Chapter 6 : Factorisation of Polynomials - Rd Sharma Solutions for Class 9 Maths CBSE

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Chapter 6 - Factorisation of Polynomials Excercise Ex. 6.1

Question 1

Solution 1

Question 2

Solution 2

Question 3
Solution 3

Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7

begin mathsize 12px style Identify space constant comma space linear comma space quadratic space and space cubic space polynomials space from space the space following space polynomials colon
open parentheses straight i close parentheses space straight f open parentheses straight x close parentheses space equals 0
open parentheses ii close parentheses space straight g space open parentheses straight x close parentheses space equals 2 straight x cubed minus 7 straight x space plus space 4
open parentheses iii close parentheses space straight h open parentheses straight x close parentheses space equals negative 3 straight x space plus 1 half
open parentheses iv close parentheses straight p open parentheses straight x close parentheses space equals space 2 straight x squared minus straight x plus 4
open parentheses straight v close parentheses space straight q open parentheses straight x close parentheses space equals space 4 straight x plus 3
open parentheses vi close parentheses straight r open parentheses straight x close parentheses space equals space 3 straight x cubed equals 4 straight x squared plus 5 straight x minus 7 end style

Solution 7

Question 8
Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Solution 8
Degree of a polynomial is the highest power of variable in the polynomial.
Binomial has two terms in it. So binomial of degree 35 can be written as x35 + 7 .  
Monomial has only one term in it. So monomial of degree 100 can be written as 7x100.
 
Concept Insight: Mono, bi and tri means one, two and three respectively. So, monomial is a polynomial having one term similarly for binomials and trinomials. Degree is the highest exponent of variable.  The answer is not unique in such problems . Remember that the terms are always separated by +ve or -ve sign and not with  .  

Chapter 6 - Factorisation of Polynomials Excercise Ex. 6.2

Question 1
If f(x) = 2x3 - 13x2 + 17x + 12, find:

(i) f(2)

(ii) f(-3)

(iii) f(0)
Solution 1
(i)

f(x) = 2x3 - 13x2 + 17x + 12

f(2) = 2(2)3 - 13(2)2 + 17(2) + 12

      = 16 - 52 + 34 + 12

      = 10

(ii)

f(-3) = 2(-3)3 - 13(-3)2 + 17(-3) + 12

       = -54 - 117 - 51 + 12

       = - 210

(iii)

f(0) = 2(0)3 - 13(0)2 + 17(0) + 12

      = 0 - 0 + 0 + 12
   
      =12
Question 2
Solution 2

Question 3
Solution 3

Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Find rational roots of the polynomial f(x) = 2x3 + x2 - 7x - 6.
Solution 7

Chapter 6 - Factorisation of Polynomials Excercise Ex. 6.3

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7

Question 8
Solution 8
Question 9

Solution 9
Question 10

If the polynomials ax3 + 3x2 - 13 and 2x3 -5x + a, when divided by (x-2) leave the same remainder, find the value of a.

Solution 10

Question 11
Find the remainder when x3 + 3x2 + 3x + 1 is divided by
 
Solution 11

Question 12

Solution 12

Chapter 6 - Factorisation of Polynomials Excercise Ex. 6.4

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6

Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12

Question 13
Solution 13

Question 14
Solution 14
Question 15
Solution 15
Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

What must be added to 3x3 + x2 - 22x + 9 so that the result is exactly divisible by 3x2 + 7x - 6?

Solution 25

Chapter 6 - Factorisation of Polynomials Excercise Ex. 6.5

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4

Using factor theorem, factorize: x4 - 7x3 + 9x2 + 7x -10

Solution 4

Question 5

Using factor theorem, factorize: 3x3 - x2 - 3x + 1

Solution 5

Question 6

Using factor theorem, factorize each of the following polynomials:

x3 - 23x2 + 142x - 120

Solution 6

Question 7

Using factor theorem, factorize: y3 - 7y + 6

Solution 7

Question 8

Using factor theorem, factorize: x3 - 10x2 - 53x - 42

Solution 8

Question 9

Using factor theorem, factorize: y3 - 2y2 - 29y - 42

Solution 9

Question 10

Using factor theorem, factorize: 2y3 - 5y2 - 19y + 42

Solution 10

Question 11

x3 + 13x2 + 32x + 20

Solution 11

         Let p(x) = x3 + 13x2 + 32x + 20
         The factors of 20 are 1, 2, 4, 5 ... ...
         By hit and trial method
         p(- 1) = (- 1)3 + 13(- 1)2 + 32(- 1) + 20
                   = - 1 + 13 - 32 + 20
                   = 33 - 33 = 0
         As p(-1) is zero, so x + 1 is a factor of this polynomial p(x).

