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

## DAV Class 8 Maths Ch 8 Brain Teasers Solutions

Question 1A.

Tick (✓) the correct option.

(i) Value of p for which (x^{2} + 3x + p) divisible by (x – 2) is

(a) 10

(b) – 10

(c) 5

(d) – 5

Solution:

(b) – 10

Let us divide x^{2} + 3x + p by x – 2. We have

Quotient = x + 5

Remainder = p + 10

If x^{2} + 3x + p is completely divisible by x – 2, then

Remainder = 0

p + 10 = 0

p = – 10.

(ii) Quotient obtained when (x^{2} + 7x – 4) is divided by (x + 7) is

(a) x

(b) x- 4

(c) x + 4

(d) x + 2

Solution:

(a) x

Quotient = x

Remainder = – 4

(iii) Polynomial which when divided by (- x^{2} + x – 1) gives a quotient (x – 2) and remainder 3 is

(a) x^{3} – x^{2} + 3x – 5

(b) – x^{3} + 3x^{2} – 3x + 5

(c) – x^{3} + 3x^{2} – 3x – 5

(d) x^{3} – 3x^{2} – 3x + 5

Solution:

(b) – x^{3} + 3x^{2} – 3x + 5

Dividend = Divisor × Quotient + Remainder

= (- x^{2} + x – 1) × (x – 2) + 3

= x(- x^{2} + x – 1) – 2 (- x^{2} + x – 1) + 3

= – x^{3} + x^{2} – x + 2x^{2} – 2x + 2 + 3

= – x^{3} + 3x^{2} – 3x + 5

(iv) What should be added to the polynomial x^{2} – 5x + 4, so that (x – 3) is factor of the resulting polynomial?

(a) 1

(b) 2

(c) 4

(d) 5

Solution:

(b) 2

Let us divide x^{2} – 5x + 4 by (x – 3)

As – 2 + 2 = 0, so to make (x – 3) a factor of x^{2} – 5x + 4, we should add (+ 2) to it.

(v) What should be subtracted from the polynomial x^{2} – 16x + 30, so that (x – 15) is factor of the resulting polynomial?

(a) 30

(b) 14

(c) 15

(d) 16

Solution:

(c) 15

Let us divide x^{2} – 16x + 30 by (x – 15)

As 15 – 15 = 0, so to make (x – 15) a factor of x^{2} – 16x + 30, we should subtract 15 from it.

Question 1B.

Answer the following questions.

(i) Find the difference between the quotient and remainder when (x^{3} – 6x^{2} + 11x – 6) is divided by (x + 1).

(ii) If (x – 3) is a factor of the polynomial (2x^{2} + x + k), find k.

(iii) Can (x – 1) be the remainder on division of a polynomial p(x) by (x + 3). Justify your answer.

(iv) Find the remainder when (x^{3} + x – 4x^{2} + 6) is divided by (x + 2). What do you conclude whether (x + 2) is a factor of (x^{3} + x – 4x^{2} + 6) or not?

(v) If (8y + 8y^{2} + 7) is divided by (1 – y), then find the remainder.

Solution:

(i) Divide x^{3} – 6x^{2} + 11x – 6 by x + 1, as shown:

Quotient = x^{2} – 7x + 18

Remainder = – 24

Difference of quotient and remainder = (x^{2} – 7x + 18) – (- 24)

= x^{2} – 7x + 18 + 24

= x^{2} – 7x + 42

(it) Dividing 2x^{2} + x + k by x – 3, we get

Remainder = k + 21

If (x – 3) is a factor of (2x^{2} + x + k), then Remainder = 0

⇒ k + 21 = 0

⇒ k = – 21.

(iii) We see that degree of both (x + 3) and (x -1) is the same, i.e., 1. We know that degree of remainder must be less than that of divisor. So, (x – 1) can’t be the remainder on division of a polynomial p(x) by (x + 3).

(iv) x^{3} + x – 4x^{2} + 6 = x^{3} – 4x^{2} + x + 6.

