# Factorisation Solution of TS & AP Board Class 8 Mathematics

#### Factorisation Solution of TS & AP Board Class 8 Mathematics

EXERCISE:12.1

Question 1.

Find the common factors of the given terms in each.

8x, 24

Answer:

∴ Given terms are 8x and 24

Prime factors of given terms are:-

8x = 2 × 2 × 2 × x

24 = 2 × 2 × 2 × 3

As the x is an undefined value,

Common factors will be

2 × 2 × 2 = 8

Question 2.

Find the common factors of the given terms in each.

3a, 21ab

Answer:

∴ Given terms are 3a and 21ab

Prime factors of given terms are:-

3a = 3 × a

21ab = a × b × 7 × 3

Common factors will be

⇒ 3 × a = 3a

Question 3.

Find the common factors of the given terms in each.

7xy, 35x2y3

Answer:

∴ Given terms are 7xy and 35x2y3

Prime factors of given terms are:-

7xy = 7 × x × y

35x2y3 = x × x × y × y × 7 × 5

Common factors will be

⇒ 7 × x × y = 7xy

Question 4.

Find the common factors of the given terms in each.

4m2, 6m2, 8m3

Answer:

∴ Given terms are 4m2,6m2 and 8m3

Prime factors of given terms are:-

4m2 = 2 × 2 × m × m

6m2 = 3 × 2 × m × m

8m3 = 2 × 2 × 2 × m × m × m

Common factors will be

⇒ 2 × m × m = 2m2

Question 5.

Find the common factors of the given terms in each.

15p, 20qr, 25rp

Answer:

∴ Given terms are 15p,20qr and 25rp

Prime factors of given terms are:-

15p = 3 × 5 × p

20qr = 2 × 2 × 5 × q × r

25rp = 5 × 5 × r × p

⇒ Common factors will be 5

Question 6.

Find the common factors of the given terms in each.

4x2, 6xy, 8y2x

Answer:

∴ Given terms are 4x2,6xy and 8y2x

Prime factors of given terms are:-

4x2 = 2 × 2 × x × x

6xy = 3 × 2 × x × y

8y2x = 2 × 2 × 2 × y × y × x

Common factors will be

⇒ 2 × x = 2x

Question 7.

Find the common factors of the given terms in each.

12x2y, 18xy2

Answer:

Given terms are 12x2yand 18xy2

Prime factors of given terms are:-

12yx2 = 3 × 2 × 2 × y × x × x

18xy2 = 3 × 2 × 3 × x × y × y

Common factors will be

⇒ 2 × 3 × x × y = 6xy

Question 8.

Factorise the following expressions

5x2 – 25xy

Answer:

In the given expression

Check the common factors for all terms;

⇒ [5 × x × x - 5 × 5 × x × y]

⇒ 5 × x[x-5 × y]

⇒ 5x[x-5y]

∴ 5x2 - 25xy = 5x[x-5y]

Question 9.

Factorise the following expressions

9a2 – 6ax

Answer:

In the given expression

Check the common factors for all terms;

⇒ [5 × a × a- 2 × 3 × x × a]

⇒ a[5 × a-2 × 3 × x]

⇒ a[5a-6x]

∴ 9a2 - 6ax = a[5a-6x]

Question 10.

Factorise the following expressions

7p2 + 49pq

Answer:

In the given expression

Check the common factors for all terms;

⇒ [7 × p × p + 7 × 7 × p × q]

⇒ 7 × p[p + 7 × q]

⇒ 7p[p + 7q]

∴ 7p2 + 49pq = 7p[p + 7q]

Question 11.

Factorise the following expressions

36a2b – 60 a2bc

Answer:

In the given expression

Check the common factors for all terms;

⇒ [2 × 2 × 3 × 3 × a × a × b - 2 × 2 × 3 × 5 × a × a × b × c]

⇒ 2 × 2 × 3 × a × a × b[3 × b-5 × c]

⇒ 12a2b[3b-5c]

∴ 36a2b - 60 a2bc = 12a2b[3b-5c]

Question 12.

Factorise the following expressions

3a2bc + 6ab2c + 9abc2

Answer:

In the given expression

Check the common factors for all terms;

⇒ [3 × a × a × b × c + 2 × 3 × a × b × b × c + 3 × 3 × a × b × c × c]

⇒ 3 × a × b × c[a + 2 × b + 3 × c]

⇒ 3abc[a + 2b + 3c]

∴ 3a2bc + 6ab2c + 9abc2 = 3abc[a + 2b + 3c]

Question 13.

Factorise the following expressions

4p2 + 5pq – 6pq2

Answer:

In the given expression

Check the common factors for all terms;

⇒ [2 × 2 × p × p + 5 × p × q - 2 × 3 × p × q × q]

⇒ p[2 × 2 × p + 5 × q - 2 × 3 × q × q]

⇒ p[4p + 5q-6q2]

∴ 4p2 + 5pq – 6pq2 = p[4p + 5q-6q2]

Question 14.

Factorise the following expressions

ut + at2

Answer:

In the given expression

Check the common factors for all terms;

⇒ [u × t + a × t × t]

⇒ t[u + a × t]

⇒ t[u + at]

∴ ut + at2 = t[u + at]

Question 15.

Factorise the following:

3ax – 6xy + 8by – 4ab

Answer:

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1st 2 terms we have,

3ax-6xy = 3x[a-2y] -------eq 1

Regrouping the last 2 terms we have,

8by-4ab = -4b[a-2y] -------eq 2

∵ we have to make common parts in both eq 1 and 2

Combining eq 1 and 2

3ax – 6xy + 8by – 4ab = 3x[a-2y] + [-4b[a-2y] ]

= 3x[a-2y] - 4b[a-2y]

= [3x-4] [a-2y]

Hence the factors of 3ax – 6xy + 8by – 4ab are [3x-4] and [a-2y]

Question 16.

Factorise the following:

x3 + 2x2 + 5x + 10

Answer:

In the given expression

Check whether there is any common factors for all terms;

None;

Regrouping the 1st 2 terms we have,

x3 + 2x2 = x2[x + 2] -------eq 1

Regrouping the last 2 terms we have,

5x + 10 = 5[x + 2] -------eq 2

Combining eq 1 and 2

x3 + 2x2 + 5x + 10 = x2[x + 2] + 5[x + 2]

= [x2 + 5][x + 2]

Hence the factors of x3 + 2x2 + 5x + 10 are [x2 + 5] and [x + 2]

Question 17.

Factorise the following:

m2 – mn + 4m – 4n

Answer:

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1st 2 terms we have,

m2 - mn= m[m - n] -------eq 1

Regrouping the last 2 terms we have,

4m – 4n = 4[m – n] -------eq 2

Combining eq 1 and 2

m2 – mn + 4m – 4n = 4[m – n] + m[m - n]

= [4 + m][m-n]

Hence the factors of m2 – mn + 4m – 4n are [m – n] and [4 + m]

Question 18.

