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Page No 167:

Question 1:

Now in each of the pairs of polynomials given below, check whether the first is a factor of the second:

(i) x + 1, x3 − 1

(ii) x − 1, x3 + 1

(iii) x + 1, x3 + 1

(iv) x2 − 1, x4 − 1

(v) x − 1, x4 − 1

(vi) x + 1, x4 − 1

(vii) x − 2, x2 − 5x + 1

(viii) x + 2, x2 + 5x + 6

(ix)

(x) 1.3x − 2.6, x2 − 5x + 6

Answer:

(i)

To find whether (x + 1) is a factor of (x3 1), we have to express (x3 1) in terms of (x + 1).

It can be done as follows:

(x3 1) = (x + 1)(ax2 + bx + c) + d

(x3 1) = ax3 + bx2 + cx + ax2 + bx + c + d

(x3 1) = ax3 + (b + a)x2 + (c + b)x + c + d

For this to hold, we need to equate the coefficients.

We have:

a = 1

b + a = 0

b = a = 1

c + b = 0

c = b = 1

c + d = 1

d = 1 c = 1 1 = 2

Thus, when (x3 1) is divided by (x + 1), the remainder is 2.

(x + 1) is not a factor of (x3 1).


 

(ii)

To find whether (x 1) is a factor of (x3 + 1), we have to express (x3 + 1) in terms of (x 1).

It can be done as follows:

(x3 + 1) = (x 1)(ax2 + bx + c) + d

(x3 + 1) = ax3 + bx2 + cx ax2 bx c + d

(x3 + 1) = ax3 + (b a)x2 + (c b)x c + d

For this to hold, we need to equate the coefficients.

We have:

a = 1

b a = 0

b = a = 1

c b = 0

c = b = 1

c + d = 1

d = 1 + c = 1 + 1 = 2

Thus, when (x3 + 1) is divided by (x 1), the remainder is 2.

(x 1) is not a factor of (x3 + 1).


 

(iii)

To find whether (x + 1) is a factor of (x3 + 1), we have to express (x3 + 1) in terms of (x + 1).

It can be done as follows:

(x3 + 1) = (x + 1)(ax2 + bx + c) + d

(x3 + 1) = ax3 + bx2 + cx + ax2 + bx + c + d

(x3 + 1) = ax3 + (b + a)x2 + (c + b)x + c + d

For this to hold, we need to equate the coefficients.

We have:

a = 1

b + a = 0

b = a = 1

c + b = 0

c = b = 1

c + d = 1

d = 1 c = 1 1 = 0

Thus, when (x3 + 1) is divided by (x +1), the remainder is 0.

(x + 1) is a factor of (x3 + 1).


 

(iv)

To find whether (x2 1) is a factor of (x4 1), we have to express (x4 1) in terms of (x2 1).

It can be done as follows:

(x4 1) = (x2 1)(ax2 + bx + c) + d

(x4 1) = ax4 + bx3 + cx2 ax2 bx c + d

(x4 1) = ax4 + bx3 + (c a)x2 bx c + d

For this to hold, we need to equate the coefficients.

We have:

a = 1

b = 0

c a = 0

c = a = 1

c + d = 1

d = 1 + c = 1 + 1 = 0

Thus, when (x4 1) is divided by (x2 1), the remainder is 0.

(x2 1) is a factor of (x4 1).


 

(v)

To find whether (x 1) is a factor of (x4 1), we have to express (x4 1) in terms of (x 1).

It can be done as follows:

(x4 1) = (x 1)(ax3 + bx2 + cx + d) + e

(x4 1) = ax4 + bx3 + cx2 + dx ax3 bx2 cx d + e

(x4 1) = ax4 + (b a)x3 + (c b)x2 + (d c)x d + e

For this to hold, we need to equate the coefficients.

We have:

a = 1

b a = 0

b = a = 1

c b = 0

c = b = 1

d c = 0

d = c = 1

d + e = 1

e = 1 + d = 1 + 1 = 0

Thus, when (x4 1) is divided by (x 1), the remainder is 0. 

(x 1) is a factor of (x4 1).


 

(vi)

To find whether (x + 1) is a factor of (x4 1), we have to express (x4 1) in terms of (x + 1).

It can be done as follows:

(x4 1) = (x + 1)(ax3 + bx2 + cx + d) + e

(x4 1) = ax4 + bx3 + cx2 + dx + ax3 + bx2 + cx + d + e

(x4 1) = ax4 + (b + a)x3 + (c + b)x2 + (d + c)x + d + e

For this to hold, we need to equate the coefficients.

