# RS Aggarwal Class 9 Solutions Chapter 2 - Polynomials Ex 2A (2.1)

## RS Aggarwal Class 9 Chapter 2 - Polynomials Ex 2A (2.1) Solutions Free PDF

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## RS Aggarwal Class 9 Ex 2A Chapter 2

Question 1:

(x3 – 7x2 + 6x + 4) is divided by (x-6)

f(x) = x3 – 7x2 + 6x + 4

Now, x – 6 = 0

x = 6

By the remainder theorem, we know that when f(x) is divide by (x – 6) the remainder is f(6)

Now, f(6) = 63 – 7 x 62 + 6 x 6 + 4

= 216 – 252 + 36 + 4

= 256 – 252

= 4

The required remainder is 4.

Question 2:

(x3 – 6x2 + 13x + 60) is divided by (x+2)

f(x) = (x3 – 6x2 + 13x + 60)

Now, x + 2 = 0

x = -2

By the remainder the theorem, we know that when f(x) is divide by (x + 2) the remainder is f(-2).

Now, f(-2) = (-2)3 – 6(-2)2 + 13(-2) + 60

= -8 – 24 – 26 + 60

= -58 + 60

= 2

The required remainder is 2.

Question 3:

(2x4 + 6x3 + 2x2 + x – 8) is divided by (x+3)

f(x) = (2x4 + 6x3 + 2x2 + x – 8)

Now, x + 3 = 0

x = -3

By the remainder the theorem, we know that when f(x) is divide by (x + 3) the remainder is f(-3).

f(-3) = 2(-3)4 + 6(-3)3 + 2(-3)2 – 3 – 8

= 162 – 162 + 18 – 3 – 8

= 18 – 11

= 7 The required remainder is 7.

Question 4:

(4x3 – 12x2 + 11x – 5) is divided by (2x-1)

f(x) = (4x3 – 12x2 + 11x – 5)

Now, 2x – 1 = 0

x=1÷2$x=1\div 2$

By the remainder theorem, we know that when f(x) is divided by (2x – 1) the remainder is  f(12)$f\left ( \frac{1}{2} \right )$

Now,

f(12)$f\left ( \frac{1}{2} \right )$= 4(12)312(12)2+11(12)5$4\left ( \frac{1}{2} \right )^{3}-12\left ( \frac{1}{2} \right )^{2}+11\left ( \frac{1}{2} \right )-5$

=4×1812×14+1125$=4\times \frac{1}{8}-12\times \frac{1}{4}+\frac{11}{2}-5$
=123+1125$=\frac{1}{2}-3+\frac{11}{2}-5$
=16+11102$=\frac{1-6+11-10}{2}$
=16+122$=\frac{-16+12}{2}$
=42$=\frac{-4}{2}$

=      -2

The required remainder is -2.

Question 5:

(81x4 + 54x3 – 9x2 – 3x + 2) is divided by (3x+2)

f(x) = (81x4 + 54x3 – 9x2 – 3x + 2)

Now, 3x + 2 = 0

x= 23$\frac{-2}{3}$

By the remainder theorem, we know that when f(x) is divided by (3x+ 2) the remainder is f(23)$f\left ( \frac{-2}{3}\right )$

Now,

f(23)$f\left ( \frac{-2}{3}\right )$=81(23)4+54(23)39(23)23(23)+2$\left ( \frac{-2}{3} \right )^{4}+54\left ( \frac{-2}{3} \right )^{3}-9\left ( \frac{-2}{3} \right )^{2}-3\left ( \frac{-2}{3} \right )+2$

=81×(1681)+54(827)9(49)+2+2$=81\times \left ( \frac{16}{81} \right )+54\left ( \frac{-8}{27} \right )-9\left ( \frac{4}{9} \right )+2+2$

=16-16-4+4

=0

The required remainder is 0.

Question 6:

(x3 – ax2 + 2x – a) is divided by (x-a)

f(x) = (x3 – ax2 + 2x – a)

Now,x-a=0

x =a

By the remainder theorem, we know that when f(x) is divided by (x – a) the remainder is f(a)

Now, f(a) = a3 – a a2 + 2 a – a

= a3 – a3 + 2a – a

=a

The required remainder is a.