Factorise the following:

a3 – a2b2 – ab + b3

Answer:

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1st 2 terms we have,

a3 – a2b2 = a2[a-b2] -------eq 1

Regrouping the last 2 terms we have,

– ab + b3 = -b[a-b2] -------eq 2

∵ we have to make common parts in both eq 1 and 2

Combining eq 1 and 2

a3 – a2b2 – ab + b3 = a2[a-b2] -b[a-b2]

= [a2 – b][a – b2]

Hence the factors of a3 – a2b2 – ab + b3 are [a2 – b] and[a – b2]

Question 19.

Factorise the following:

p2q – pr2 – pq + r2

Answer:

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1st 2 terms we have,

p2q – pr2 = p[pq-r2] -------eq 1

Regrouping the last 2 terms we have,

– pq + r2 = -1[pq-r2] -------eq 2

∵ we have to make common parts in both eq 1 and 2

Combining eq 1 and 2

p2q – pr2 – pq + r2 = p[pq-r2] -1[pq-r2]

= [p – 1][pq – r2]

Hence the factors of p2q – pr2 – pq + r2 are [p – 1] and [pq – r2]

EXERCISE:12.2

Question 1.

Factories the following expression-

a2 + 10a + 25

Answer:

In the given expression

1st and last terms are perfect square

⇒ a2 = a × a

⇒ 25 = 5 × 5

And the middle expression is in form of 2ab

10a = 2 × 5 × a

∴ a × a + 2 × 5 × a + 5 × 5

Gives (a + b)2 = a2 + 2ab + b2

⇒ In a2 + 10a + 25

a = a and b = 5;

∴ a2 + 10a + 25 = (a + 5)2

Hence the factors of a2 + 10a + 25 are (a + 5) and (a + 5)

Question 2.

Factories the following expression-

l2 – 16l + 64

Answer:

In the given expression

1st and last terms are perfect square

⇒ l2 = l × l

⇒ 64 = 8 × 8

And the middle expression is in form of 2ab

16l = 2 × 8 × l

∴ l × l + 2 × 8 × l + 8 × 8

Gives (a-b)2 = a2-2ab + b2

⇒ In l2 + 16l + 64

a = l and b = 8;

∴ l2 + 16l + 64 = (l + 8)2

Hence the factors of l2 + 16l + 64 are (l + 8) and (l + 8)

Question 3.

Factories the following expression-

36x2 + 96xy + 64y2

Answer:

In the given expression

1st and last terms are perfect square

⇒ 36x2 = 6x × 6x

⇒ 64y2 = 8y × 8y

And the middle expression is in form of 2ab

96xy = 2 × 6x × 8y

∴ 6x × 6x + 2 × 8y × 6x + 8y × 8y

Gives (a + b)2 = a2 + 2ab + b2

⇒ In 36x2 + 96xy + 64y2

a = 6x and b = 8y;

∴ 36x2 + 96xy + 64y2 = (6x + 8y)2

Hence the factors of 36x2 + 96xy + 64y2 are (6x + 8y) and (6x + 8y)

Question 4.

Factories the following expression-

25x2 + 9y2 – 30xy

Answer:

In the given expression

1st and last terms are perfect square

⇒ 25x2 = 5x × 5x

⇒ 9y2 = 3y × 3y

And the middle expression is in form of 2ab

30xy = 2 × 5x × 3y

∴ 5x × 5x + 2 × 3y × 5x + 3y × 3y

Gives (a-b)2 = a2-2ab + b2

⇒ In 25x2 – 30xy + 9y2

a = 5x and b = 3y;

∴ 25x2 - 30xy + 9y2 = (5x-3y)2

Hence the factors of 25x2 - 30xy + 9y2 are (5x-3y) and (5x-3y)

Question 5.

Factories the following expression-

25m2 – 40mn + 16n2

Answer:

In the given expression

1st and last terms are perfect square

⇒ 25m2 = 5m × 5m

⇒ 16n2 = 4n × 4n

And the middle expression is in form of 2ab

40mn = 2 × 5m × 4n

∴ 5m × 5m - 2 × 4n × 5m + 4n × 4n

Gives (a-b)2 = a2-2ab + b2

⇒ In 25m2 – 40mn + 16n2

a = 5m and b = 4n;

∴ 25m2 – 40mn + 16n2 = (5m-4n)2

Hence the factors of 25m2 – 40mn + 16n2 are (5m-4n)and (5m-4n)

Question 6.

Factories the following expression-

81x2– 198 xy + 121y2

Answer:

In the given expression

1st and last terms are perfect square

⇒ 81x2 = 9x × 9x

⇒ 121y2 = 11y × 11y

And the middle expression is in form of 2ab

198xy = 2 × 9x × 11y

∴ 9x × 9x - 2 × 11y × 9x + 11y × 11y

Gives (a-b)2 = a2-2ab + b2

⇒ In 81x2 – 198xy + 121y2

a = 9x and b = 11y;

∴ 81x2 – 198xy + 121y2 = (9x-11y)2

Hence the factors of 81x2 – 198xy + 121y2 are (9x-11y)and (9x-11y)

Question 7.

Factories the following expression-

(x + y)2 – 4xy

(Hint: first expand (x + y)2

Answer:

If (a + b)2 = a2 + 2ab + b2

Then (x + y)2 – 4xy

= x2 + 2xy + y2-4xy

= x2 + y2-2xy

In given expression

1st and last terms are perfect square

⇒ x2 = x × x

⇒ y2 = y × y

And the middle expression is in form of 2ab

2xy = 2 × x × y

∴ x × x - 2 × y × x + y × y

Gives (a-b)2 = a2-2ab + b2

⇒ In x2 – 2xy + y2

a = x and b = y;

∴ x2 – 2xy + y2 = (x-y)2

Hence the factors of (x + y)2 – 4xy are (x-y)and (x-y)

Question 8.

Factories the following expression-

l4 + 4l2m2 + 4m4

Answer:

In given expression

1st and last terms are perfect square

⇒ l4 = l2 × l2

⇒ m4 = m2 × m2

And the middle expression is in form of 2ab

4l2m2 = 2 × l2 × m2

∴ l2 × l2 + 2 × m2 × l2 + m2 × m2

Gives (a + b)2 = a2 + 2ab + b2

⇒ In l4 + 4l2m2 + m4

a = l2 and b = m2;

∴ l4 – 4l2m2 + m4 = (l2-m2)2

Hence the factors of l4 + 4l2m2 + 4m4 are (l2-m2)and (l2-m2)

Question 9.

