The DAV Class 8 Maths Solutions and **DAV Class 8 Maths Chapter 8 Worksheet 3 **Solutions of Polynomials offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 8 Worksheet 3 Solutions

Question 1.

Using factor method, divide the following polynomials by a binomial:

(i) x^{2} + 3x + 2 by x + 1

Solution:

x^{2} + 3x + 2 ÷ x + 1

\(\frac{x^2+3 x+2}{x+1}\) = \(\frac{x^2+2 x+x+2}{x+1}\)

= \(\frac{x(x+2)+1(x+2)}{x+1}\)

= \(\frac{(x+1)(x+2)}{x+1}\)

= x + 2

Hence the quotient is (x + 2)

(ii) x^{2} – 7x – 18 by x – 9

Solution:

x^{2} – 7x – 18 ÷ x – 9

\(\frac{x^2-7 x-18}{x-9}\) = \(\frac{x^2-9 x+2 x-18}{x-9}\)

= \(\frac{x(x-9)+2(x-9)}{x-9}\)

= \(\frac{(x+2)(x-9)}{x-9}\)

= x + 2

Hence the quotient is (x + 2)

(iii) p^{2} + 6p + 8 by p + 2

Solution:

p^{2} + 6p + 8 ÷ p + 2

\(\frac{p^2+6 p+8}{p+2}\) = \(\frac{p^2+4 p+2 p+8}{p+2}\)

= \(\frac{p(p+4)+2(p+4)}{p+2}\)

= \(\frac{(p+2)(p+4)}{(p+2)}\)

= (p + 4)

Hence the quotient is (p + 4)

(iv) p^{2} – p – 42 by p + 6

Solution:

p^{2} – p – 42 ÷ p + 6

\(\frac{p^2-p-42}{p+6}\) = \(\frac{p^2-7 p+6 p-42}{p+6}\)

= \(\frac{p(p-7)+6(p-7)}{p+6}\)

= \(\frac{(p+6)(p-7)}{p+6}\)

= (p – 7)

Hence the quotient is (p – 7).

(v) y^{2} + 12y + 35 by y + 7

Solution:

y^{2} + 12y + 35 ÷ y + 7

\(\frac{y^2+12 y+35}{y+7}\) = \(\frac{y^2+7 y+5 y+35}{y+7}\)

= \(\frac{y(y+7)+5(y+7)}{y+7}\)

= \(\frac{(y+7)(y+5)}{(y+7)}\)

= y + 5

Hence the quotient is (y + 5).

(vi) z^{2} – 10z + 16 by z – 2

Solution:

z^{2} – 10z + 16 ÷ z – 2

\(\frac{z^2-10 z+16}{z-2}\) = \(\frac{z^2-8 z-2 z+16}{z-2}\)

= \(\frac{z(z-8)-2(z-8)}{z-2}\)

= \(\frac{(z-8)(z-2)}{(z-2)}\)

= z – 8

Hence the quotient is (z – 8).

(vii) x^{4} + 3x^{2} – 10 by x^{2} + 5 (Hint: put x^{2} = y)

Solution:

x^{4} + 3x^{2} – 10 ÷ x^{2} + 5

\(\frac{x^4+3 x^2-10}{x^2+5}\) = \(\frac{y^2+3 y-10}{y+5}\)

= \(\frac{y^2+5 y-2 y-10}{y+5}\)

= \(\frac{y(y+5)-2(y+5)}{(y+5)}\)

= \(\frac{(y+5)(y-2)}{(y+5)}\)

= y – 2

= x^{2} – 2

Hence the quotient is (x^{2} – 2).

(viii) x^{2} – 7x + 12 by x – 4

Solution:

x^{2} – 7x + 12 ÷ x – 4

\(\frac{x^2-7 x+12}{x-4}\) = \(\frac{x^2-4 x-3 x+12}{x-4}\)

= \(\frac{x(x-4)-3(x-4)}{x-4}\)

= \(\frac{(x-4)(x-3)}{x-4}\)

= (x – 3)

Hence the quotient is (x – 3).

Question 2.

Using long division method, divide the following polynomials by a binomial and check your answer.