Question 7:

The polynomials (ax3 + 3x2 – 3) and (2x3 – 5x + a) when divided by (x-4) leave the same remainder. Find the value of a.

Let f(x) = ax3 + 3x2 – 3

and g(x) = 2x3 – 5x + a

f(4) = a x 43 + 3 x 42 – 3

= 64a + 48 – 3

= 64a + 45

g(4) = 2 x 43 – 5 x 4 + a

= 128 – 20 + a

= 108 + a

It is given that:

f(4) = g(4)

64a + 45 = 108 + a

64a – a = 108 – 45

63a = 63

a=6363$a=\frac{63}{63}$ =  1

The value of a is 1.

Question 8:

The polynomial f(x) = (x4 – 2x3 + 3x2 – ax + b) when divided by (x-1) and (x+1) leaves the remainders 5 and 19 respectively. Find the values of a and b. Hence, find the remainder when f(x) is divided by (x-2).

Let f(x) = (x4 – 2x3 + 3x2 – ax + b)

From the given information,

f(1) = 14– 2(1)3 + 3(1)2 – a (1 ) + b = 5

1 – 2 + 3 – a + b = 5

2 – a + b = 5 ….(i)

And,

f(-1) = (-1)4 – 2(4)3 + 3(4)2 – a(-1) + b = 19

1 + 2 + 3 +a+b= 19

6 + a + b = 19 ….(ii)

Adding (i) and (ii), we get

8 + 2b = 24

2b = 24 – 8 = 16

b = 162$\frac{16}{2}$

Substituting the value of b = 8 in (i), we get

2 – a + 8 = 5

-a + 10 = 5

-a = -10 + 5

a = 5

a = 5 and b = 8

f(x) = x4 – 2x3 + 3x2 – ax + b

= x4 – 2x3 + 3x2 – 5x + 8

f(2) = (2)4 – 2(2)3 + 3(2)2 – 5(2) + 8

= 16 – 16 + 12 – 10 + 8

= 20 – 10

= 10

The required remainder is 10.

Question 9:

(x-2) is a factor of (x3 – 8)

f(x) = (x3 – 8)

By the Factor Theorem, (x – 2) will be a factor of f(x) if f(2) = 0.

Here, f(2) = (2)3 – 8

= 8 – 8

= 0

Therefore,  (x – 2) is a factor of (x3 – 8).

Question 10:

(x-3) is a factor of  (2x3 + 7x2 – 24x – 45)

f(x) = (2x3 + 7x2 – 24x – 45)

By the Factor Theorem, (x – 3) will be a factor of f(x) if f(3) = 0.

Here, f(3) = 2 x 33 + 7 x 32 – 24 x 3 – 45

= 54 + 63 – 72 – 45

= 117 – 117

= 0

Therefore,  (x – 3) is a factor of (2x3 + 7x2 – 24x – 45).

Question 11:

(x-1) is a factor of  (2x4 + 9x4 + 6x2 – 11x – 6)

f(x) = (2x4 + 9x4 + 6x2 – 11x – 6)

By the Factor Theorem, (x – 1) will be a factor of f(x) if f(1) = 0.

Here, f(1) = 2 x 14 + 9 x 13 +6 x 12 – 11 x 1 – 6

= 2 + 9 + 6 – 11 – 6

= 17 – 17

= 0

Therefore,  (x – 1) is factor of (2x4 + 9x3 + 6x2 – 11x – 6).

Question 12:

(x+2) is a factor of  (x4 – x2 – 12)

f(x) = (x4 – x2 – 12)

By the Factor Theorem, (x + 2) will be a factor of f(x) if f(-2) = 0.

Here, f(-2) = (-2)4 – (-2)2 – 12

= 16 – 4 – 12

= 16 – 16

= 0

Therefore,  (x + 2) is a factor of (x4 – x2 – 12).

