WBBSE Solutions For Class 8 Maths Chapter 4 Multipilcation And Division Of Polynomials

Multipilcation And Division Of Polynomials

Today we have decided to make some funny things and hang them in our classroom with the half of colourful charts. So we have made paper cuttings of various colours and sizes. My friends have written various numbers and expressions on some coloured papers and stuck them on the charts.

Let’s draw the pictures in the given blank spaces from the pictutres above.

Let’s draw cards showing constants.

Let’s draw cards showing monomial algebraic expressions.

Let’s find out the sum the monomials and write it in the blank space.

-6y+4y+3x²+7x²=-2y+10x²

Read and Learn More WBBSE Solutions For Class 8 Maths

Let’s write the binomial algebraic expressions and draw those cards in the blank space.

Let’s write the binomial algebraic expressions and draw those cards in the blank space.

5x+2y+(-7y-3x)

= 5x+2y-7y-3x

= 2x-5y

5x+2y+(-7y-3x) = 2x-5y

(a + b – c) of 3. It is a trinomial expression, but (2x⁴+5y³-10y²-8) is an algebraic expression having 4 terms. It is a tetranomial expression.

What do we say if there are many terms in such an algebraic expression?

The algebraic expression having one or more terms is called a Polynomial expression.

Put Rupa made different types of funny things. She separated all the rectangular cards. She attached these to a big cardboard. Sakil wrote the area, length or breadth of these rectangular coloured cards. See the pictures and try to write the length, breadth or area of these rectangular cards of different colours.

1.4 Area of the yellow coloured rectangular card

= (7x – 18 – 3x² + x3) x (5 – x²) Sq. m.

= x³ – 3x² + 7x – 18 x – x² + 5 Sq. m.

= {(x³ – 3x² + 7x – 18) x (-x²) + (x3- 3x² + 7x – 18) x 5} Sq. m.

⇒ -x⁵ + 3x⁴ – 7x³ + 18x² + 5x³ – 15×2 + 35x – 90 Sq. m.

⇒ -x⁵ + 3x⁴ – 2x³ + 3x² + 35x – 90 Sq. m.

Multiplication And Division Of Polynomials Exercise

Question 1. Let’s find the product by successive multiplication

1. (x⁵+ 1) X (3 -x⁴), (4 + x³ +x⁶)

Solution:

Given

(x⁵ + 1), (3-x⁴), (4 + x³+x⁶)

= {x⁵(3-x⁴)+1(3-x⁴)}(4+x³+x⁶)

=(3x⁵ – x⁹ + 3 – x⁴) (4 + x³ + x⁶) .

= 3x⁵(4 + x³ +x⁶) – x⁹(4 + x³,+ x⁶) +3(4 + x³ + x⁶) – x⁴(4 + x³ + x⁶)

= 12x⁵ + 3x⁸ + 3x¹¹ -4x⁹-x¹²-x15 + 12 + 3x³ + 3x⁶ – 4x⁴ – x⁷– x¹⁰

= -x¹⁵– x¹²+ 3x¹¹– x¹⁰ – 4x⁹+ 3x⁸– x⁷ + 3x⁶ + 12×5- 4x⁴ + 3x³ + 12

(x⁵ + 1), (3-x⁴), (4 + x³+x⁶) = -x¹⁵– x¹²+ 3x¹¹– x¹⁰ – 4x⁹+ 3x⁸– x⁷ + 3x⁶ + 12×5- 4x⁴ + 3x³ + 12

2. (2a³ – 3b⁵), (2a³ + 3b⁵), (2a⁴ – 3a²b² + b⁴)

Solution:

Given

(2a³ – 3b⁵) x (2a³ + 3b⁵) (2a⁴ – 3a²b² + b⁴)

= {2a³(2a³ + 3b⁵) – 3b⁵ (2a³ + 3b⁵)} (2a⁴ – 3a²b² + b⁴)

= (4a⁶ + 6a³b⁵ – 6a³b⁵ – 9b¹⁰) (2a⁴ – 3a²b² + b⁴)

= (4a⁶ – 9b¹⁰) (2a⁴ – 3a²b² + b⁴)

= 4a⁶ (2a⁴ – 3a²b² + b⁴) – 9b¹⁰ (2a⁴ – 3a²b² + b⁴)