Factories the following

x2 – 36

Answer:

In given expression

Both terms are perfect square

⇒ x2 = x × x

⇒ 36 = 6 × 6

∴ x2-62

Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = x and b = 6;

x2 – 36 = (x + 6)(x-6)

Hence the factors of x2 – 36 are (x + 6) and (x-6)

Question 10.

Factories the following

49x2 – 25y2

Answer:

In given expression

Both terms are perfect square

⇒ 49x2 = 7x × 7x

⇒ 25y2 = 5y × 5y

∴ 49x2-25y2 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = 7x and b = 5y;

49x2 – 25y2 = (7x + 5y)(7x-5y)

Hence the factors of 49x2 – 25y2 are (7x + 5y) and (7x-5y)

Question 11.

Factories the following

m2 – 121

Answer:

In given expression

Both terms are perfect square

⇒ m2 = m × m

⇒ 121 = 11 × 11

∴ m2-121 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = m and b = 11;

m2 – 121 = (m + 11)(m-11)

Hence the factors of m2 – 121 are (m + 11) and (m-11)

Question 12.

Factories the following

81 – 64x2

Answer:

In given expression

Both terms are perfect square

⇒ 64x2 = 8x × 8x

⇒ 81 = 9 × 9

∴ 81-64x2 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = 9 and b = 8x;

81-64x2 = (9-8x)(9 + 8x)

Hence the factors of 81-64x2are(9-8x) and (9 + 8x)

Question 13.

Factories the following

x2y2 – 64

Answer:

In given expression

Both terms are perfect square

⇒ y2x2 = xy × xy

⇒ 64 = 8 × 8

∴ x2y2 – 64 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = xy and b = 8;

x2y2 – 64 = (xy-8)(xy + 8)

Hence the factors of x2y2 – 64 are (xy-8) and (xy + 8)

Question 14.

Factories the following

6x2 – 54

Answer:

In given expression

Take out the common factor,

[2 × 3 × x × x-2 × 3 × 3 × 3]

⇒ 2 × 3[x × x-3 × 3]

⇒ 6[x2-9]

Both terms are perfect square

⇒ x2 = x × x

⇒ 9 = 3 × 3

∴ x2– 9 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = x and b = 3;

x2– 9 = (x-3)(x + 3)

Hence the factors of 6x2 – 54 are 6,(x-3) and (x + 3)

Question 15.

Factories the following

x2– 81

Answer:

In given expression

Both terms are perfect square

⇒ x2 = x × x

⇒ 81 = 9 × 9

∴ x2 – 81 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = x and b = 9;

x2 – 81 = (x-9)(x + 9)

Hence the factors of x2 – 81 are (x-9) and (x-9)

Question 16.

Factories the following

2x – 32x5

Answer:

In given expression

Take out the common factor,

[2 × x - 2 × 2 × 2 × 2 × 2 × x × x × x × x × x]

⇒ 2 × x[1 - 2 × 2 × 2 × 2 × x × x × x × x]

⇒ 2x [1-16x4] = 2x [1-(2x)4]

⇒ In the term 1-(2x)4

= 1-(4x2)2

Both terms are perfect square

⇒ (4x2 )2 = 4x2 × 4x2

⇒ 1 = 1 × 1

∴ 1-(4x2)2 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = 1 and b = 4x2;

1-16x4 = (1-4x2)(1 + 4x2)

→ 1-4x2 = 1-(2x)2

∴ 1-4x2 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = 1 and b = 2x;

1-4x2 = (1-2x)(1 + 2x)

∴ 1-16x2 = (1-2x)(1 + 2x) (1 + 4x2)

Hence the factors of 2x – 32x5 are 2x,(1-2x),(1 + 2x) and (1 + 4x2)

Question 17.

Factories the following

81x4 – 121x2

Answer:

In given expression

Take out the common factor,

[3 × 3 × 3 × 3 × x × x × x × x - 11 × 11 × x × x]

⇒ x × x[3 × 3 × 3 × 3 × x × x - 11 × 11]

⇒ x2[81x2 – 121]

In expression 81x2 - 121

Both terms are perfect square

⇒ 81x2 = 9x × 9x

⇒ 121 = 11 × 11

∴ 81x2 – 121 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = 9x and b = 11;

81x2 – 121 = (9x-11)(9x + 11)

Hence the factors of 81x4 – 121x2 are x2,(9x-11) and (9x + 11)

Question 18.

Factories the following

(p2 – 2pq + q2) – r2

Answer:

In the given expression p2 – 2pq + q2

1st and last terms are perfect square

⇒ p2 = p × p

⇒ q2 = q × q

And the middle expression is in form of 2ab

2pq = 2 × p × q

∴ p × p - 2 × p × q + q × q

Gives (a-b)2 = a2-2ab + b2

⇒ In p2 – 2pq + q2

a = p and b = q;

∴ p2 – 2pq + q2 = (p-q)2

Now the given expression is (p-q)2– r2

Both terms are perfect square

⇒ (p-q)2 = (p-q) × (p-q)

⇒ r2 = r × r

∴ (p-q)2– r2 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = (p-q) and b = r;

(p-q)2– r2 = (p-q-r) (p-q + r)

Hence the factors of (p2 – 2pq + q2) – r2 are (p-q-r) and (p-q + r)

Question 19.

Factories the following

(x + y)2 – (x – y)2

Answer:

In the given expression

We know that

(a + b)2 = a2 + 2ab + b2

(a-b)2 = a2-2ab + b2

Hence

If a = x and b = y

(x + y)2 – (x – y)2 = x2 + y2 + 2xy – [x2 + y2-2xy]

= x2 + y2 + 2xy -x2-y2 + 2xy

= 4xy

Question 20.

Factories the expressions-

lx2 + mx

Answer:

In the given expression

Take out the common in all the terms,

⇒ lx2 + mx

⇒ x[lx + m]

Question 21.

Factories the expressions-

7y2 + 35z2

Answer:

In the given expression

Take out the common in all the terms,

⇒ 7y2 + 35z2

⇒ 7[y2 + 5z2]

Question 22.

Factories the expressions-

3x4 + 6x3y + 9x2z

Answer:

In the given expression

Take out the common in all the terms,

⇒ 3x4 + 6x3y + 9x2z

⇒ 3x2[x2 + 2xy + 3z]

Question 23.

Factories the expressions-

x2 – ax – bx + ab

Answer:

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1st 2 terms we have,

x2 - ax= x[x - a] -------eq 1

Regrouping the last 2 terms we have,

-bx + ab = -b[x – a] -------eq 2

Combining eq 1 and 2

x2 – ax – bx + ab = x[x - a] - b[x – a]

= [x - b][x - a]

Hence the factors of[ x2 – ax – bx + ab] are [x - b]and [x - a]

Question 24.

