel{(5\mathrm{y}+7)}}\\ =3\mathrm{y}(5\mathrm{y}-7)\\ \\ \end{array}$

Q.12

\begin{array}{l}\text{Find and correct the errors in the following mathematical}\\ \text{statements}\text{.}\\ \text{1}\text{. 4(x -5) = 4x- 5 2}{\text{. x(3x + 2) = 3x}}^{\text{2}}\text{+ 2 3}\text{. 2x + 3y = 5xy}\\ \text{4}\text{. x + 2x + 3x = 5x 5}\text{. 5y + 2y + y- 7y = 0 6}{\text{. 3x + 2x = 5x}}^{\text{2}}\\ \text{7}{\text{. (2x)}}^{\text{2}}{\text{+ 4(2x) + 7 = 2x}}^{\text{2}}\text{+ 8x + 7}\\ \text{8}{\text{. (2x)}}^{\text{2}}\text{+ 5x = 4x + 5x = 9x}\\ \text{9}{\text{. (3x + 2)}}^{\text{2}}{\text{= 3x}}^{\text{2}}\text{+ 6x + 4}\end{array}

Ans

1. 4(x – 5) = 4x – 5

L.H.S.

4(x − 5)

= 4 × x − 4 × 5

= 4x − 20 ≠ R.H.S.

The correct statement is:

4(x − 5) = 4x – 20

2. x(3x + 2) = 3x² + 2

L.H.S.

= x(3x + 2)

= x × 3x + x × 2

= 3x² + 2x ≠ R.H.S.

The correct statement is:

x(3x + 2) = 3x² + 2x

3. 2x + 3y = 5xy

L.H.S.

= 2x + 3y

≠ R.H.S.

The correct statement is :

2x + 3y = 2x + 3y

4. x + 2x + 3x = 5x

L.H.S

= x + 2x + 3x

=1x + 2x + 3x

= x(1 + 2 + 3)

= 6x ≠ R.H.S.

The correct statement is :

x + 2x + 3x = 6x

5. 5y + 2y + y – 7y = 0

L.H.S.

= 5y + 2y + y − 7y

= 8y − 7y

= y ≠ R.H.S

The correct statement is:

5y + 2y + y − 7y = y

6. 3x + 2x = 5x²

L.H.S.

= 3x + 2x

= 5x ≠ R.H.S

The correct statement is:

3x + 2x = 5x

7. (2x)² + 4(2x) + 7 = 2x² + 8x + 7

L.H.S

= (2x)² + 4(2x) + 7

= 4x² + 8x + 7 ≠ R.H.S

The correct statement is:

(2x)² + 4(2x) + 7 = 4x² + 8x + 7

8. (2x)² + 5x = 4x + 5x = 9x

L.H.S

= (2x)² + 5x

= 4x² + 5x ≠ R.H.S.

The correct statement is:

(2x)² + 5x = 4x² + 5x

9. (3x + 2)² = 3x² + 6x + 4

L.H.S.

= (3x + 2)²