Now, dividing x^{3} – 4x^{2} + x + 6 by x + 2, we get

Remainder = – 20

We conclude that (x + 2) is not a factor of x^{3} – 4x^{2} + x + 6 as remainder ≠ 0.

(v) Dividend = 8y + 8y^{2} + 7 – 8y – 16

Dividend in standard form = 8y^{2} + 8y – 16

Divisor = 1 – y = – y + 1

Quotient = – 8y – 16

Remainder = 23.

Question 2.

Divide:

(i) – 39x^{4} by \(\sqrt{13}\) x^{2}

(ii) \(\sqrt{125}\)y^{2} by 5y^{2}

(iii) 49p^{3} by – 7\(\sqrt{7}\) p^{2}

(iv) 14z^{6} by \(\frac{14}{3}\) z^{3}

(v) 6x^{3} – 4x^{2} + 8x by \(\frac{2}{3}\) x

(vi) y^{2} – 5y + 1 by \(\frac{-1}{3}\) y

(vii) 18a^{3} – 12a^{2} + 9a by 3a

Solution:

(i) \(\frac{-39 x^4}{\sqrt{13} x^2}\) = \(\frac{-39 \times \sqrt{13}}{\sqrt{13} \sqrt{13}}\) x^{4 – 2}

= \(\frac{-39 \sqrt{13} x^2}{13}\)

= – 3\(\sqrt{13}\) x^{2}

(ii) \(\frac{\sqrt{125} y^2}{5 y^2}=\frac{\sqrt{5 \times 5 \times 5}}{5} \cdot \frac{y^2}{y^2}\)

= \(\frac{5 \sqrt{5}}{5}\)

= \(\sqrt{5}\)

(iii) \(\frac{49 p^3}{-7 \sqrt{7} p^2}=\frac{7 \times 7}{-7 \sqrt{7}} p^{3-2}\)

= \(\frac{7 \times \sqrt{7}}{-\sqrt{7} \sqrt{7}}\) p

= – \(\sqrt{7}\) p

(iv) \(\frac{14 z^6}{\frac{14}{3} z^3}\) = \(\frac{14 \times 3}{14}\) . z^{6 – 3}

= 3z^{3}

(v) \(\frac{6 x^3-4 x^2+8 x}{\frac{2}{3} x}\) = \(\frac{6 x^3}{\frac{2}{3} x}-\frac{4 x^2}{\frac{2}{3} x}+\frac{8 x}{\frac{2}{3} x}\)

= \(\frac{6 \times 3}{2} x^{3-1}-\frac{3 \times 4}{2} \cdot x^{2-1}+\frac{8 \times 3}{2}\)

= 9x^{2} – 6x + 12

(vi) \(\frac{y^2-5 y+1}{\frac{-1}{3} y}\) = \(\frac{y^2}{\frac{-1}{3} y}-\frac{5 y}{\frac{-1}{3} y}+\frac{1}{\frac{-1}{3} y}\)

= – 3y^{2-1} + 5 3y^{1-1} – \(\frac{3}{y}\)

= – 3y + 15 – \(\frac{3}{y}\)

(vii) \(\) = \(\)

= 6a

Question 3.

Divide and write down the quotient and remainder. Check your answer.

(i) 4x^{2} + 7x – 11 by 2x – 4

(ii) 125 – 225x + 135x^{2} – 27x^{3} by 5 – 3x

(iii) y^{3} + 5y^{2} + 12y + 9 by y + 2

(iv) 8p^{3} – 729 – 108p^{2} + 486p by 2p – 9

(v) q^{4} + 3q^{2} – 4 by q^{2} – 1

(vi) p^{4} + \(\frac{63}{4}\) p^{2} – 5 by p^{2} – \(\frac{1}{4}\)

Solution:

(i) 4x^{2} + 7x – 11 ÷ 2x – 4

Quotient = 2x + \(\frac{15}{2}\)

and Remainder = 19

Check:

Dividend = Divisor × Quotient + Remainder

4x^{2} + 7x – 11 = (2x – 4) (2x + \(\frac{15}{2}\)) + 19

= 2x (2x + \(\frac{15}{2}\)) – 4 (2x + \(\frac{15}{2}\)) + 19

= 4x^{2} + 15x – 8x – 30 + 19

= 4x^{2} + 7x – 11

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

Hence verified.