Factories the expressions-

3ax – 6ay – 8by + 4bx

Answer:

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1st 2 terms we have,

3ax – 6ay= 3a[x - 2y] -------eq 1

Regrouping the last 2 terms we have,

-8by + 4bx = 4b[x – 2y] -------eq 2

Combining eq 1 and 2

3ax – 6ay – 8by + 4bx = 3a[x - 2y] + 4b[x – 2y]

= [x – 2y][3a + 4b]

Hence the factors of[3ax – 6ay – 8by + 4bx] are [x – 2y] and [3a + 4b]

Question 25.

Factories the expressions-

mn + m + n + 1

Answer:

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1st 2 terms we have,

mn + m = m[n + 1] -------eq 1

Regrouping the last 2 terms we have,

n + 1 = 1[n + 1] -------eq 2

Combining eq 1 and 2

mn + m + n + 1 = m[n + 1] + 1[n + 1]

= [m + 1][n + 1]

Hence the factors of[mn + m + n + 1] are [m + 1] and [n + 1]

Question 26.

Factories the expressions-

6ab – b2 + 12ac – 2bc

Answer:

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1st 2 terms we have,

6ab – b2 = b[6a - b] -------eq 1

Regrouping the last 2 terms we have,

12ac – 2bc = 2c[6a - b] -------eq 2

Combining eq 1 and 2

6ab – b2 + 12ac – 2bc = b[6a - b] + 2c[6a - b]

= [6a - b][b + 2c]

Hence the factors of[6ab – b2 + 12ac – 2bc] are [6a - b] and[b + 2c]

Question 27.

Factories the expressions-

p2q – pr2 – pq + r2

Answer:

In the given expression

Check weather there is any common factors for all terms;

None;

Regrouping the 1st 2 terms we have,

p2q – pr2 = p[pq – r2] -------eq 1

Regrouping the last 2 terms we have,

– pq + r2 = -1[pq – r2] -------eq 2

∵ we have to make common parts in both eq 1 and 2

Combining eq 1 and 2

p2q – pr2 – pq + r2 = p[pq – r2] -1[pq – r2]

= [pq – r2][p - 1]

Hence the factors of[p2q – pr2 – pq + r2] are [pq – r2] and [p - 1]

Question 28.

Factories the expressions-

x (y + z) – 5 (y + z)

Answer:

In the given expression

Take out the common in all the terms,

⇒ x (y + z) – 5 (y + z)

⇒ (y + z)(x - 5)

Hence the factors of x (y + z) – 5 (y + z) are (y + z) and (x - 5)

Question 29.

Factories the following

x4 – y4

Answer:

In expression x4 – y4

Both terms are perfect square

⇒ x4 = x2 × x2

⇒ y4 = y2 × y2

∴ x4 – y4 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = x2 and b = y2;

x4 – y4 = (x2 – y2)( x2 + y2),

∴ x2 – y2 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = x and b = y;

x2 – y2 = (x– y)( x+ y),

⇒ x4 – y4 = (x– y)( x+ y), ( x2 + y2)

Hence the factors of x4 – y4 are (x– y),( x+ y) and ( x2 + y2)

Question 30.

Factories the following

a4 – (b + c)4

Answer:

In expression a4 – (b + c)4

Both terms are perfect square

⇒ a4 = a2 × a2

⇒ (b + c)4 = (b + c)2 × (b + c)2

∴ a4 – (b + c)4 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = a2 and b = (b + c)2;

a4 – (b + c)4 = (a2 – (b + c)2)( a2 + (b + c)2),

∴ a2 – (b + c)2 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = a and b = (b + c);

a2 – (b + c)2 = (a– (b + c))( a+ (b + c)),

⇒ a4 – (b + c)4 = (a– (b + c))( a+ (b + c)), ( a2 + (b + c)2)

⇒ a4 – (b + c)4 = (a–b–c)(a + b + c), (a2 + b2 + c2 + 2bc)

Hence the factors of a4 – (b + c)4 are (a–b–c),(a + b + c),( a2 + b2 + c2 + 2bc)

Question 31.

Factories the following

l2 – (m – n)2

Answer:

In the given expression l2 – (m – n)2

Both terms are perfect square

⇒ l2 = l × l

⇒ (m – n)2 = (m – n) × (m – n)

∴ l2 – (m - n)2 Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = a and b = (m - n);

∴ l2 – (m – n)2 = (l + m-n)(l-m + n)

Hence the factors of l2–(m–n)2are (l + m-n)(l-m + n)

Question 32.

Factories the following

49x2 –

Answer:

In the given expression 49x2 –

Both terms are perfect square

⇒ 49x2 = 7x × 7x

⇒ ()2 =

49x2 – Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = 7x and b = ;

∴ (7x)2 –()2 = ( 7x – ) (7x +  )

Hence the factors of 49x2 – are ( 7x – ) and (7x +  )

Question 33.

Factories the following

x4 – 2x2y2 + y4

Answer:

In the given expression

1st and last terms are perfect square

⇒ x4 = x2 × x2

⇒ y4 = y2 × y2

And the middle expression is in form of 2ab

2x2y2 = 2 × x2 × y2

∴ x2 × x2 - 2 × x2 × y2 + y2 × y2

Gives (a-b)2 = a2-2ab + b2

⇒ x4 – 2x2y2 + y4

a = x2 and b = y2;

∴ x4 – 2x2y2 + y4 = (x2 – y2) (x2 + y2)

Hence the factors of x4 – 2x2y2 + y4 are (x2 – y2) and (x2 + y2)

Question 34.

Factories the following

4 (a + b)2 – 9 (a – b)2

Answer:

In the given expression

We know that

(a + b)2 = a2 + 2ab + b2

(a-b)2 = a2-2ab + b2

Hence

4[a2 + 2ab + b2] – 9[a2-2ab + b2]

4a2 + 8ab + 4b2 - 9a2 + 18ab - 9b2

26ab – 5a2 - 5b2

25ab + ab – 5a2 – 5b2

[25ab – 5a2] + [ab – 5b2]

5a[5b – a] – b[5b – a]

[5a – b][5b – a]

Hence the factors 4 (a + b)2 – 9 (a – b)2 are [5a – b] and [5b – a]

Question 35.

Factories the following expressions

a2 + 10a + 24

Answer:

The given expression looks as

x2 + (a + b)x + ab

where a + b = 10; and ab = 24;

factors of 24 their sum

1 × 24 1 + 24 = 25

12 × 2 2 + 12 = 14

6 × 4 6 + 4 = 10

∴ the factors having sum 10 are 6 and 4

a2 + 10a + 24 = a2 + (6 + 4)a + 24

= a2 + 6a + 4a + 24

= a(a + 6) + 4(a + 6)

= (a + 6)(a + 4)

Hence the factors of a2 + 10a + 24 are (a + 6) and (a + 4)

Question 36.