Question 13:

(x+5) is a factor of  2x3 + 9x2 – 11x – 30

f(x) = 2x3 + 9x2 – 11x – 30

By the Factor Theorem, (x + 5) will be a factor of f(x) if f(-5) = 0.

Here, f(-5) = 2(-5)3 + 9(-5)2 – 11(-5) – 30

= -250 + 225 + 55 – 30

= -280 + 280

= 0

Therefore,  (x + 5) is a factor of (2x3 + 9x2 – 11x – 30).

Question 14:

(2x-3) is a factor of  (2x4 + x3 – 8x2 – x + 6)

f(x) = (2x4 + x3 – 8x2 – x + 6)

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.

Here, 2x – 3 = 0

X = 32$\frac{3}{2}$

f(32)=2(32)4+(32)38(32)2(32)+6$f\left ( \frac{3}{2} \right )=2\left ( \frac{3}{2} \right )^{4}+\left ( \frac{3}{2} \right )^{3}-8\left ( \frac{3}{2} \right )^{2}-\left ( \frac{3}{2} \right )+6$
2×8116+2788×9432+6$2\times \frac{81}{16}+\frac{27}{8}-8\times \frac{9}{4}-\frac{3}{2}+6$
=818+2781832+6$=\frac{81}{8}+\frac{27}{8}-18-\frac{3}{2}+6$
=81+2714412+488$=\frac{81+27-144-12+48}{8}$
=1561568$=\frac{156-156}{8}$

= 0

Therefore,  (2x – 3) is a factor of (2×4 + x3 – 8×2 – x + 6).

Question 15:

(x-2$\sqrt{2}$) is a factor of (7x2 – 42×$\sqrt{2} \times$ – 6 = 0)

f(x) = (7x2 – 42×$\sqrt{2} \times$ – 6 = 0)

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.

Here,     f(2)=7(2)242×26$f\left ( \sqrt{2} \right ) = 7\left ( \sqrt{2} \right )^{2}-4\sqrt{2}\times \sqrt{2}-6$

= 14 – 8 – 6

= 14 – 14

= 0

Therefore, (x –2$\sqrt{2}$ ) is a factor of (7 – 42$4\sqrt{2}$ x – 6 = 0).

Question 16:

(x+2$\sqrt{2}$) is a factor of  (42x2$4\sqrt{2} x^{2}$+ 5x +2$\sqrt{2}$ = 0)

f(x) = (42x2$4\sqrt{2} x^{2}$+ 5x +2$\sqrt{2}$ = 0)

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.

f(2)=22(2)2+5(2)+2$f\left ( -\sqrt{2} \right )=2\sqrt{2}\left ( -\sqrt{2} \right )^{2}+5\left ( -\sqrt{2} \right )+\sqrt{2}$
=22×252+2$=2\sqrt{2}\times 2-5\sqrt{2}+\sqrt{2}$
=4252+2$=4\sqrt{2}-5\sqrt{2}+\sqrt{2}$
=5252=0$=5\sqrt{2}-5\sqrt{2}=0$

Here,

(x + =2$=\sqrt{2}$) is a factor of (4=2$=\sqrt{2}$ + 5x +2$\sqrt{2}$= 0).

Question 17:

Find the value of k for which (x-1) is a factor of (2x3 + 9x2+ x + k)

f(x) = (2x3 + 9x2+ x + k)

x – 1 = 0

= x = 1

f(1) = 2 x 13+ 9 x 12 + 1 + k

= 2 + 9 + 1 + k

= 12 + k

Given that (x – 1) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(1) = 0.

f(1) = 12 + k = 0

k = -12.