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

– 27x^{3} + 135x^{2} – 225x + 125 and – 3x + 5

Quotient = 9x^{2} – 30x + 25

and Remainder = 0

Check:

Dividend = Divisor × Quotient + Remainder

– 27x^{3} + 135x^{2} – 225x + 125 = (- 3x + 5) (9x^{2} – 30x + 25) + 0

= – 3x (9x^{2} – 30x + 25) + 5 (9x^{2} – 30x + 25)

= – 27x^{3} +90x^{2} – 75x + 45x^{2} – 150x + 125

= – 27x^{3} + 135x^{2} – 225x + 125

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

Hence verified.

(iii)

Quotient = y^{2} + 3y + 6

and Remainder = – 3

Check:

Dividend = Divisor × Quotient + Remainder

y^{3} + 5y^{2} + 12y + 9 = (y + 2) x (y^{2} + 3y + 6) + (- 3)

= y (y^{2} + 3y + 6) + 2(y^{2} + 3y + 6) – 3

= y^{3} + 3y^{2} + 6y + 2y^{2} + 6y + 12 – 3

= y^{3} + 5y^{2} + 12y + 9

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

Hence verified.

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

8p^{3} – 108p^{2} + 486p – 729

Check:

Dividend = Divisor × Quotient + Remainder

8p^{3} – 108p^{2} + 486p – 729 = (2p – 9) × (4p^{2} – 36p + 81) + 0

= 2p (4p^{2} – 36p + 81) – 9 (4p^{2} – 36p + 81)

= 8p^{3} – 72p^{2} + 162p – 36p^{2} + 324p – 729

= 8p^{3} – 108p^{2} + 486p – 729

L.H.S. = R.H.S. Hence verified.

(v)

Quotient = q^{2} + 4

and Remainder = 0

Check:

Dividend = Divisor × Quotient + Remainder

q^{4} + 3q^{2} – 4 = (q^{2} – 1) (q^{2} + 4) + 0

= q^{2} (q^{2} + 4) – 1 (q^{2} + 4)

= q^{4} + 4q^{2} – q^{2} – 4

= q^{4} + 3q^{2} – 4

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

Hence verified.

(vi)

Quotient = p^{2} + 16

and Remainder = – 1

Check:

Dividend = Divisor × Quotient + Remainder

p^{4} + \(\frac{63}{4}\) p^{2} – 5 = (p^{2} – \(\frac{1}{4}\)) (p^{2} + 16) (- 1)

= p^{2} (p^{2} + 16) – \(\frac{1}{4}\) (p^{2} + 16) – 1

= p^{4} + 16p^{2} – \(\frac{1}{4}\) p^{2} – 4 – 1

= p^{4} + \(\frac{63}{4}\) p^{2} – 5

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

Hence verified.

Question 4.

Find whether or not, the first polynomial is a factor of the second polynomial.

(i) 3x + 2, 3x^{4} + 5x^{3} – x^{2} + 13x + 10

(ii) x^{2} + 1, x^{4} – 3x^{3} – 4x^{2} + 3x + 2

Solution:

(i)

Quotient = x^{3} + x^{2} – x + 5

and Remainder = 0

∴ 3x + 2 is the factor of 3x^{4} + 5x^{2} – x^{2} + 13x + 10.

(ii)

Quotient = (x^{2} – 3x – 5)

and Remainder = (6x + 7)

∴ (x^{2} + 1) is not the factor of x^{4} – 3x^{3} – 4x^{2} + 3x + 2.

### DAV Class 8 Maths Chapter 8 HOTS

Question 1.

The sum of the remainders obtained when (x^{3} + (k + 8) x + k) is divided by (x – 2) or when it is divided by (x + 1) is zero. Find the value of k.