Factories the following expressions

x2 + 9x + 18

Answer:

The given expression looks as

x2 + (a + b)x + ab

where a + b = 9; and ab = 18;

factors of 18 their sum

1 × 18 1 + 18 = 19

9 × 2 2 + 9 = 11

6 × 3 6 + 3 = 9

∴ the factors having sum 9 are 6 and 3

x2 + 9x + 18 = x2 + (6 + 3)x + 18

= x2 + 6x + 3x + 18

= x(x + 6) + 3(x + 6)

= (x + 6)(x + 3)

Hence the factors of x2 + 9x + 18 are (x + 6) and (x + 3)

Question 37.

Factories the following expressions

p2 – 10p + 21

Answer:

The given expression looks as

x2 + (a + b)x + ab

where a + b = -10; and ab = 21;

factors of 21 their sum

-1 × -21 -1-18 = -19

-7 × -3 -7-3 = -10

∴ the factors having sum -10 are -7 and -3

p2 + 9p + 18 = p2 + (-7-3)p + 21

= p2 -7p-3p + 21

= p(p-7) -3(p-7)

= (p-7)(p-3)

Hence the factors of p2 + 9p + 18 are (p-7) and (p-3)

Question 38.

Factories the following expressions

x2 – 4x – 32

Answer:

The given expression looks as

x2 + (a + b)x + ab

where a + b = -4; and ab = -32;

factors of -32 their sum

1 × -32 1-32 = -31

-16 × 2 2 -16 = - 14

-8 × 4 4 -8 = -4

∴ the factors having sum -4 are -8 and 4

x2 – 4x – 32 = x2 + (4 -8)x - 32

= x2 + 4x - 8x - 32

= x(x + 4) -8(x + 4)

= (x + 4)(x-8)

Hence the factors of x2 – 4x – 32 are (x + 4) and (x-8)

Question 39.

The lengths of the sides of a triangle are integrals, and its area is also integer. One side is 21 and the perimeter is 48. Find the shortest side.

Answer:

A = √ s(s-a)(s-b)(s-c)

If the area is an integer

Then [s(s-a)(s-b)(s-c)] should be proper square

If s =  Then s =  = 24

Hence ;

A = √ 24(24-a)(24-b)(24-c)

If side of triangle are

a = 21 and b + c = 27

let c be smallest side

then b = 27-c

∴ √ 24(24-21)(24-27 + c)(24-c)

⇒ √ 24 × 3 × (c-3)(24-c)

⇒ √ 2 × 2 × 2 × 3 × 3 × (c-3)(24-c)

⇒ 2 × 3√2(c-3)(24-c)

⇒ 6√2(c-3)(24-c)

∴ the value of [2(c-3)(24-c)] must be a perfect square for area to be a integer

For getting square 2(c-3) should be equal to (24-c)

2(c-3) = (24-c)

2c-6 = 24-c

2c + c = 24 + 6

3c = 10

c = 10; b = 27-c = 27-10 = 17

Hence the size of smallest size is 10.

Question 40.

Find the values of ‘m’ for which x2 + 3xy + x + my –m has two linear factors in x and y, with integer coefficients.

Answer:

For the given 2 degree equation

That must be equal to(ax + by + c)(dx + e)

= ad.x2 + bd.xy + cd.x + ea.x + be.y + ec

= ad.x2 + bd.xy + (cd + ea).x + be.y + ec

x2 + 3xy + x + my–m = ad.x2 + bd.xy + (cd + ea).x + be.y + ec

compare the equation

and take out the coefficient of every term

a.d = 1 ----------1

b.d = 3 ----------2

c.d + e.a = 1 ----------3

b.e = m ----------4

e.c = -m ----------5

⇒ from eq 1; a = d = 1 ∵ all coefficient are integers

After putting result in eq 3; c + e = 1 -------6

After putting result in eq 2; b = 3 --------7

⇒ divide eq 4 and 5

∴ that implies b = -c = -3 ∵ eq 7

Put value of c in eq 6

-3 + e = 1

e = 1 + 3 = 4

Putting value of b and e in eq 4

m = b × e

m = 3 × 4 = 12

EXERCISE:12.3

Question 1.

Carry out the following divisions

48a3 by 6a

Answer:

In the given term

Dividend = 48a3 = 2 × 2 × 2 × 2 × 3 × a × a × a

Divisor = 6a = 2 × 3 × a

= 2 × 2 × 2 × a × a

= 8a2

Hence dividing 48a3 by 6a gives 8a2

Question 2.

Carry out the following divisions

14x3 by 42x2

Answer:

In the given term

Dividend = 14x3 = 2 × 7 × x × x × x

Divisor = 42x2 = 2 × 3 × 7 × x × x

Hence dividing 14x3 by 42x2 gives

Question 3.

Carry out the following divisions

72a3b4c5 by 8ab2c3

Answer:

In the given term

Dividend = 72a3b4c5 = 2 × 2 × 2 × 3 × 3 × a × a × a × b × b × b × b × c × c × c × c × c

Divisor = 8ab2c3 = 2 × 2 × 2 × a × b × b × c × c × c

= 3 × 3 × a × a × b × b × c × c

= 9a2b2c2

Hence dividing 72a3b4c5 by 8ab2c3 gives 9a2b2c2

Question 4.

Carry out the following divisions

11xy2z3 by 55xyz

Answer:

In the given term

Dividend = 11xy2z3 = 11 × x × y × y × z × z × z

Divisor = 55xyz = 5 × 11 × x × y × z

Hence dividing 11xy2z3 by 55xyz gives

Question 5.

Carry out the following divisions

–54l4m3n2 by 9l2m2n2

Answer:

In the given term

Dividend = -54l4m3n2 = (-1) × 2 × 3 × 3 × 3 × l × l × l × l × m × m × m × n × n

Divisor = 9l2m2n2 = 3 × 3 × l × l × m × m × n × n

= (-1) × 3 × 2 × l × l × m

= -6l2m

Hence dividing –54l4m3n2 by 9l2m2n2 gives -6l2m

Question 6.

Divide the given polynomial by the given monomial

(3x2– 2x) ÷ x

Answer:

In the given term

Dividend = (3x2– 2x)

Take out the common part in binomial term

= (3 × x × x– 2 × x)

= x(3x-2)

Divisor = x

= 3x-2

Hence dividing (3x2– 2x) by x gives out 3x-2

Question 7.

