# 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.