Sol.

x^{3} + (k + 8) x + k = x^{3} + kx + 8x + k

Case I:

Case II:

According to question,

Remainder of case I + Remainder of case II = 0

3k + 24 + (- 9) = 0

⇒ 3k + 15 = 0

⇒ k = – 5

Additional Questions:

Question 1.

State whether the given expression are polynomial or not. If not give reason.

(i) √3x + 5√x + 3

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

(iii) √5x^{\(\frac{5}{2}\)} – √2x^{\(\frac{3}{2}\)}+ √x

(iv) x + \(\frac{1}{x}\) – 3

(v) 5x – \(\frac{2}{x}\) + \(\frac{3}{x^2}\) – 5

(vi) p^{2} + p – √p

Solution:

(i) √3x + 5√x + 3

The given expression is not a polynomial as the power of x is a rational number, i.e. \(\frac{1}{2}\)

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

The given expression is a polynomial.

(iii) √5x^{\(\frac{5}{2}\)} – √2x^{\(\frac{3}{2}\)}+ √x

The given expression is not a polynomial as the degree of the polynomial is rational.

(iv) 5x – \(\frac{2}{x}\) + \(\frac{3}{x^2}\) – 5

The given expression is not a polynomial as the power of x is an integer.

(v) p^{2} + p – √p

The given expression is not a polynomial as the power of x is a negative integer.

Question 2.

Divide 5x^{4} – 3x^{3} + 2x^{2} + 6x – 11 by 5x – 1. Find the remainder and the quotient then verify the answer.

Solution:

Quotient = x^{3} – \(\frac{2}{5} x^2+\frac{8}{25} x+\frac{158}{125}\)

and Remainder = \(\frac{-1217}{125}\)

Verification:

Dividend = Divisor × Quotient + Remainder

5x^{4} – 3x^{3} + 2x^{2} + 6x – 11 = (5x – 1) × \(\left(x^3-\frac{2}{5} x^2+\frac{8}{25} x+\frac{158}{125}\right)-\frac{1217}{125}\)

= 5x \(\left(x^3-\frac{2}{5} x^2+\frac{8}{25} x+\frac{158}{125}\right)\) – 1 \(\left(x^3-\frac{2}{5} x^2+\frac{8}{25} x+\frac{158}{125}\right)\) – \(\frac{1217}{125}\)

= 5x^{4} – 2x^{3} + \(\frac{8}{5} x^2+\frac{158}{25} x-x^3+\frac{2}{5} x^2-\frac{8}{25} x-\frac{158}{125}-\frac{1217}{125}\)

= 5x^{4} – 3x^{3} + 2x^{2} + 6x – 11

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

Hence verified.

Question 3.

Divide x^{6} – 1 by x – 1 and verify the answer.

Solution:

Quotient = x^{5} + x^{4} + x^{3} + x^{2} + x + 1

and Remainder = 0

Verification:

Dividend = Divisor × Quotient + Remainder

x^{6} – 1 = (x – 1) (x^{5} + x^{4} + x^{3} + x^{2} + x + 1 )+ 0

= x (x^{5} + x^{4} + x^{3} + x^{2} + x + 1 ) – 1 (x^{5} + x^{4} + x^{3} + x^{2} + x + 1 )

= x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x – x^{5} – x^{4} – x^{3} – x^{2} – x – 1

= x^{6} – 1

L.H.S. – R.H.S.

Hence verified.

Question 4.

Using division, show that (3x + 1) is a factor of 12x^{3} – 2x^{2} + x + 1.

Solution:

Remainder = 0

Hence (3x + 1) is a factor of 12x^{3} + 2x^{2} + x + 1.

Question 5.

Using factor method, divide the following polynomial by the given binomial.

(i) x^{4} – 1 by x – 1

(ii) x^{2} – x – 42 by x – 7

Solution:

(i) \(\frac{x^4-1}{x-1}\)

⇒ \(\frac{\left(x^2+1\right)\left(x^2-1\right)}{(x-1)}\)

⇒ \(\frac{\left(x^2+1\right)(x+1)(x-1)}{(x-1)}\)

= (x^{2} + 1) (x + 1)

= x^{3} + x^{2} +x + 1

Hence the quotient is x^{3} + x^{2} + x + 1.