Divide the given polynomial by the given monomial

(5a3b – 7ab3) ÷ ab

Answer:

In the given term

Dividend = (5a3b – 7ab3)

Take out the common part in binomial term

= (5 × a × a × a × b – 7 × a × b × b × b)

= ab(5a2 – 7b2)

Divisor = ab

= (5a2 – 7b2)

Hence dividing (5a3b – 7ab3) by ab gives out (5a2 – 7b2)

Question 8.

Divide the given polynomial by the given monomial

(25x5 – 15x4) ÷ 5x3

Answer:

In the given term

Dividend = (25x5 – 15x4)

Take out the common part in binomial term

= (5 × 5 × x × x × x × x × x – 3 × 5 × x × x × x × x)

= (5x – 3)5x4

Divisor = 5x3

= (5x – 3)x

= 5x2 – 3x

Hence dividing (25x5 – 15x4) by 5x3 gives out 5x2 – 3x

Question 9.

Divide the given polynomial by the given monomial

4(l5 – 6l4 + 8l3) ÷ 2l2

Answer:

In the given term

Dividend = (4l5 – 6l4 + 8l3)

Take out the common part in binomial term

= (2 × 2 × l × l × l × l × l– 3 × 2 × l × l × l × l + 2 × 2 × 2 × l × l × l )

= (2l2 – 3l + 4)2l3

Divisor = 2l2

= (2l2 – 3l + 4)l

= (2l3 –2l2 + 4l)

Hence dividing 4(l5 – 6l4 + 8l3) by 2l2 gives out (2l3 –2l2 + 4l)

Question 10.

Divide the given polynomial by the given monomial

15 (a3b2c2– a2b3c2 + a2b2c3) ÷ 3abc

Answer:

In the given term

Dividend = 15 (a3b2c2– a2b3c2 + a2b2c3)

Take out the common part in binomial term

= 3 × 5(a × a × a × b × b × c × c– a × a × b × b × b × c × c + a × a × b × b × c × c × c )

= 15 a2b2c2(a – b + c)

Divisor = 3abc

= 5abc[a-b + c]

= [5a2bc– 5ab2c + 5abc2]

Hence dividing 15 (a3b2c2– a2b3c2 + a2b2c3) by 3abc gives out [5a2bc– 5ab2c + 5abc2]

Question 11.

Divide the given polynomial by the given monomial

(3p3– 9p2q - 6pq2) ÷ (–3p)

Answer:

In the given term

Dividend = (3p3– 9p2q - 6pq2)

Take out the common part in binomial term

= (3 × p × p × p– 3 × 3 × p × p × q - 2 × 3 × p × q × q )

= 3 × p(p2– 3pq - 2q2)

Divisor = (–3p)

= (-1) (p2– 3pq - 2q2)

= (2q2 + 3pq - p2)

Hence dividing (3p3– 9p2q - 6pq2) by (–3p) gives out (2q2 + 3pq - p2)

Question 12.

Divide the given polynomial by the given monomial

( a2b2c2 + ab2c2) ÷abc

Answer:

In the given term

Dividend = ( a2b2c2 + ab2c2)

Take out the common part in binomial term

= ( × a × a × b × b × c × c +  × 2 × a × b × b × c × c )

× a × b × b × c × c (a + 2)

ab 2c2(a + 2)

Divisor = abc

bc(a + 2)

Hence dividing ( a2b2c2 + ab2c2) by abc gives out  bc(a + 2)

Question 13.

Workout the following divisions:

(49x – 63) ÷ 7

Answer:

In the given term

Dividend = (49x – 63)

Take out the common part in binomial term

= (7 × 7 × x - 7 × 9)

= 7(7 × x - 9)

= 7(7x - 9)

Divisor = 7

= (7x - 9)

Hence dividing (49x – 63) by 7 gives out (7x - 9)

Question 14.

Workout the following divisions:

12x (8x – 20) ÷ 4(2x – 5)

Answer:

In the given term

Dividend = 12x (8x – 20)

Take out the common part in binomial term

= 2 × 2 × 3 × x(2 × 2 × 2 × x - 2 × 2 × 5 )

= 2 × 2 × 2 × 2 × 3 × x(2 × x - 5 )

= 48x (2x - 5 )

Divisor = 4(2x – 5)

= 12x

Hence divides 12x (8x – 20) by 4(2x – 5) gives out 12x

Question 15.

Workout the following divisions:

11a3b3(7c – 35) ÷ 3a2b2(c – 5)

Answer:

In the given term

Dividend = 11a3b3(7c – 35) ÷ 3a2b2(c – 5)

Take out the common part in binomial term

= 11 × a × a × a × b × b × b (7 × c - 5 × 7 )

= 11 × a × a × a × b × b × b × 7 (c - 5 )

= 77a3b3(c - 5)

Divisor = 3a2b2(c – 5)

=  ab

Hence dividing 11a3b3(7c – 35) by 3a2b2(c – 5) gives out  ab

Question 16.

Workout the following divisions:

54lmn (l + m) (m + n) (n + l) ÷ 81mn (l + m) (n + l)

Answer:

In the given term

Dividend = 54lmn (l + m) (m + n) (n + l)

Divisor = 81mn (l + m) (n + l)

l(m + n)

Hence dividing 54lmn (l + m) (m + n) (n + l) by 81mn (l + m)(n + l) gives out  l(m + n)

Question 17.

Workout the following divisions:

36 (x + 4) (x2 + 7x + 10) ÷ 9 (x + 4)

Answer:

In the given term

Dividend = 36 (x + 4) (x2 + 7x + 10)

Divisor = 9 (x + 4)

= 4(x2 + 7x + 10)

= (4x2 + 27x + 40)

Hence dividing 36 (x + 4) (x2 + 7x + 10) by 9 (x + 4) gives out

(4x2 + 27x + 40)

Question 18.

Workout the following divisions:

a (a + 1) (a + 2) (a + 3) ÷ a (a + 3)

Answer:

In the given term

Dividend = a (a + 1) (a + 2) (a + 3)

Divisor = a (a + 3)

=

= (a + 1) (a + 2)

Hence dividing a (a + 1) (a + 2) (a + 3) by a (a + 3) gives out

(a + 1) (a + 2)

Question 19.

Factorize the expressions and divide them as directed:

(x2 + 7x + 12) ÷ (x + 3)

Answer:

In the given term

Dividend = (x2 + 7x + 12)

The given expression looks as

x2 + (a + b)x + ab

where a + b = 7; and ab = 12;

factors of 12 their sum

1 × 12 1 + 12 = 13

6 × 2 2 + 6 = 8

4 × 3 4 + 3 = 7

∴ the factors having sum 7 are 4 and 3

x2 + 7x + 12 = x2 + (4 + 3)x + 12

= x2 + 4x + 3x + 12

= x(x + 4) + 3(x + 4)

= (x + 4)(x + 3)

Divisor = (x + 3)

= (x + 4)

Hence dividing (x2 + 7x + 12) by (x + 3) gives out (x + 4)

Question 20.