(i) 4p^{3} – 4p^{2} + 6p – \(\frac{5}{2}\) by 2p – 1

(ii) x – 6 + 15x^{2} by 2 + 3x

(iii) 4x^{3} – 37x^{2} + 52x – 15 by 4x – 5

(iv) y^{3} – 8 by y – 2

(v) 29z – 6z^{2} – 28 by 3z – 4

(vi) p^{4} + p^{3} – p^{2} + 1 by p – 1

(vii) 12z^{3} + 4z + 3z^{2} + 1 by 4z + 1

(viii) 8x^{2} – 2 – 3x + 12x^{3} by 4x^{2} – 1

Solution:

(i) 4p^{3} – 4p^{2} + 6p – \(\frac{5}{2}\) ÷ 2p – 1

Check:

Dividend = Divisor × Quotient + Remainder

4p^{3} – 4p^{2} + 6p – \(\frac{5}{2}\) = (2p – 1) × (2P^{2} – p + \(\frac{5}{2}\)) + o

= 2p (2p^{2} – p + \(\frac{5}{2}\)) – 1 (2P^{2} – p + \(\frac{5}{2}\))

= 4p^{3} – 2p^{2} + 5p – 2p^{2} + p – \(\frac{5}{2}\)

= 4p^{3} – 4p^{2} + 6p – \(\frac{5}{2}\)

⇒ L.H.S. = R.H.S.

Hence verified.

(ii) Let us write dividend and divisor in standard form.

Dividend = 15x^{2} + x – 6

and Divisor = 3x + 2

Check:

Dividend = Divisor × Quotient + Remainder

15x^{2} + x – 6 = (3x + 2) × (5x – 3) + 0

= 3x (5x – 3) + 2 (5x – 3)

= 15x^{2} – 9x + 10x – 6

= 15x^{2} + x – 6

⇒ L.H.S. = R.H.S.

Hence verified.

(iii) 4x^{3} – 37x^{2} + 52x – 15 ÷ 4x – 5

Check:

Dividend = Divisor × Quotient + Remainder

4x^{3} – 37x^{2} + 52x – 15 = (4x – 5) x (x^{2} – 8x + 3) + 0

= 4x (x^{2} – 8x + 3) – 5 (x^{2} – 8x + 3)

= 4x^{3} – 32x^{2} + 12x – 5x^{2} + 40x – 15

= 4x^{3} – 37x^{2} + 52x – 15

⇒ L.H.S. = R.H.S.

Hence Verified.

(iv) y^{3} – 8 ÷ y – 2

Check:

Dividend = Divisor × Quotient + Remainder

L.H.S. = y^{3} – 8

R.H.S. = (y – 2) (y^{2} + 2y + 4) + 0

= y (y^{2} + 2y + 4) – 2 (y^{2} + 2y + 4)

= 9 + 2y^{2} + 4y – 2y^{2} – 4y – 8

= y^{3} – 8

⇒ L.H.S. = R.H.S.

Hence Verified.

(v) Let us write the dividend in standard form.

We have 29 – 6z^{2} – 28 = – 6z^{2} + 29z – 28

Check:

Dividend = Divisor × Quotient + Remainder

– 6z^{2} + 29z – 28 = (3z – 4) × (- 2z + 7) + 0

= 3z (- 2z + 7) – 4(- 2z + 7)

= – 6z^{2} + 21z + 8z – 28

= – 6z^{2} + 29z – 28

⇒ L.H.S. = R.H.S.

Hence verified.

(vi) p^{4} + p^{3} – p^{2} + 1 ÷ p – 1

Quotient = p^{3} + 2p^{2} + p + 1

Remainder = 2

Check:

Dividend = Divisor × Quotient + Remainder

p^{4} + p^{3} – p^{2} + 1 = (p – 1) (p3 + 2p2 +p + 1)

= p (p^{3} + 2p^{2} + p + 1) – 1 (p^{3} + 2p^{2} + p + 1)

= p^{4} + 2p^{3} + p^{2} + p – p^{3} – 2p^{2} – p – 1

= p^{4} + p^{3} – p^{2} – 1

∴ L.H.S. = R.H.S.

Hence verified.

(vii) Let us write the dividend in standard form:

Dividend = 12z^{3} + 3z^{2} + 4z + 1

Check:

Dividend = Divisor × Quotient + Remainder

12z^{3} + 3z^{2} + 4z + 1 = (4z + 1) (3z^{2} + 1) + 0

= 4z (3z^{2} + 1) + 1(3z^{2} + 1)

= 12z^{3} + 4z + 3z^{2} + 1

= 12z^{3} + 3z^{2} + 4z + 1

L.H.S. = R.H.S.

Hence verified.