         Let us find the quotient while dividing x3 + 13x2 + 32x + 20 by (x + 1)
          By long division

 
        
 
         We know that
         Dividend = Divisor  Quotient + Remainder
         x3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20) + 0
                                         = (x + 1) (x2 + 10x + 2x + 20)
                                         = (x + 1) [x (x + 10) + 2 (x + 10)]
                                         = (x + 1) (x + 10) (x + 2)
                                         = (x + 1) (x + 2) (x + 10)
Question 12

Factorise:

x3 - 3x2 - 9x - 5

Solution 12

        Let p(x) = x3 - 3x2 - 9x - 5
        Factors of 5 are 1,  5.
        By hit and trial method
        p(- 1) = (- 1)3 - 3(- 1)2 - 9(- 1) - 5
           = - 1 - 3 + 9 - 5 = 0
        So x + 1 is a factor of this polynomial
        Let us find the quotient while dividing x3 + 3x2 - 9x - 5 by x + 1
        By long division

 
      
 
        Now, Dividend = Divisor  Quotient + Remainder
         x3 - 3x2 - 9x - 5 = (x + 1) (x2 - 4 x - 5) + 0
                                     = (x + 1) (x2 - 5 x + x - 5)
                                     = (x + 1) [(x (x - 5) +1 (x - 5)]
                                     = (x + 1) (x - 5) (x + 1)
                                     = (x - 5) (x + 1) (x + 1)
 
Question 13

Factorise:

2y3 + y2 - 2y - 1

Solution 13

         Let p(y) = 2y3 + y2 - 2y - 1

         By hit and trial method
         p(1) = 2 ( 1)3 + (1)2 - 2( 1) - 1
                = 2 + 1 - 2 - 1= 0
         So, y - 1 is a factor of this polynomial
         By long division method,
        
 
          p(y) = 2y3 + y2 - 2y - 1
                 = (y - 1) (2y2 +3y + 1)
                 = (y - 1) (2y2 +2y + y +1)
                 = (y - 1) [2y (y + 1) + 1 (y + 1)]
                 = (y - 1) (y + 1) (2y + 1)

 
 
Question 14

Using factor theorem, factorize: x3 - 2x2 - x + 2

Solution 14

Question 15

Solution 15

Question 16

Using factor theorem, factroize : x4 - 2x3 - 7x2 + 8x + 12

Solution 16

Question 17

Using factor theorem, factroize : x4 + 10x3 + 35x2 + 50x + 24

Solution 17

Question 18

Using factor theorem, factorize : 2x4 - 7x3 - 13x2 + 63x - 45

Solution 18

Chapter 6 - Factorisation of Polynomials Excercise 6.34

Question 1

If x - 2 is a factor of x2 + 3ax - 2a, then a =

(a) 2

(b) -2

(c) 1

(d) -1

Solution 1

Let p(x) = x2 + 3ax - 2a be the given polynomial.

x - 2 is a factor of p(x).

Thus, p(2) = 0

(2)2 + 3a × 2 - 2a = 0

4 + 4a = 0

a = -1

Hence, correct option is (d).

Question 2

If x3 + 6x2 + 4x + k is exactly divisible by x + 2, then k =

(a) -6

(b) -7

(c) -8

(d) -10

Solution 2

Since, p(x) = x3 + 6x2 + 4x + k is exactly divisible by x + 2,

(x + 2) is a factor of p(x).

So, p(-2) = 0

i.e (-2)3 + 6(-2)2 + 4(-2) + k = 0

     -8 + 24 - 8 + k = 0

     24 - 16 + k = 0

     8 + k = 0

     k = -8

Hence, correct option is (c).