Factorize the expressions and divide them as directed:

(x2 – 8x + 12) ÷ (x – 6)

Answer:

In the given term

Dividend = (x2 - 8x + 12)

The given expression looks as

x2 + (a + b)x + ab

where a + b = -8; and ab = 12;

factors of 12 their sum

-1 × -12 -1-12 = -13

-6 × -2 -2-6 = -8

-4 × -3 -4-3 = -7

∴ the factors having sum 7 are 4 and 3

x2 - 8x + 12 = x2 + (-6-2)x + 12

= x2 - 6x - 2x + 12

= x(x - 6) -2(x - 6)

= (x - 6)(x - 2)

Divisor = (x - 6)

= (x – 2)

Hence dividing (x2 – 8x + 12) by (x – 6) gives out (x – 2)

Question 21.

Factorize the expressions and divide them as directed:

(p2 + 5p + 4) ÷ (p + 1)

Answer:

In the given term

Dividend = (p2 + 5p + 4)

The given expression looks as

x2 + (a + b)x + ab

where a + b = 5; and ab = 4;

factors of 4 their sum

1 × 4 1 + 4 = 5

2 × 2 2 + 2 = 4

∴ the factors having sum 5 are 4 and 1

(p2 + 5p + 4) = p2 + (4 + 1)p + 4

= p2 + 4p + p + 4

= p(p + 4) + 1(p + 4)

= (p + 1)(p + 4)

Divisor = (p + 1)

= (p + 4)

Hence dividing (p2 + 5p + 4) by (p + 1) gives out (p + 4)

Question 22.

Factorize the expressions and divide them as directed:

15ab (a2–7a + 10) ÷ 3b (a – 2)

Answer:

In the given term

Dividend = 15ab (a2–7a + 10)

The given expression (a2–7a + 10) looks as

x2 + (a + b) x + ab

where a + b = -7; and ab = 10;

factors of 10 their sum

-1 × -10 -1-10 = -11

-2 × -5 -2-5 = -7

∴ the factors having sum -7 are -2 and -5

(a2–7a + 10) = a2 + (-2-5)a + 10

= a2–5a – 2a + 10

= a(a – 5) – 2(a – 5)

= (a – 5)(a – 2)

Divisor = 3b (a – 2)

= 5a(a – 5)

Hence dividing 15ab (a2–7a + 10) by 3b (a – 2) gives out 5a(a – 5)

Question 23.

Factorize the expressions and divide them as directed:

15lm (2p2–2q2) ÷ 3l (p + q)

Answer:

In the given term

Dividend = 15lm (2p2–2q2)

In given expression (2p2–2q2)

Take out the common factor in binomial term

⇒ (2 × p × p – 2 × q × q)

→ 2(p2 – q2)

Both terms are perfect square

⇒ p2 = p × p

⇒ q2 = q × q

∴ (p2 – q2) Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = p and b = q;

p2 – q2 = (p + q)(p – q)

Hence the factors of p2 – q2 are (p + q) and (p – q)

Divisor = 3l (p + q)

=

=

= 10m(p – q)

Hence dividing 15lm (2p2–2q2) by 3l (p + q) gives out 10m(p – q)

Question 24.

Factorize the expressions and divide them as directed:

26z3(32z2–18) ÷ 13z2(4z – 3)

Answer:

In the given term

Dividend = 26z3(32z2–18)

Take out the common factor in binomial term

⇒ 2 × 13 × z × z × z (2 × 2 × 2 × 2 × 2 × z × z – 2 × 3 × 3)

⇒ 2 × 2 × 13 × z × z × z (2 × 2 × 2 × 2 × z × z – 3 × 3)

⇒ 52z3(16z2 – 9)

In given expression (16z2 – 9)

Both terms are perfect square

⇒ 16z2 = 4z × 4z

⇒ 9 = 3 × 3

∴ (16z2 – 9) Seems to be in identity a2-b2 = (a + b)(a-b)

Where a = 4z and b = 3;

(16z2 – 9) = (4z + 3)(4z – 3)

Hence the factors of (16z2 – 9)are (4z + 3) and (4z – 3)

Divisor = 13z2(4z – 3)

=

=

= 4z(4z + 3)

Hence dividing 26z3(32z2–18) by 13z2(4z – 3) gives out 4z(4z + 3)

EXERCISE:12.4

Question 1.

Find the errors and correct the following mathematical sentences

3(x – 9) = 3x – 9

Answer:

If LHS is

3(x – 9)

Then RHS would be

⇒ 3(x – 9)

= 3 × x – 3 × 9

= 3x – 27

The error is 27 instead of 9

Hence 3(x – 9) = 3x – 27

Question 2.

Find the errors and correct the following mathematical sentences

x(3x + 2) = 3x2 + 2

Answer:

If LHS is

x(3x + 2)

Then RHS would be

⇒ x(3x + 2)

= 3 × x × x – 2 × x

= 3x2 – 2x

The error is 2x instead of 2

Hence x(3x + 2) = 3x2 + 2x

Question 3.

Find the errors and correct the following mathematical sentences

2x + 3x = 5x2

Answer:

If LHS is

2x + 3x

Then RHS would be

⇒ 2x + 3x

= x(2 + 3)

= 5x

The error is 5x instead of 5x2

Hence 2x + 3x = 5x

Question 4.

Find the errors and correct the following mathematical sentences

2x + x + 3x = 5x

Answer:

If LHS is

2x + x + 3x = 5x

Then RHS would be

⇒ 2x + x + 3x

= x(2 + 1 + 3)

= 6x

The error is 6x instead of 5x

Hence 2x + x + 3x = 6x

Question 5.

Find the errors and correct the following mathematical sentences

4p + 3p + 2p + p – 9p = 0

Answer:

If LHS is

4p + 3p + 2p + p – 9p

Then RHS would be

⇒ 4p + 3p + 2p + p – 9p

= p(4 + 3 + 2 + 1–9)

= p(10–9)

= p

The error is p instead of 0

Hence 4p + 3p + 2p + p–9p = p

Question 6.

Find the errors and correct the following mathematical sentences

3x + 2y = 6xy

Answer:

If RHS is

6xy

Then LHS would be

⇒ 6xy

= 2 × 3 × x × y

= 3 × x × 2 × y

= 3x × 2y

The error is sign of multiplication instead of sign of addition

Hence 3x × 2y = 6xy

Question 7.