= 8a¹⁰ – 12a⁸ b² + 4a⁶b⁴ – 18 a⁴b¹⁰ + 27a²b12 – 9b¹⁴

(2a³ – 3b⁵) x (2a³ + 3b⁵) (2a⁴ – 3a²b² + b⁴) = 8a¹⁰ – 12a⁸ b² + 4a⁶b⁴ – 18 a⁴b¹⁰ + 27a²b12 – 9b¹⁴

3. (ax + by), (ax – by), (a⁴x⁴ + a²b²x²y² + b⁴y⁴)

Solution:

Given

(ax + by), (ax – by), (a⁴x⁴ + a²b²x²y² + b⁴y⁴)

= (ax + by) x (ax – by) (a⁴x⁴ + a²b²x²y² + b⁴y⁴)

= {ax (ax – by) + by (ax – by) (a⁴x⁴ + a²b²x²y² + b⁴y⁴)

= (a²x² – abxy + abxy – b²y²) (a⁴x⁴ + a²b²x²y² + b⁴y⁴)

= (a²x² – b²y²) (a⁴x⁴ + a²b²x²y²+ b⁴y⁴)

= a²x² (a⁴x⁴ + a²b²x²y² + b⁴y⁴) – b²y² (a⁴x⁴ + a²b²x²y² + b⁴y⁴)

= a⁶x⁶ + a⁴b²x⁴y² + a²b⁴x²y⁴ – a⁴b²x⁴y² – a²b⁴x²y⁴– b⁶y⁶

= a⁶x⁶ – b⁶y⁶

(ax + by), (ax – by), (a⁴x⁴ + a²b²x²y² + b⁴y⁴)= a⁶x⁶ – b⁶y⁶

4. (a + b + c), (a – b + c), (a + b – c)

Solution:

Given

(a + b + c), (a – b + c), (a + b – c)

(a + b + c) x (a – b + c), (a + b – c)

= {a (a – b + c) + b (a – b + c) + c (a – b + c)} (a + b – c)

= (a²– ab + ac + ab – b² + bc + ac – ba + c²) (a + b – c)

= (a² + 2ac – b² + c²) (a + b – c)

= a (a² + 2ac – b² + c²) + b (a² + 2ac – b²+c²) – c (a² + 2ac – b² + c²)

= a³ + 2a²c -ab²+ ac² + a²b + 2abc – b³ + bc² – a²c – 2ac²+ b²c – c3

= a³+ a²c – ab²– ac² + a²b + 2abc + be² + b²c – b³ – c³

= a³ – b³ – c³+ 2abc-+ a²b + a²c – ab² – ac² + b²c + bc²

(a + b + c), (a – b + c), (a + b – c)= a³ – b³ – c³+ 2abc-+ a²b + a²c – ab² – ac² + b²c + bc²

5. \(\left(\frac{2 p^2}{q^2}+\frac{5 q^2}{p^2}\right)\left(\frac{2 p^2}{q^2}-\frac{5 q^2}{p^2}\right)\)

Solution:

Given

⇒ \(\left(\frac{2 p^2}{q^2}+\frac{5 q^2}{p^2}\right) \times\left(\frac{2 p^2}{q^2}-\frac{5 q^2}{p^2}\right)\)

⇒ \(\frac{2 p^2}{q^2}\left(\frac{2 p^2}{q^2}-\frac{5 q^2}{p^2}\right)+\frac{5 q^2}{p^2}\left(\frac{2 p^2}{q^2}-\frac{5 q^2}{p^2}\right)\)

⇒ \(\frac{4 p^4}{q^4}-\frac{10 p^2 q^2}{p^2 q^2}+\frac{10 p^2 q^2}{p^2 q^2}-\frac{25 q^4}{p^4}\)

⇒ \(\frac{4 p^4}{q^4}-\frac{25 q^4}{p^4}\)

6. \(\left(\frac{x^2}{y^2}+\frac{y^2}{z^2}\right),\left(\frac{y^2}{z^2}+\frac{z^2}{x^2}\right),\left(\frac{z^2}{x^2}+\frac{x^2}{y^2}\right)\)

Solution.