Question 3

If x - a is a factor of x3 - 3x2a + 2a2x + b, then the value of b is

(a) 0

(b) 2

(c) 1

(d) 3

Solution 3

Let p(x) = x3 - 3x2a + 2a2x + b

(x - a) is a factor of p(x). 

So, p(a) = 0

a3 - 3a2.a + 2a2.a + b = 0

a3 - 3a3 + 2a3 + b = 0

3a3 - 3a3 + b = 0

b = 0

Hence, correct option is (a).

Question 4

If x140 + 2x151 + k is divisible by x + 1, then the value of k is

(a) 1

(b) -3

(c) 2

(d) -2

Solution 4

Let p(x) = x140 + 2x151 + k

Since p(x) is divisible by (x + 1),  

(x + 1) is a factor of p(x).

So, p(-1) = 0

(-1)140 + 2(-1)151 + k = 0

1 + 2(-1) + k = 0

1 - 2 + k = 0

k - 1 = 0

k = 1

Hence, correct option is (a).

Question 5

If x + 2 is a factor of x2 + mx + 14, then m =

(a) 7

(b) 2

(c) 9

(d) 14

Solution 5

If x + 2 is a factor of x2 + mx + 14,

then at x = -2,

x2 + mx + 14 = 0

i.e. (-2)2 + m(-2) + 14 = 0

4 - 2m + 14 = 0

2m = 18

m = 9

Hence, correct option is (c).

Question 6

If x - 3 is a factor of x2 - ax - 15, then a =

(a) -2

(b) 5

(c) -5

(d) 3

Solution 6

x - 3 is a factor of x2 - ax - 15,

then at x = 3,

x2 - ax - 15 = 0

i.e. (3)2 - a(3) - 15 = 0

9 - 3a - 15 = 0

a = -2

Hence, correct option is (a).

Question 7

If x51 + 51 is divided by x + 1, the remainder is

(a) 0

(b) 1

(c) 49

(d) 50

Solution 7

begin mathsize 12px style When space straight a space polynomial space straight p left parenthesis straight x right parenthesis space is space divided space by space straight q left parenthesis straight x right parenthesis space straight i. straight e. space left parenthesis straight x space plus-or-minus straight alpha right parenthesis space then space straight p left parenthesis minus-or-plus space straight alpha right parenthesis space is space the space remainder.
If space straight x space plus-or-minus space straight alpha space is space the space factor space of space polynomial comma space then space remainder space is space apostrophe 0 apostrophe.
So space if space straight x to the power of 51 space plus space 51 space is space divided space by space straight x space plus space 1 comma
remainder space equals space left parenthesis negative 1 right parenthesis to the power of 51 space plus space 51 space equals negative 1 space plus space 51 space equals space 50
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

 

Question 8

If x + 1 is a factor of the polynomial 2x2 + kx, then k =

(a) -2

(b) -3

(c) 4

(d) 2

Solution 8

x + 1 is a factor of p(x) = 2x2 + kx

Then, p(-1) = 0

i.e. 2(-1)2 + k(-1) = 0

2 - k = 0

k = 2

Hence, correct option is (d).

Question 9

If x + α is a factor of x4 - a2x2 + 3x - 6a, then a =

(a) 0

(b) -1

(c) 1

(d) 2

Solution 9

begin mathsize 12px style straight x space plus space straight a space is space straight a space factor space of space polynomial comma space straight p left parenthesis straight x right parenthesis space equals space straight x to the power of 4 space end exponent minus space straight a squared straight x squared space plus space 3 straight x space minus space 6 straight a space
Then comma space at space straight x equals negative straight a comma space straight p left parenthesis straight x right parenthesis space equals space 0 space
rightwards double arrow left parenthesis negative straight a right parenthesis to the power of 4 space minus space straight a squared left parenthesis negative straight a right parenthesis squared space plus space 3 left parenthesis negative straight a right parenthesis space minus space 6 straight a space equals space 0
rightwards double arrow up diagonal strike straight a to the power of 4 end strike space minus space up diagonal strike straight a to the power of 4 end strike space minus space 3 straight a space minus space 6 straight a space equals space 0
rightwards double arrow negative 9 straight a space equals space 0
rightwards double arrow straight a space equals space 0
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