Find the errors and correct the following mathematical sentences

(3x)2 + 4x + 7 = 3x2 + 4x + 7

Answer:

If LHS is

(3x)2 + 4x + 7

Then RHS would be

⇒ (3x)2 + 4x + 7

= 32 × x2 + 4x + 7

= 9x2 + 4x + 7

The error is 9x2 instead of 3x2

Hence (3x)2 + 4x + 7 = 9x2 + 4x + 7

Question 8.

Find the errors and correct the following mathematical sentences

(2x)2 + 5x = 4x + 5x = 9x

Answer:

If LHS is

(2x)2 + 5x

Then RHS would be

⇒ (2x)2 + 5x

= 22 × x2 + 5x

= 4x2 + 5x

The error is 4x2 instead of 4x

Hence (2x)2 + 5x = 4x2 + 5x

Question 9.

Find the errors and correct the following mathematical sentences

(2a + 3)2 = 2a2 + 6a + 9

Answer:

If LHS is

(2a + 3)2

Then RHS would be

⇒ (2a + 3)2

= (2a)2 + 32 + 2 × 2a × 3

= 4a2 + 9 + 12a

= 4a2 + 12a + 9

The error is 4a2 instead of 2a2 and 12a instead of 6a

Hence = (2a + 3)2 = 4a2 + 9 + 12a

Question 10.

Find the errors and correct the following mathematical sentences

Substitute x = – 3 in

(a) x2 + 7x + 12 = (–3)2 + 7 (–3) + 12 = 9 + 4 + 12 = 25

Answer:

If LHS is

x2 + 7x + 12

Then RHS would be

⇒ x2 + 7x + 12

Putting x = (-3)

= (–3)2 + 7 (–3) + 12

= 9 + (-21) + 12

= 21-21

= 0

The error is (-21) instead of 4 and end result 0 instead of 25

Hence putting x = (-3) in x2 + 7x + 12 results to 0

Question 11.

Find the errors and correct the following mathematical sentences

Substitute x = – 3 in

(b) x2– 5x + 6 = (–3)2 –5 (–3) + 6 = 9 – 15 + 6 = 0

Answer:

If LHS is

x2– 5x + 6

Then RHS would be

⇒ x2– 5x + 6

Putting x = (-3)

= (–3)2 –5 (–3) + 6

= 9 + 15 + 6

= 30

The error is + 15 instead of (-15) and end results to 30 instead of 0

Hence putting x = (-3) in x2– 5x + 6 results to 30

Question 12.

Find the errors and correct the following mathematical sentences

Substitute x = – 3 in

(c) x2 + 5x = (–3)2 + 5 (–3) + 6 = – 9 – 15 = –24

Answer:

If LHS is

x2 + 5x

Then RHS would be

⇒ x2 + 5x

Putting x = (-3)

= (–3)2 + 5 (–3)

= 9 + (-15)

= -6

The error is ( + 9) instead of (-9) and end results to (-6) instead of (-24)

Hence putting x = (-3) in x2 + 5x results to (-6)

Question 13.

Find the errors and correct the following mathematical sentences

(x – 4)2 = x2 – 16

Answer:

If LHS is

(x – 4)2

Then RHS would be

⇒ (x – 4)2

= (x)2 + 42 – 2 × x × 4

= x2 + 16 – 8x

The error is x2 + 16 – 8x instead of x2 – 16

Hence (x – 4)2 = x2 + 13 – 8x

Question 14.

Find the errors and correct the following mathematical sentences

(x + 7)2 = x2 + 49

Answer:

If LHS is

(x + 7)2

Then RHS would be

⇒ (x + 7)2

= (x)2 + 72 + 2 × x × 7

= x2 + 49 + 14

The error is x2 + 14x + 49 instead of x2 + 49

Hence (x + 7)2 = x2 + 14x + 49

Question 15.

Find the errors and correct the following mathematical sentences

(3a + 4b) (a – b) = 3a2 – 4a2

Answer:

For getting in the equation

(a2 – b2 ) = (a + b)(a-b)

RHS would be

3a2 – 4b2

Then LHS would be

⇒ 3a2 – 4b2

= (3a – 4b)(3a + 4b)

The error is (a – b) instead of (3a – 4b)

3a2 – 4b2 instead of 3a2 – 4a2

Hence 3a2 – 4b2 = (3a – 4b)(3a + 4b)

Question 16.

Find the errors and correct the following mathematical sentences

(x + 4) (x + 2) = x2 + 8

Answer:

If LHS is

(x + 4) (x + 2)

Then RHS would be

⇒ (x + 4) (x + 2)

= x2 + 4 × x + 2 × x + 2 × 4

= x2 + 4x + 2x + 8

= x2 + 6x + 8

The error is x2 + 6x + 8 instead of x2 + 8

Hence (x + 4) (x + 2) = x2 + 6x + 8

Question 17.

Find the errors and correct the following mathematical sentences

(x – 4) (x – 2) = x2 – 8

Answer:

If LHS is

(x – 4) (x – 2)

Then RHS would be

⇒ (x – 4) (x – 2)

= x2 – 4 × x – 2 × x + (-2) × (-4)

= x2 – 4x – 2x + 8

= x2 – 6x + 8

The error is x2 – 6x + 8 instead of x2 – 8

Hence (x – 4) (x – 2) = x2 – 6x + 8

Question 18.

Find the errors and correct the following mathematical sentences

5x3 ÷ 5x3 = 0

Answer:

If LHS is

5x3 ÷ 5x3

Then RHS would be

⇒ 5x3 ÷ 5x3

= 1

The error is1 instead of 0

Hence 5x3 ÷ 5x3 = 1

Question 19.

Find the errors and correct the following mathematical sentences

2x3 + 1 ÷ 2x3 = 1

Answer:

If LHS is

(2x3 + 1) ÷ 2x3

Then RHS would be

⇒ (2x3 + 1) ÷ 2x3

The error is  instead of 1

Hence (2x3 + 1) ÷ 2x3 =

Question 20.

Find the errors and correct the following mathematical sentences

3x + 2 ÷ 3x =

Answer:

If LHS is

(3x + 2) ÷ 3x

Then RHS would be

⇒ (3x + 2) ÷ 3x

The error is  instead of

Hence (3x + 2 )÷ 3x =

Question 21.

Find the errors and correct the following mathematical sentences

3x + 5 ÷ 3 = 5

Answer:

If LHS is

For the complete and perfect division

There must be 3x instead of x

(3x + 5)÷3x

Then RHS would be

⇒ (3x + 5)÷3x

The error is instead of 5 and 3x instead of x

Hence = (3x + 5)÷3x =

Question 22.

Find the errors and correct the following mathematical sentences

= x + 1

Answer:

If LHS is

Then RHS would be

⇒

x +

x + 1

The error is x + 1 instead of x + 1

Hence x + 1

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