⇒ \(\left(\frac{x^2}{y^2}+\frac{y^2}{z^2}\right) \times\left(\frac{y^2}{z^2}+\frac{z^2}{x^2}\right)\left(\frac{z^2}{x^2}+\frac{x^2}{y^2}\right)\)

⇒ \(\left\{\frac{x^2}{y^2}\left(\frac{y^2}{z^2}+\frac{z^2}{x^2}\right)+\frac{y^2}{z^2}\left(\frac{y^2}{z^2}+\frac{z^2}{x^2}\right)\right\}\left(\frac{z^2}{x^2}+\frac{x^2}{y^2}\right)\)

⇒ \(\left(\frac{x^2 y^2}{y^2 z^2}+\frac{x^2 z^2}{x^2 y^2}+\frac{y^4}{z^4}+\frac{y^2 z^2}{x^2 z^2}\right)\left(\frac{z^2}{x^2}+\frac{x^2}{y^2}\right)\)

⇒ \(\left(\frac{x^2}{z^2}+\frac{z^2}{y^2}+\frac{y^4}{z^4}+\frac{y^2}{x^2}\right) \cdot\left(\frac{z^2}{x^2}+\frac{x^2}{y^2}\right)\)

⇒ \(\frac{x^2}{z^2}\left(\frac{z^2}{x^2}+\frac{x^2}{y^2}\right)+\frac{z^2}{y^2}\left(\frac{z^2}{x^2}+\frac{x^2}{y^2}\right)+\frac{y^4}{z^4}\left(\frac{z^2}{x^2}+\frac{x^2}{y^2}\right)+\frac{y^2}{x^2}\left(\frac{z^2}{x^2}+\frac{x^2}{y^2}\right)\)

⇒ \(\frac{x^2 z^2}{x^2 z^2}+\frac{x^4}{y^2 z^2}+\frac{z^4}{x^2 y^2}+\frac{x^2 z^2}{y^4}+\frac{y^4 z^2}{x^2 z^4}+\frac{x^2 y^4}{y^2 z^4}+\frac{y^2 z^2}{x^4}+\frac{x^2 y^2}{x^2 y^2}\)

⇒ \(1+\frac{x^4}{x^2 z^2}+\frac{z^4}{x^2 y^2}+\frac{x^2 z^2}{y^4}+\frac{y^4}{x^2 z^2}+\frac{x^2 y^2}{z^4}+\frac{y^2 z^2}{x^4}+1\)

⇒ \(2+\frac{x^4}{y^2 z^2}+\frac{y^4}{x^2 z^2}+\frac{z^4}{x^2 y^2}+\frac{y^2 z^2}{x^4}+\frac{x^2 z^2}{y^4}+\frac{x^2 y^2}{z^4}\)

Question 2. Simplify :

1. (x + y) (x² – xy + y²) + (x – y) (x² + xy + y²)

Solution:

Given

(x + y) (x² – xy + y²) + (x – y) (x² + xy + y²)

= x (x² – xy + y²) + y (x² – xy + y²) + x (x² + xy + y²) – y (x² + xy + y²)

= x³ – x²y + xy² + x²y – xy² + y³ + x³ + x2y + xy² – x²y – xy² – y³

= x³ + y³ + x³ – y³

= 2x³

(x + y) (x² – xy + y²) + (x – y) (x² + xy + y²)= 2x³

2. a² (b² – c²) + b² (c² – a²) + c² (a² – b²)

Solution:

Given

a² (b² – c²) + b² (c² – a²) + c² (a² – b²)

= a²b² – a²c² + b²c² – a²b² + a²c² – b²c²

= 0

a² (b² – c²) + b² (c² – a²) + c² (a² – b²) = 0

Question 3. 

1. If a = x² + xy + y², b= y²+ yz + z², c = z² + xz + x² then find the value of (x – y)a + (y – z)b + (z – x) c.

Solution:

Given

a = x² + xy + y², b= y²+ yz + z², c = z² + xz + x²

= (x – y)a + (y – z)b + (z – x) c

= (x – y) (x² + xy + y²) + (y – z) (y² + yz + z²) + (z – x) (z² + xz + x²)

= x³ – y³ + y³ – z³ + z³ – x³

= 0 . . .

The value of (x – y)a + (y – z)b + (z – x) c = 0 . . .

2. If a = lx + my + n, b = mx + ny + I, c = nx + ly + m, then find the value of a (m + n) + b(n +1) + c (I + m).

Solution:

Given

a = lx + my + n, b = mx + ny + I, c = nx + ly + m,

= a (m + n) + b(n + I) + c (I + m)

= (lx + my + n) (m + n) + (mn + ny +1) (n + I) + (nx + ly + m) (I + m)

= Imx + Inx + m²y + mny + mn + n² + mnx + Imx + n²y + Iny + In + I² + Inx + mnx + l²y + Imy + Im + m²

= l²+ m² + n² + 2lmx + 2lnx + 2mnx + l²y + m²y + n²y + Imy + Iny + mny + Im + In + mn

Mrinal and Sraboni have made very colourful cards. There are many algebraic expressions on these cards.

