DAV Class 8 Maths Chapter 8 Worksheet 3 Solutions

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) x² + 3x + 2 by x + 1
Solution:
x² + 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) x² – 7x – 18 by x – 9
Solution:
x² – 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) p² + 6p + 8 by p + 2
Solution:
p² + 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) p² – p – 42 by p + 6
Solution:
p² – 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) y² + 12y + 35 by y + 7
Solution:
y² + 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) z² – 10z + 16 by z – 2
Solution:
z² – 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) x⁴ + 3x² – 10 by x² + 5 (Hint: put x² = y)
Solution:
x⁴ + 3x² – 10 ÷ x² + 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
= x² – 2
Hence the quotient is (x² – 2).

(viii) x² – 7x + 12 by x – 4
Solution:
x² – 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) 4p³ – 4p² + 6p – \(\frac{5}{2}\) by 2p – 1
(ii) x – 6 + 15x² by 2 + 3x
(iii) 4x³ – 37x² + 52x – 15 by 4x – 5
(iv) y³ – 8 by y – 2
(v) 29z – 6z² – 28 by 3z – 4
(vi) p⁴ + p³ – p² + 1 by p – 1
(vii) 12z³ + 4z + 3z² + 1 by 4z + 1
(viii) 8x² – 2 – 3x + 12x³ by 4x² – 1
Solution:
(i) 4p³ – 4p² + 6p – \(\frac{5}{2}\) ÷ 2p – 1

Check:
Dividend = Divisor × Quotient + Remainder
4p³ – 4p² + 6p – \(\frac{5}{2}\) = (2p – 1) × (2P² – p + \(\frac{5}{2}\)) + o
= 2p (2p² – p + \(\frac{5}{2}\)) – 1 (2P² – p + \(\frac{5}{2}\))
= 4p³ – 2p² + 5p – 2p² + p – \(\frac{5}{2}\)
= 4p³ – 4p² + 6p – \(\frac{5}{2}\)
⇒ L.H.S. = R.H.S.
Hence verified.

(ii) Let us write dividend and divisor in standard form.
Dividend = 15x² + x – 6
and Divisor = 3x + 2

Check:
Dividend = Divisor × Quotient + Remainder
15x² + x – 6 = (3x + 2) × (5x – 3) + 0
= 3x (5x – 3) + 2 (5x – 3)
= 15x² – 9x + 10x – 6
= 15x² + x – 6
⇒ L.H.S. = R.H.S.
Hence verified.

(iii) 4x³ – 37x² + 52x – 15 ÷ 4x – 5

Check:
Dividend = Divisor × Quotient + Remainder
4x³ – 37x² + 52x – 15 = (4x – 5) x (x² – 8x + 3) + 0
= 4x (x² – 8x + 3) – 5 (x² – 8x + 3)
= 4x³ – 32x² + 12x – 5x² + 40x – 15
= 4x³ – 37x² + 52x – 15
⇒ L.H.S. = R.H.S.
Hence Verified.

(iv) y³ – 8 ÷ y – 2

Check:
Dividend = Divisor × Quotient + Remainder
L.H.S. = y³ – 8
R.H.S. = (y – 2) (y² + 2y + 4) + 0
= y (y² + 2y + 4) – 2 (y² + 2y + 4)
= 9 + 2y² + 4y – 2y² – 4y – 8
= y³ – 8
⇒ L.H.S. = R.H.S.
Hence Verified.

(v) Let us write the dividend in standard form.
We have 29 – 6z² – 28 = – 6z² + 29z – 28

Check:
Dividend = Divisor × Quotient + Remainder
– 6z² + 29z – 28 = (3z – 4) × (- 2z + 7) + 0
= 3z (- 2z + 7) – 4(- 2z + 7)
= – 6z² + 21z + 8z – 28
= – 6z² + 29z – 28
⇒ L.H.S. = R.H.S.
Hence verified.