Question 18:

Find the value of a for which (2x3 – 3x2 – 18x + a) is divisible by (x-4).

f(x) = (2x3 – 3x2 – 18x + a)

x – 4 = 0

x = 4

f(4) = 2(4)3 – 3(4)2 – 18 x 4 + a

= 128 – 48 – 72 + a

= 128 – 120 + a

= 8 + a

Given that (x – 4) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(4) = 0.

f(4) = 8 + a = 0

a = -8

Question 19:

Find the value of a for which (x4 – x3 – 11x2 – x + a) is divisible by (x+3).

f(x) = x4 – x3 – 11x2 – x + a

x + 3 = 0

x = -3

f(-3) = (-3)4 – (-3)3 -11 (-3)2 – (-3) + a

= 81 + 27 – 11 x 9 + 3 + a

= 81 + 27 – 99 + 3 + a

= 111 – 99 + a

= 12 + a

Given that f(x) is divisible by (x + 3), that is (x+3) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(-3) = 0

. f(-3) = 12 + a =0

a = -12.

Question 20:

For what value of a is the polynomial (2x3 + ax2 + 11x + a + 3) exactly divisible by (2x-1)?

f(x) = (2x3 + ax2 + 11x + a + 3)

2x – 1 = 0

X= 12$\frac{1}{2}$

Given that f(x) is exactly divisible by (2x – 1), that is (2x – 1) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0  therefore  f(12)0$f\left ( \frac{1}{2} \right )\neq 0$

Therefore, we have

f(12)=2(12)3+a(12)2+11×12+a+3=0$f\left ( \frac{1}{2} \right )=2\left ( \frac{1}{2} \right )^{3}+a\left ( \frac{1}{2} \right )^{2}+11\times \frac{1}{2}+a+3=0$
=2(18)+a(14)+112+a+3=0$=2\left ( \frac{1}{8} \right )+a\left ( \frac{1}{4} \right )+\frac{11}{2}+a+3=0$
14+14a+112+a+3=0$\Rightarrow \frac{1}{4}+\frac{1}{4}a+\frac{11}{2}+a+3=0$
1+a+22+4a+124=0$\Rightarrow \frac{1+a+22+4a+12}{4}= 0$
5a+354=0$\Rightarrow \frac{5a+35}{4}= 0$

$\Rightarrow$ 5a+35= 0

$\Rightarrow$ 5a = -35

$\Rightarrow$   a = 355$\frac{-35}{5}$  = -7

Therefore, The value of a = -7.

Question 21:

Find the values of a and b so that the polynomial (x3 – 10x2 + ax + b) is exactly divisible by (x-1) as well as (x-2).

Let f(x) = (x3 – 10x2 + ax + b), then by factor theorem

(x – 1) and (x – 2) will be factors of f(x) if f(1) = 0 and f(2) = 0.

f(1)= 13 – 10x 12 +a x 1 +b= 0

1 – 10 + a + b = 0

a + b = 9 ….(i)

And f(2) = 23 – 10 x 22 + a x 2 + b = 0

8 – 40 + 2a + b = 0

2a + b = 32 ….(ii)

Subtracting (i) from (ii), we get

a = 23

Substituting the value of a = 23 in (i), we get

23 + b = 9

b = 9 – 23

b = -14

a= 23 and b = -14.

Question 22:

Find the values of a and b so that the polynomial (x4 + ax3 – 7x2 – 8x + b) is exactly divisible by (x+2) as well as (x+3).

Let f(x) = (x4 + ax3 – 7x2 – 8x + b)

Now, x + 2 = 0 x = -2 and x + 3 = 0 x = -3

By factor theorem, (x + 2) and (x + 3) will be factors of f(x) if

f(-2) = 0 and f(-3) = 0

f(-2) = (-2)4 + a (-2)3 – 7 (-2)2 – 8 (-2) + b = 0

16 – 8a – 28 + 16 + b = 0

-8a + b = -4

8a – b = 4 …(i)

And, f(-3) = (-3)4 + a (-3)3– 7 (-3)2 – 8 (-3) + b = 0

81 – 27a – 63 + 24 + b = 0

-27a + b = -42

27a – b = 42 …. (ii)

Subtracting (i) from (ii), we get,

19a = 38

So, a = 2 S

substituting the value of a = 2 in (i), we get

8(2) – b = 4

16 – b = 4

-b = -16 + 4

-b = -12

b = 12

a = 2 and b = 12.

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Motion of a motorcyclist is shown in the figure. Which of the following shows the distance-time graph for this motion?