Multiplying the algebraic expressions of their cards, I have written the product on another card. My friend Niladri is trying to find the multiplicant or multiplier from the product written on cards.

Product of two numbers -s- One of them = The other one

The value of a (m + n) + b(n +1) + c (I + m) = l²+ m² + n² + 2lmx + 2lnx + 2mnx + l²y + m²y + n²y + Imy + Iny + mny + Im + In + mn

Question 4. Let’s multiply and then verify the quotient dividing the product by the multiplier or the muliplicant.

Solution:

(1 + 5x) x (4 – 3x) = (4 + 17x – 15x²)

= (4 + 17x – 15x²) ÷ (1 + 5x) = 4-3x

1. (a²-3a -2) x (2a- 1) = 2a³-7a²-a+2

Solution:

Given

(a²-3a -2) x (2a- 1) = 2a³-7a²-a+2

(2a³-7a²-a+2) ÷ ?(a²-3a+2) = 2a-1

2a-1

Quotient = 2a-1 and Remainder = 0

2. 27p³+9p²+3p+1×3p-1=81p⁴-1

Solution:

Given

27p³+9p²+3p+1×3p-1=81p⁴-1

Quotient =27p³+9p2+3p+1 and Remainder = 0

Multiplication And Division Of Polynomials Exercise

Question 1. Let’s divide by arranging in decreasing powers of variables.

1. (x²– 13x + 22) by (x – 11)

Solution: 

Given

(x²– 13x + 22) and (x – 11)

Divided by arranging in decreasing powers of variables (x²– 13x + 22) by (x – 11) we get

Quotient = (x-2) and Remainder = 0

2. (a²-5a+6) by (a-2)

Solution: 

Given

(a²-5a+6) and (a-2)

Divided by arranging in decreasing powers of variables (a²-5a+6) by (a-2) we get

Quotient = (a-3) and Remainder = 0

3. (2a³-7a²-a+2) by (a²-3a-2)

Solution:

Given

(2a³-7a²-a+2) and (a²-3a-2)

Quotient = (2a-1) and Remainder = 0

4. (4a²-9b²) by (2a+3b)

Solution:

Given

(4a²-9b²) and (2a+3b)

Quotient = (2a-3b) and Remainder = 0

Question 2. Let’s find divisor and remainder of the expression in the red card.

Solution:

Divisor =6x³y- x²y² – 7xy³ + 12y⁴

Quotient = 2x + 3y

Dividend = (3x2y – 2x- 4)and Remainder= 3x-4

Question 3. Let’s find divisor and remainder of the expression in the blue card.

Solution:

The dividend = 12x⁴+5x³-33x²-3x+16 and divisor = 4x²-x-5

Dividend = 3x²+2x-4 and remainder = 3x-4

Question 4. Let’s find divisor and remainder of the expression in the green cards.

Solution:

Dividend = x and Remainders = = x

Multiplication And Division Of Polynomials Exercise 4.2

Question 1. The product of 2 numbers is 3x² + 8x + 4 and one number is 3x +2. Let’s find the other number.

Solution:

Given

The product of 2 numbers is 3x² + 8x + 4 and one number is 3x +2.

∴ Other number = (3x² + 8x + 4) – (3x + 2)

∴ Other number = (x+2)

Question 2. The area of a rectangle is (24x² – 65xy + 21 y²) sq. cm and length is (8x – 3y). Let’s find the breadth of it.

Solution:

Given

The area of a rectangle is (24x² – 65xy + 21 y²) sq. cm and length is (8x – 3y).

The area of a rectangle is (24x² – 65xy + 21 y²) sq. cm.

Length (8x – 3y) cm.

Breadth = (24x² – 65xy + 21 y²) -(8x – 3y)

∴ Breadth of the rectangle = (3x-7y)cm.

Question 3. If in a division problem, the dividend is x⁴+x³y+xy³-y⁴ and division is x²+xy-y². Then find the quotient and remainder.

Solution:

Given

If in a division problem, the dividend is x⁴+x³y+xy³-y⁴ and division is x²+xy-y².