(ii) \(\frac{x^2-x-42}{x-7}\)

⇒ \(\frac{x^2-7 x+6 x-42}{x-7}\)

⇒ \(\frac{x(x-7)+6(x-7)}{(x-7)}\)

⇒ \(\frac{(x-7)(x+6)}{(x-7)}\)

= (x + 6)

Hence the quotient is (x + 6).

Question 6.

Divide 3x (5x^{2} + 10x^{3} + 5) – (x^{2} – 4 – 2x) by (- 3 + 5x).

Solution:

Dividend:

3x (5x^{2} + 10x^{3} + 5) – (x^{2} – 4 – 2x) = 15x^{3} + 30x^{4} + 15x – x^{2} + 4 + 2x

= 30x^{4} + 15x^{3} – x^{2} + 17x + 4

Hence quotient = 6x^{3} + \(\frac{33}{5} x^2-\frac{94}{25} x+\frac{707}{125}\)

Remainder = \(\frac{2621}{125}\)

Question 7.

Divide: x^{4} + \(\frac{63}{4}\) x^{2} – 5 by x^{2} – \(\frac{1}{4}\) and check the answer.

Solution:

Quotient = x^{2} + 16

and the remainder = – 1

Check:

Dividend = Divisor × Quotient + Remainder

x^{2} + 16 = (x^{2} – \(\frac{1}{4}\)) × (x^{2} + 16) – 1

= x^{2} (x^{2} + 16) – \(\frac{1}{4}\) (x^{2} + 16) – 1

= x^{4} + 16x^{2} – \(\frac{1}{4}\) x^{2} – 4 – 1

= x^{4} + \(\frac{63}{4}\) x^{2} – 5

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

Hence verified.

Question 8.

Find the value of k if x^{2} – kx – 42 is divisible by x – 7.

Solution:

If the given expression is completely divisible by (x – 7), then remainder = 0.

∴ – 7 (k – 7) – 42 = 0

Dividing both side by – 7,

(k – 7) + 6 = 0

∴ k = 1

Hence the value of k = 1.

Question 9.

Find the values of‘a’ and ‘b’ if the polynomial x^{3} + ax^{2} + bx – 12 is completely divisible by (x – 1) and (x + 3).

Solution:

Case I:

Case II:

If (x – 1) and (x + 3) are factors of x^{3} + ax^{2} + bx – 12, then remainders should be zero.

In Case I,

a + b – 11 = 0

⇒ a + b = 11 ………………(1)

In Case II,

9a – 36 – 39 = 0

⇒ 3a – b = 13

On solving equations (1) and (2),

a = 6 and b = 5 ……………..(2)

Question 10.

Find the value of ‘k’ if (x + k) is a factor of (x^{2} + 5x + 6).

Solution:

If (x + k) is a factor of x^{2} + 5x + 6, then

Remainder = 0

k^{2} – 5k + 6 = 0

k^{2} – 3k – 2k + 6 = 0

= k (k – 3) – 2 (k – 3) = 0

(k – 3) (k – 2) = 0

If k – 3 = 0

⇒ k = 3

and if k – 2 = 0

⇒ k = 2.

Hence the values of k are 2 and 3.

Question 11.

Divide the following polynomials by a monomial.

(i) 8p^{3} + 16p^{2} + 14p + 1 by 4p

(ii) – 3x^{2} + √3x by √3x

Solution:

(i) \(\frac{-8 p^3+16 p^2+14 p+1}{4 p}\) = \(\frac{-8 p^3}{4 p}+\frac{16 p^2}{4 p}+\frac{14 p}{4 p}+\frac{1}{4 p}\)

= – 2p^{2} + 4p + \(\frac{7}{2}+\frac{1}{4 p}\)

(ii) \(\frac{-3 x^2+\sqrt{3} x}{\sqrt{3} x}\) = \(\frac{-3 x^2}{\sqrt{3} x}+\frac{\sqrt{3} x}{\sqrt{3} x}\)

= – √3x + 1

Question 12.