We have:

a = 1

b + a = 0

b = a = 1

c + b = 0

c = b = 1

d + c = 0

d = c = 1

d + e = 1

e = 1 d = 1 + 1 = 0

Thus, when (x4 1) is divided by (x + 1), the remainder is 0.

(x + 1) is a factor of (x4 1).


 

(vii)

To find whether (x 2) is a factor of (x2 5x + 1), we have to express (x2 5x + 1) in terms of (x 2).

It can be done as follows:

(x2 5x + 1) = (x 2)(ax + b) + c

(x2 5x + 1) = ax2 + bx 2ax 2b + c 

(x2 5x + 1) = ax2 + (b 2a)x 2b + c

For this to hold, we need to equate the coefficients.

We have:

a = 1

b 2a = 5

b = 2a 5 = 2 5 = 3

2b + c = 1

c = 1 + 2b = 1 6 = 5

Thus, when (x2 5x + 1) is divided by (x 2), the remainder is 5.

(x 2) is not a factor of (x2 5x + 1).


 

(viii)

To find whether (x + 2) is a factor of (x2 + 5x + 6), we have to express (x2 + 5x + 6) in terms of (x + 2).

It can be done as follows:

(x2 + 5x + 6) = (x + 2)(ax + b) + c

(x2 + 5x + 6) = ax2 + bx + 2ax + 2b + c 

(x2 + 5x + 6) = ax2 + (b + 2a)x + 2b + c

For this to hold, we need to equate the coefficients.

We have:

a = 1

b + 2a = 5

b = 2a +5 = 2 + 5 = 3

2b + c = 6

c = 6 2b = 6 6 = 0

Thus, when (x2 + 5x + 6) is divided by (x + 2), the remainder is 0.

(x + 2) is a factor of (x2 + 5x + 6).


 

(ix)

To find whether is a factor of (x2 5x + 6), we have to express (x2 5x + 6) in terms of .

It can be done as follows:

(x2 5x + 6) = (ax + b) + c

(x2 5x + 6) =

(x2 5x + 6)

For this to hold, we need to equate the coefficients.

We have:

a = 3

= 5

b = 2a + 15 = 6 + 15 = 21

= 6

3c = 18 + 2b = 18 + 42 = 60

Thus, when (x2 5x + 6) is divided by the remainder is 60.

is not a factor of (x2 5x + 6).


 

(x)

To find whether (1.3x 2.6) is a factor of (x2 5x + 6), we have to express (x2 5x + 6) in terms of (1.3x + 2.6).

It can be done as follows:

(x2 5x + 6) = (1.3x + 2.6)(ax + b) + c

(x2 5x + 6) = 1.3ax2 + 1.3bx + 2.6ax + 2.6b + c 

(x2 5x + 6) = 1.3ax2 + (1.3b + 2.6a)x + 2.6b + c

For this to hold, we need to equate the coefficients.

We have:

1.3a = 1

a =

1.3b + 2.6a = 5

1.3b = 2.6a

= 2

= 7

b =

2.6b + c = 6

c = 6 2.6b = 6 + 14 = 20

Thus, when (x2 5x + 6) is divided by (1.3x 2.6), the remainder is 20.

(1.3x 2.6) is not a factor of (x2 5x + 6).



Page No 170:

Question 1:

Check whether each of the polynomials listed below is a factor of 3x32x2 − 3x + 2; if not, find the remainder.

(i) x − 1

(ii) 3x − 2

(iii) 2x − 3

(iv) x + 1

(v) 3x + 2

(vi) 2x + 3

Answer:

Factor theorem says that for the polynomial p(x) and for the number a, if we have p(a) = 0 then (x a) is a factor of p(x).

Given: Polynomial 3x3 2x2 3x + 2

(i)

Divisor = (x 1)

Putting x = 1 in the given polynomial: 

3(1)3 2(1)2 3(1) + 2

= 3 2 3 + 2

= 0

(x 1) is a factor of the polynomial 3x3 2x2 3x + 2.


 

(ii)

Divisor = (3x 2)

To check whether (3x 2) is a factor of 3x3 2x2 3x + 2, we have to convert (3x 2) in a form that is suitable for the application of the Factor theorem.

(3x 2) = 3

Putting x = in the given polynomial: 

3 2 3 + 2

Thus, is a factor of the given polynomial. 