(viii) Let us write the dividend in standard form:

Dividend = 12x^{3} + 8x^{2} – 3x – 2

Check:

Dividend = Divisor × Quotient + Remainder

12x^{3} + 8x^{2} – 3x – 2 = (4x^{2} – 1) × (3x + 2) + 0

= 4x^{2} (3x + 2) – 1 (3x + 2)

= 12x^{3} + 8x^{2} – 3x – 2.

Question 3.

Using long division method, check whether the second polynomial is the factor of the first polynomial.

(i) x^{4} – 16 ; x + 2

(ii) z^{4} – z^{3} + 3z^{2} – 2z + 2 ; z^{2} + 2

(iii) x^{3} – 4x^{2} – 3x + 5 ; x – 3

(iv) 4p^{3} – 12p^{2} – 37p – 15 ; 2p + 1

(v) 4q^{3} – 6q^{2} – 4q + 3 ; 2q – 1

(vi) 20x^{2} – 32x – 16 ; 5x + 2

(vii) 12a^{2} – 7a – 2 ; 4a – 1

(viii) 6x^{3} + 19x^{2} + 13x + (- 3) ; 2x + 3

Solution:

(i) x^{4} – 16 ; x + 2

Here, remainder = 0

Hence (x + 2) is a factor of x^{4} – 16.

(ii) z^{4} – z^{3} + 3z^{2} – 2z + 2 ; z^{2} + 2

Here, the remainder = 0

Hence z^{2} + 2 is a factor of z^{4} – z^{3} + 3z^{2} – 2z + 2.

(iii) x^{3} – 4x^{2} – 3x + 5 ; x – 3

As remainder = – 13 which is not equal to 0.

So, x – 3 is not a factor of x^{3} – 4x^{2} – 3x + 5.

(iv) 4p^{3} – 12p^{2} – 37p – 15 ; 2p + 1

Here, remainder = 0

Thus, (2p + 1) is a factor of 4p^{3} – 12p^{2} – 37p – 15.

(v) 4q^{3} – 6q^{2} – 4q + 3 ; 2q – 1

Here, remainder = 0

Thus, (2q – 1) is a factor of 4q^{3} – 6q^{2} – 4q + 3.

(vi) 20x^{2} – 32x – 16 ; 5x + 2

Here, remainder = 0

Thus, (5x + 2) is a factor of 20x^{2} – 32x – 16.

(vii) 12a^{2} – 7a – 2 ; 4a – 1

As remainder = – 3 which is not equal to 0.

So 4a – 1 is not a factor of 12a^{2} – 7a – 2.

(viii) 6x^{3} + 19x^{2} + 13x + (- 3) ; 2x + 3

Here, remainder = 0

Thus, (2x + 3) is a factor of 6x^{3} + 19x^{2} + 13x – 8.

### DAV Class 8 Maths Chapter 8 Value Based Questions

Question 1.

Mr. Lalit donated (x^{2} + 12x + 35) for the education of (x + 7) children.

(i) Find the amount received by each child, if each child got the equal amount.

(ii) Why it is necessary to educate children? Give one reason.

Solution:

(i) Total donation = (x^{2} + 12x + 35)

= [x^{2} + (7 + 5) x + 7 × 5]

= [x^{2} + 7x + 5x + 7 × 5]

= [x (x + 7) + 5 (x + 7)]

= (x + 7) (x + 5)

Number of children = (x + 7)

Amount received by each child = \(\frac{(x+7)(x+5)}{(x+7)}\) = (x + 5)

(ii) Children is the future asset of the nation as they have to represent the nation.

So, it is necessary to educate children.

Question 2.

Every morning Abhay goes for Jogging in a park.

(i) Find the time taken by him to cover a distance of (2t^{4} + 3t^{3} – 2t^{2} – 9t – 12) km at the speed of (t^{2} – 3) km/hr.

(ii) Write the importance of physical exercises like jogging, walking, etc.

Solution:

(i) Distance = (2t^{4} + 3t^{3} – 2t^{2} – 9t – 12) km

Speed = (t^{2} – 3) km/hr

Time = \(\frac{\text { Distance }}{\text { Speed }}\)

= \(\frac{\left(2 t^4+3 t^3-2 t^2-9 t-12\right) \mathrm{km}}{\left(t^2-3\right) \mathrm{km} / \mathrm{hr}}\)

= (2t^{2} + 3t + 4) hr

(ii) Physical exercises keeps the body fit and sound.