Divide the following by using long division method and check your answer.

(i) x^{3} – 3x^{2} + 3x – 1 by x – 1

(ii)12x + 10x^{3} + 8x^{4} + 15 by 5 + 4x

Solution:

Check:

Dividend = Divisor × Quotient + Remainder

x^{3} – 3x^{2} + 3x – 1 = (x – 1) x (x^{2} – 2x + 1) + 0

= x (x^{2} – 2x + 1) – 1 (x^{2} – 2x + 1)

= x^{3} – 2x^{2} + x – x^{2} + 2x – 1

= x^{3} – 3x^{2} + 3x – 1

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

Hence verified.

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

Dividend = 8x^{4} + 10x^{3} + 12x + 15

and Divisor = 4x + 5

Check:

Dividend = Divisor × Quotient + Remainder

8x^{4} + 10x^{3} + 12x + 15 = (4x + 5) × (2x^{3} + 3) + 0

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

= 8x^{4} + 12x + 10x^{3} + 15

= 8x^{4} +10x^{3} + 12x + 15

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

Hence verified.

Question 13.

Check whether the second polynomial is a factor of first polynomial. Use long division method to check,

(i) 2y^{3} – 14y + 12; y + 3

(ii) 3x^{2} – 11x + 6; x – 3

Solution:

Here, remainder = 0

Hence, y + 3 is a factor of 2y^{3} – 14y + 12.

(ii)

Here, remainder = 0

Hence, x – 3 is a factor of 3x^{2} – 11x + 6.

Question 14.

Degree of the polynomial q^{10} + q^{6} – q^{4} + q^{8} is

(a) 4

(b) 6

(c) 10

(d) 8

Solution:

(c) 10

Degree of the given polynomial is 10.

Question 15.

Using division method, show that (3x + 1) is a factor of the polynomial 12x^{3} – 2x^{2} + x + 1.

Solution:

Quotient = 4x^{2} – 2x + 1

and Remainder = 0

Hence (3x + 1) is a factor of the polynomial 12x^{3} – 2x^{2} + x + 1.

Question 16.

Divide: (y^{2} + 2y – 35) by (y + 7) using factor method.

Solution:

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

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

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

= y – 5

Hence Quotient = y – 5

and Remainder = 0

Question 17.

Find the quotient and remainder, when – 6x^{4} + 5x^{2} + 11x + 1 is divided by 2x^{2} + 1.

Solution:

Quotient = – 3x^{2} + 4

and the Remainder = 11x – 3

Question 18.

Divide (- 1 + x^{4}) by (- 1 + x) and verify your answer.

Solution:

Verification:

Dividend = Divisor × Quotient + Remainder

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

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

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

= x^{4} – 1

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

Hence verified.

Multiple Choice Questions:

Question 1.

If x – 1 is a factor of x^{2} – 2kx + 1, then the value of k is

(a) 0

(b) 2

(c) 1

(d) – 1

Solution:

(c) 1

Question 2.

If 6x^{2} – 13x + 6 is divisible by 2x – 3, then the remainder is

(a) 0

(b) 3x – 2

(c) 1

(d) None of these

Solution:

(a) 0

Question 3.

If 8x^{3} – 6x^{2} + 10x + 3 is completely divisible by (4x + 1), then the quotient is

(a) 2x^{2} + 2x – 3

(b) 2x^{2} – 3x + 3

(c) 2x^{2} – 2x – 3

(d) 2x^{2} – 2x + 3

Solution:

(d) 2x^{2} – 2x + 3

Question 4.

The value of k for which x^{6} – k is completely divisible by x – 1 is

(a) – 1

(b) 0

(c) – 2

(d) 1

Solution:

(d) 1

Question 5.

If x^{4} – a^{4} is divisible by x^{2} + a^{2}, then the quotient is

(a) x^{2} – a^{2}

(b) x – a

(c) x + a

(d) 0

Solution:

(a) x^{2} – a^{2}