(vi) p⁴ + p³ – p² + 1 ÷ p – 1

Quotient = p³ + 2p² + p + 1
Remainder = 2

Check:
Dividend = Divisor × Quotient + Remainder
p⁴ + p³ – p² + 1 = (p – 1) (p3 + 2p2 +p + 1)
= p (p³ + 2p² + p + 1) – 1 (p³ + 2p² + p + 1)
= p⁴ + 2p³ + p² + p – p³ – 2p² – p – 1
= p⁴ + p³ – p² – 1
∴ L.H.S. = R.H.S.
Hence verified.

(vii) Let us write the dividend in standard form:
Dividend = 12z³ + 3z² + 4z + 1

Check:
Dividend = Divisor × Quotient + Remainder
12z³ + 3z² + 4z + 1 = (4z + 1) (3z² + 1) + 0
= 4z (3z² + 1) + 1(3z² + 1)
= 12z³ + 4z + 3z² + 1
= 12z³ + 3z² + 4z + 1
L.H.S. = R.H.S.
Hence verified.

(viii) Let us write the dividend in standard form:
Dividend = 12x³ + 8x² – 3x – 2

Check:
Dividend = Divisor × Quotient + Remainder
12x³ + 8x² – 3x – 2 = (4x² – 1) × (3x + 2) + 0
= 4x² (3x + 2) – 1 (3x + 2)
= 12x³ + 8x² – 3x – 2.

Question 3.
Using long division method, check whether the second polynomial is the factor of the first polynomial.
(i) x⁴ – 16 ; x + 2
(ii) z⁴ – z³ + 3z² – 2z + 2 ; z² + 2
(iii) x³ – 4x² – 3x + 5 ; x – 3
(iv) 4p³ – 12p² – 37p – 15 ; 2p + 1
(v) 4q³ – 6q² – 4q + 3 ; 2q – 1
(vi) 20x² – 32x – 16 ; 5x + 2
(vii) 12a² – 7a – 2 ; 4a – 1
(viii) 6x³ + 19x² + 13x + (- 3) ; 2x + 3
Solution:
(i) x⁴ – 16 ; x + 2

Here, remainder = 0
Hence (x + 2) is a factor of x⁴ – 16.

(ii) z⁴ – z³ + 3z² – 2z + 2 ; z² + 2

Here, the remainder = 0
Hence z² + 2 is a factor of z⁴ – z³ + 3z² – 2z + 2.

(iii) x³ – 4x² – 3x + 5 ; x – 3

As remainder = – 13 which is not equal to 0.
So, x – 3 is not a factor of x³ – 4x² – 3x + 5.

(iv) 4p³ – 12p² – 37p – 15 ; 2p + 1

Here, remainder = 0
Thus, (2p + 1) is a factor of 4p³ – 12p² – 37p – 15.

(v) 4q³ – 6q² – 4q + 3 ; 2q – 1

Here, remainder = 0
Thus, (2q – 1) is a factor of 4q³ – 6q² – 4q + 3.

(vi) 20x² – 32x – 16 ; 5x + 2

Here, remainder = 0
Thus, (5x + 2) is a factor of 20x² – 32x – 16.

(vii) 12a² – 7a – 2 ; 4a – 1

As remainder = – 3 which is not equal to 0.
So 4a – 1 is not a factor of 12a² – 7a – 2.

(viii) 6x³ + 19x² + 13x + (- 3) ; 2x + 3

Here, remainder = 0
Thus, (2x + 3) is a factor of 6x³ + 19x² + 13x – 8.

DAV Class 8 Maths Chapter 8 Value Based Questions

Question 1.
Mr. Lalit donated (x² + 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 = (x² + 12x + 35)
= [x² + (7 + 5) x + 7 × 5]
= [x² + 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 (2t⁴ + 3t³ – 2t² – 9t – 12) km at the speed of (t² – 3) km/hr.
(ii) Write the importance of physical exercises like jogging, walking, etc.
Solution:
(i) Distance = (2t⁴ + 3t³ – 2t² – 9t – 12) km
Speed = (t² – 3) km/hr
Time = \(\frac{\text { Distance }}{\text { Speed }}\)
= \(\frac{\left(2 t^4+3 t^3-2 t^2-9 t-12\right) \mathrm{km}}{\left(t^2-3\right) \mathrm{km} / \mathrm{hr}}\)

= (2t² + 3t + 4) hr

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