 

Question 10

The value of k for which x - 1 is a factor of 4x3 + 3x2 - 4x + k, is

(a) 3

(b) 1

(c) -2

(d) -3

Solution 10

begin mathsize 12px style Let space straight p left parenthesis straight x right parenthesis space equals space 4 straight x cubed space plus space 3 straight x squared space minus space 4 straight x space plus space straight k
Now comma space if space left parenthesis straight x space minus space 1 right parenthesis space is space straight a space facor space of space straight p left parenthesis straight x right parenthesis comma space then space at space straight x space equals space 1 comma space straight p left parenthesis straight x right parenthesis equals space 0
So comma space straight p left parenthesis 1 right parenthesis space equals space 0
rightwards double arrow space 4 left parenthesis 1 right parenthesis cubed space plus space 3 left parenthesis 1 right parenthesis squared space minus space 4 left parenthesis 1 right parenthesis space plus space straight k space equals space 0
rightwards double arrow space up diagonal strike 4 space plus space 3 space minus up diagonal strike space 4 end strike space plus space straight k space equals space 0
rightwards double arrow space straight k space equals space minus 3
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 11

If x + 2 and x - 1 are the factors of x3 + 10x2 + mx + n, then the values of m and n are respectively

(a) 5 and -3

(b) 17 and -8

(c) 7 and -18

(d) 23 and -19

Solution 11

If (x + 2) and (x - 1) are factors of polynomial x3 + 10x2 + mx + n,

then x = -2, x = +1 will satisfy the polynomial.

Let p(x) = x3 + 10x2 + mx + n

Then, p(-2) = 0

(-2)3 + 10(-2)2 + m(-2) + n = 0

-8 + 40 - 2m + n = 0

32 - 2m + n = 0            ....(1)

And, p(1) = 0

(1)3 + 10(1)2 + m(1) + n = 0

1 + 10 + m + n = 0

11 + m + n = 0                   ....(2)

Substracting equation (1) from equation (2), we get

-21 + 3m = 0

3m = 21

m = 7       

Substituting m = 7 in equation (2),

11 + 7 + n = 0

18 + n = 0

n = -18

Hence, correct option is (c).

Chapter 6 - Factorisation of Polynomials Excercise 6.35

Question 1

begin mathsize 11px style Let space straight f left parenthesis straight x right parenthesis space be space straight a space polynomial space such space that space straight f open parentheses negative 1 half close parentheses equals 0 comma space then space straight a space factor space of space straight f left parenthesis straight x right parenthesis space is
left parenthesis straight a right parenthesis space 2 straight x space minus space 1
left parenthesis straight b right parenthesis space 2 straight x space plus space 1
left parenthesis straight c right parenthesis space straight x space minus space 1
left parenthesis straight d right parenthesis space straight x space plus space 1 end style

Solution 1

begin mathsize 11px style If space straight f left parenthesis straight x right parenthesis space is space straight a space polynomial space and space straight f left parenthesis straight alpha right parenthesis space equals space 0 comma space then space left parenthesis straight x space minus straight alpha right parenthesis space is space straight a space factor space of space straight f left parenthesis straight x right parenthesis space or
vice space versa space if space left parenthesis straight x space minus space straight alpha right parenthesis space is space straight a space factor space of space straight f left parenthesis straight x right parenthesis space then space straight f left parenthesis straight alpha right parenthesis space equals space 0.
Now comma space straight f open parentheses fraction numerator negative 1 over denominator 2 end fraction close parentheses equals 0
So comma space at space straight x space equals space fraction numerator negative 1 over denominator 2 end fraction comma space straight f open parentheses straight x close parentheses space equals space 0
or space at space 2 straight x space equals space minus 1 comma space straight f open parentheses straight x close parentheses space equals space 0
or space at space 2 straight x space plus space 1 space equals space 0 comma space straight f left parenthesis straight x right parenthesis space equals space 0
rightwards double arrow open parentheses 2 straight x space plus space 1 close parentheses space is space straight a space factor space of space straight f open parentheses straight x close parentheses
Hence comma space straight c orrect space option space is space left parenthesis straight b right parenthesis. end style