Dividend x⁴+x³y+xy³-y⁴

Divisor x²+xy-y²

∴ Quotient = x²+y² and Remainder = 0

Question 4. Divide

1. (m² + 4m – 21) by (m – 3)

Solution:

Given

(m² + 4m – 21) and (m – 3)

Quotient = (m-7) and Remainder = 0

2. (6c²-7c+2) by Bold (3c-2)

Solution:

Given

(6c²-7c+2) and Bold (3c-2)

Quotient = (2c-2) and Remainder = 0

3. (2a⁴-a³-2a²+5a-1) by (2a²+a-3)

Solution:

Given

(2a⁴-a³-2a²+5a-1) and (2a²+a-3)

Quotient = (a²-a+1) and Remainder = (a+2)

4. (m⁴ – 2m³ – 7m² + 8m + 12) by (m² – m – 6)

Solution:

Given

(m⁴ – 2m³ – 7m² + 8m + 12) and (m² – m – 6)

Quotient = m²-m-2 and Remainder = 0

Question 5. 1 (6x²a³ – 4x³a² + 8x⁴a²) ÷ 2a²x²

Solution:

Given

(6x²a³ – 4x³a² + 8x⁴a²) and 2a²x²

⇒ \(\frac{6 x^2 a^3-4 x^3 a^2+8 x^4 a^2}{2 a^2 x^2}\)

⇒ \(\frac{6 x^2 a^3}{2 a^2 a^2}-\frac{4 x^3 a^2}{2 a^2 x^2}+\frac{8 x^4 a^2}{2 a^2 x^2}\)

= 3a-2x+4x²

2. \(\frac{2 y^9 x^5}{5 x^2} \times \frac{125 x y^5}{16 x^4 y^{10}}\)

Solution.

Given \(\frac{2 y^9 x^5}{5 x^2} \times \frac{125 x y^5}{16 x^4 y^{10}}\)

⇒ \(\frac{7 a^4 y^2}{9 a^2} \times \frac{729 a^6}{42 y^6}\)

⇒ \(\frac{27 a^{10} y^2}{2 a^2 y^6}=\frac{27 a^8}{2 y^4}\)

3. (p2q2r5 – p3q5r2 + p5q3r2) – p2q2r2

⇒ \(\frac{p^2 q^2 r^5-p^3 q^5 r^2+p^5 q^3 r^2}{p^2 q^2 r^2}\)

⇒ \(\frac{p^2 q^2 r^5}{p^2 q^2 r^2}-\frac{p^3 q^5 r^2}{p^2 q^2 r^2}+\frac{p^5 q^3 r^2}{p^2 q^2 r^2}\)

= r³-pq³+p³q

Question 6. In a division problem the Divisor is (x – 4), Quotient is (x² + 4x + 4) and Remainder is 3. Let’s find the Dividend = Divisor × + Remainder ]

Solution:

Given

In a division problem the Divisor is (x – 4), Quotient is (x2 + 4x + 4) and Remainder is 3.

Divisor (x – 4), Quotient (x2 + 4x + 4)

Remainder = 3

∴ Dividend = Divisor x Quotient + Remainder

= (x – 4) x (x² + 4x + 4) + 3 .

= x (x² + 4x + 4) -4 (x² + 4x + 4) + 3

= x³ + 4x² + 4x – 4x² – 16x – 16 + 3

= x³ – 12x – 13

Question 7. In a division problem, the divisor is (a² + 2a – 1), Quotient is 5a -14 and the remainder is 35a-17. Let’s find out and write the dividend.

Solution:

Given

In a division problem, the divisor is (a² + 2a – 1), Quotient is 5a -14 and the remainder is 35a-17.

Divisor is (a² + 2a – 1), Quotient is (5a – 14)

Remainder = 35a – 17

Divisor x Quotient + Remainder

= (a² + 2a – 1) (5a – 14) + (35a- 17)

= 5a (a² + 2a – 1) -14 (a² + 2a – 1) + 35a – 17

= 5a³ + 10a² – 5a – 14a² – 28a + 14 + 35a – 17

= 5a³– 4a²+ 2a – 3

Question 8. Let’s write the quotient and the remainder.

1. (x² + 11x + 27) ÷ (x+6)

Solution:

Given

(x² + 11x + 27) and (x+6)

Quotient = (x+5), Remainder = -3

2. (81 x⁴+ 2) + (3x – 1)

Solution:

Given

(81 x⁴+ 2) and (3x – 1)

Quotient = (27×3+9x²+3x+1), Remainder = 3

3. (63x² – 19x – 20) + (9x² +5)

Solution:

Given

(63x² – 19x – 20) and (9x² +5)

Quotient = 7, Remainder = -19x-55

4. (x³-x²-8x-13) ÷ (x²+3x+3)

Solution:

Given

(x³-x²-8x-13) and (x²+3x+3)

Quotient = (x-4),Remainder = (x-1)