Question 2

When x3 - 2x2 + ax - b is divided by x2 - 2x - 3, then remainder is x - 6. The values of a and b are respectively

(a) -2, -6

(b) 2 and -6

(c) -2 and 6

(d) 2 and 6

Solution 2

Let p(x) = x3 - 2x2 + ax - b, r(x) = x - 6 and q(x) = x2 - 2x - 3

Then q(x) is a factor of [p(x) - r(x)] 

{because if p(x) is divided by q(x), remainder is r(x). So, [p(x) - r(x)] will be exactly divided by q(x)}

Now, q(x) = x2 - 2x - 3 = (x - 3)(x + 1)

If q(x) is a factor of [p(x) - r(x)] then (x - 3) and (x + 1) are also factors of [p(x) - r(x)]

So, at x = 3 and x = -1, p(x) - r(x) will be zero.

Now p(3) - r(3) = 0

i.e. (3)3 - 2(3)2 + a(3) - b - (3 - 6) = 0

i.e. 27 - 18 + 3a - b + 3 = 0

i.e. 3a - b + 12 = 0  ....(1)

And, p(-1) - r(-1) = 0

i.e. (-1)3 - 2(-1)2 + a(-1) - b - (-1 - 6) = 0

i.e. -1 - 2 - a - b + 7 = 0

i.e -a - b + 4 = 0    ....(2)

Subtracting equation (2) from equation (1), we get

4a + 8 = 0

a = -2

From (2), -(-2) - b + 4 = 0

b = 6

Hence, correct option is (c).

Question 3

One factor of x4 + x2 - 20 is x2 + 5. The other factor is

(a) x2 - 4

(b) x - 4

(c) x2 - 5

(d) x + 2

Solution 3

x4 + x2 - 20

= x4 + 5x2 - 4x2 - 20

=x2(x2 + 5) - 4(x2 + 5)

= (x2 + 5)(x2 - 4)

So, other factor is x2 - 4.

Hence, correct option is (a).

Question 4

If (x - 1) is a factor of polynomial f(x) but not of g(x), then it must be a factor of

(a) f(x) g(x)

(b) -f(x) + g(x)

(c) f(x) - g(x)

(d) {f(x) + g(x)} g(x)

Solution 4

If space straight x space minus space 1 space is space straight a space factor space of space straight f left parenthesis straight x right parenthesis space then space definitely space straight f left parenthesis 1 right parenthesis space equals space 0
And comma space straight x space minus space 1 space is space not space straight a space factor space of space straight g left parenthesis straight x right parenthesis comma space then space straight g left parenthesis 1 right parenthesis space not equal to space 0 space space
So comma space at space straight x space equals space 1 space space
option space left parenthesis straight a right parenthesis space straight f left parenthesis 1 right parenthesis space straight g left parenthesis 1 right parenthesis space equals space 0 space cross times space straight g left parenthesis 1 right parenthesis equals space 0
option space left parenthesis straight b right parenthesis space minus straight f left parenthesis 1 right parenthesis space plus space straight g left parenthesis 1 right parenthesis space equals space 0 space plus space straight g left parenthesis 1 right parenthesis space equals space straight g left parenthesis 1 right parenthesis space not equal to space 0
option space left parenthesis straight c right parenthesis space straight f left parenthesis 1 right parenthesis space minus space straight g left parenthesis 1 right parenthesis space equals space 0 space minus space straight g left parenthesis 1 right parenthesis space equals space minus straight g left parenthesis 1 right parenthesis space not equal to space 0
option space left parenthesis straight d right parenthesis space left curly bracket straight f left parenthesis 1 right parenthesis space plus space straight g left parenthesis 1 right parenthesis right curly bracket space straight g left parenthesis 1 right parenthesis space equals space left curly bracket 0 space plus space straight g left parenthesis 1 right parenthesis right curly bracket space straight g left parenthesis 1 right parenthesis equals left curly bracket straight g left parenthesis 1 right parenthesis right curly bracket squared space not equal to space 0
So space at space straight x space equals space 1 space only comma space straight f left parenthesis straight x right parenthesis space straight g left parenthesis straight x right parenthesis space equals space 0
Thus comma space left parenthesis straight x space minus space 1 right parenthesis space is space factor space of space straight f left parenthesis straight x right parenthesis space straight g left parenthesis straight x right parenthesis space too.
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. space space

 

Question 5

(x + 1) is a factor of xn + 1 only if

(a) n is an odd integer

(b) n is an even integer

(c) n is a negative integer

(d) n is a positive integer

Solution 5

If x + 1 is a factor of xn + 1,

then, at x = -1, xn + 1 = 0

(-1)n + 1 = 0

(-1)n = -1

(-1)n will be equal to -1 if and only if n is an odd integer.

If n is even, then (-1)n = 1

So, n should be an odd integer.

Hence, correct option is (a).

Question 6

If x2 + x + 1 is a factor of the polynomial 3x3 + 8x2 + 8x + 3 + 5k, then the value of k is

(a) 0

(b) 2/5

(c) 5/2

(d) -1

Solution 6

begin mathsize 12px style Let space straight p open parentheses straight x close parentheses space equals space 3 straight x cubed space plus space 8 straight x squared space plus space 8 straight x space plus space 3 space plus space 5 straight k space and space straight q open parentheses straight x close parentheses space equals space straight x squared space plus space straight x space plus space 1
Now comma space if space straight q open parentheses straight x close parentheses space is space straight a space factor space of space straight p open parentheses straight x close parentheses comma space then space arranging space straight p open parentheses straight x close parentheses space in space order space to space have space straight q open parentheses straight x close parentheses space in space common comma
straight p open parentheses straight x close parentheses space equals space 3 straight x cubed space plus space 3 straight x squared space plus space 3 straight x space plus space 5 straight x squared space plus space 5 straight x space plus space 3 space plus space 2 space minus space 2 space plus space 5 straight k space space space space space space left square bracket adding space plus 2 comma space minus 2 space in space straight p left parenthesis straight x right parenthesis right square bracket
space space space space space space space space space equals 3 straight x left parenthesis straight x squared space plus space straight x space plus space 1 right parenthesis space plus space 5 left parenthesis straight x squared space plus space straight x space plus space 1 right parenthesis space plus space 5 straight k space minus space 2
straight p left parenthesis straight x right parenthesis space equals space left parenthesis straight x squared space plus space straight x space plus space 1 right parenthesis left parenthesis 3 straight x space plus space 5 right parenthesis space plus space 5 straight k space minus space 2 space space space space space.... left parenthesis 1 right parenthesis
From space equation space left parenthesis 1 right parenthesis comma space we space can space see space if space we space divide space straight p left parenthesis straight x right parenthesis space by space straight q left parenthesis straight x right parenthesis comma
then space quotient space will space be space left parenthesis 3 straight x space plus space 5 right parenthesis space and space remainder space will space be space left parenthesis 5 straight k space minus space 2 right parenthesis
But space straight q space left parenthesis straight x right parenthesis space is space straight a space factor space of space straight p left parenthesis straight x right parenthesis.
So comma space remainder space equals space 0 space rightwards double arrow space 5 straight k space minus space 2 space equals space 0 space space rightwards double arrow space straight k equals 2 over 5
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 7

If (3x - 1)7 = a7x7 + a6x6 + a5x5 + ... + a1x + a0, then a7 + a6 + a5 + ... + a1 + a0 =

(a) 0

(b) 1

(c) 128

(d) 64

Solution 7

Correct option (c)

(3x - 1)7 = a7x7 + a6x6 + ......... + a1x + a0  ....(1)

Putting x = 1 in equation (1), we have

[3(1) - 1]7 = a7 + a6 + ..... + a1 + a0

So, a7 + a6 + a5 + ..... + a1 + a0 = 2= 128

Hence, correct option is (c).

Question 8

begin mathsize 11px style If space both space straight x space minus space 2 space and space straight x space minus space 1 half space are space factors space of space px squared space plus space 5 straight x space plus space straight r comma space space then
left parenthesis straight a right parenthesis space straight p space equals space straight r
left parenthesis straight b right parenthesis space straight p space plus space straight r space equals space 0
left parenthesis straight c right parenthesis space 2 straight p space plus space straight r space equals space 0
left parenthesis straight d right parenthesis space straight p space plus space 2 straight r space equals space 0 end style

Solution 8

begin mathsize 12px style Let space straight f left parenthesis straight x right parenthesis space equals space px squared space plus space 5 straight x space plus space straight r
Now comma space if space straight x space minus space 2 space and space straight x space minus 1 half space are space factors space of space straight f left parenthesis straight x right parenthesis comma
then space at space straight x space equals space 2 space and space straight x equals 1 half comma space straight f left parenthesis straight x right parenthesis space equals space 0.
So comma space straight f left parenthesis 2 right parenthesis space equals space 0 comma space space straight f open parentheses 1 half close parentheses equals 0
rightwards double arrow straight p open parentheses 2 close parentheses squared space plus space 5 open parentheses 2 close parentheses space plus space straight r space equals space 0 space space space space space space space space space space space space space space space and space space space space space space space space straight p open parentheses 1 half close parentheses squared plus 5 open parentheses 1 half close parentheses plus straight r space equals space 0
rightwards double arrow 4 straight p space plus space straight r space plus space 10 space equals space 0 space space space space.... left parenthesis 1 right parenthesis space space space space space space space space and space space space space space space space space space 4 straight r space plus space straight p space plus space 10 space equals space 0 space space space space space space.... left parenthesis 2 right parenthesis
Subtracting space equation space left parenthesis 2 right parenthesis space from space left parenthesis 1 right parenthesis comma space we space have
3 straight p space minus space 3 straight r space equals space 0
rightwards double arrow straight p space equals space straight r
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 9

If x2 - 1 is a factor of ax4 + bx3 + cx2 + dx + e, then

(a)  a + c + e = b + d

(b) a + b + e = c + d

(c) a + b + c = d + e

(d) b + c + d = a + e

Solution 9

begin mathsize 11px style If space straight x squared space minus space 1 space is space factor space of space straight p left parenthesis straight x right parenthesis space equals space ax to the power of 4 space plus space bx cubed space plus space cx squared space plus space dx space plus space straight e comma
then space left parenthesis straight x space minus space 1 right parenthesis space and space left parenthesis straight x space plus space 1 right parenthesis space will space also space be space factors space of space straight p left parenthesis straight x right parenthesis.
straight B ecause space straight x squared space minus space 1 space equals space left parenthesis straight x space minus space 1 right parenthesis left parenthesis straight x space plus space 1 right parenthesis
straight T hen comma space at space straight x space equals space 1 space and space straight x space equals space minus 1 comma space straight p left parenthesis straight x right parenthesis space equals space 0
rightwards double arrow space straight p left parenthesis 1 right parenthesis space equals space 0 space and space straight p left parenthesis negative 1 right parenthesis space equals 0
rightwards double arrow space straight a space plus space straight b space plus space straight c space plus space straight d space plus space straight e space equals space 0 space space space.... left parenthesis 1 right parenthesis space space space space space space space space space space and space space space space space space straight a space minus space straight b space plus space straight c space minus space straight d space plus space straight e space equals space 0 space space space space.... left parenthesis 2 right parenthesis
Adding space equations space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma
2 straight a space plus space 2 straight c space plus space 2 straight e space equals space 0
rightwards double arrow space straight a space plus space straight c space plus space straight e space equals space 0 space space space space.... left parenthesis 3 right parenthesis
Substracting space equation space left parenthesis 2 right parenthesis space from space left parenthesis 1 right parenthesis
2 straight b space plus space 2 straight d space equals space 0
rightwards double arrow straight b space plus space straight d space equals space 0 space space space.... left parenthesis 4 right parenthesis
From space equations space left parenthesis 3 right parenthesis space and space left parenthesis 4 right parenthesis comma space we space get space
straight a space plus space straight c space plus space straight e space equals space straight b space plus space straight d
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

CBSE Class 9 Maths Homework Help

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