(3x 2) is a factor of the polynomial 3x3 2x2 3x + 2.


 

(iii)

Divisor = (2x 3)

To check whether (2x 3) is a factor of 3x3 2x2 3x + 2, we have to convert (2x 3) in a form that is suitable for the application of the Factor theorem.

(2x 3) = 2

Putting x = in the given polynomial:

3 2 3 + 2

= 

=

=

Thus, is a factor of the given polynomial. 

(2x 3) is not a factor of the polynomial 3x3 2x2 3x + 2.


 

(iv)

Divisor = (x + 1)

To check whether (x + 1) is a factor of 3x3 2x2 3x + 2, we have to convert (x + 1) in a form that is suitable for the application of the Factor theorem.

(x + 1) = {x (1)} 

Putting x = 1 in the given polynomial:

3(1)3 2(1)2 3(1) + 2

= 3 2 + 3 + 2

= 0

(x + 1) is a factor of the polynomial 3x3 2x2 3x + 2.


 

(v)

Divisor = (3x + 2)

To check whether (3x + 2) is a factor of 3x3 2x2 3x + 2, we have to convert (3x + 2) in a form that is suitable for the application of the Factor theorem.

(3x + 2) = 3

= 3

Putting x = in the given polynomial:

3 2 3 + 2

=  

=

=

Thus, is a factor of the given polynomial.

(3x + 2) is not a factor of the expression 3x3 2x2 3x + 2.


 

(vi)

Divisor = (2x + 3)

To check whether (2x + 3) is a factor of 3x3 2x2 3x + 2, we have to convert (2x + 3) in a form that is suitable for the application of the Factor theorem.

(2x + 3) = 2

= 2

Putting x = in the given polynomial:

3 2 3 + 2

=

=

=

Thus, is not a factor of the given polynomial.

(2x + 3) is not a factor of the polynomial 3x3 2x2 3x + 2.



Page No 171:

Question 1:

What is the remainder on dividing the polynomial p(x) by ax + b? What is condition under which ax + b is a factor of the polynomial p(x)?

Answer:

Remainder theorem states that when a polynomial p(x) is divided by (xa), the remainder obtained is p(a).

Divisor = ax + b

ax + b =

=

The remainder obtained on dividing p(x) by is.

Thus, the remainder obtained on dividing p(x) by ax + b is.

Factor theorem says that for the polynomial p(x) and for the number a, if we have p(a) = 0, then (x a) is a factor of p(x).

For (ax + b) to be a factor of the polynomial p(x), p should be equal to 0.

Page No 171:

Question 2:

Is x − 1 a factor of x100 − 1? What about x + 1?

Answer:

Factor theorem says that for the polynomial p(x) and for the number a, if we have p(a) = 0, then (x a) is a factor of p(x).

Given: Polynomial x100 1.

Divisor = (x 1)

Putting x = 1 in the given polynomial: 

(1)100 1

= 1 − 1 

= 0

( x 1) is a factor of the polynomial x100 1.

Similarly, checking for (x + 1).

To check whether (x + 1) is a factor of x100 1, we have to convert (x + 1) in a form that is suitable for the application of the Factor theorem. 

(x + 1) = {x − (1)}

Putting x = 1 in the given polynomial: 

(1)100 1

= 1 − 1 

= 0

( x + 1) is also a factor of the polynomial x100 1.

Page No 171:

Question 3:

Is x − 1 a factor of x101 − 1? What about x + 1?

Answer:

Factor theorem says that for the polynomial p(x) and for the number a, if we have p(a) = 0, then (x a) is a factor of p(x).

Given: Polynomial x101 1

Divisor = (x 1)

Putting x = 1 in the given polynomial:

(1)101 1

= 0

( x 1) is a factor of the polynomial x101 1.

Similarly, checking for x + 1.

To check whether (x + 1) is a factor of x101 1, we have to convert (x + 1) in a form that is suitable for the application of the Factor theorem. 

(x + 1) = {x − (1)}

Putting x = 1 in the given polynomial:

(1)101 1

= 1 1

= 2

( x + 1) is not a factor of the polynomial x101 1.

Page No 171:

Question 4:

Prove that x − 1 is a factor of xn − 1 for every natural number n.

Answer:

Factor theorem says that for the polynomial p(x) and for the number a, if we have p(a) = 0, then (x a) is a factor of p(x).

Given: Polynomial xn 1

Divisor = (x 1)

Putting x = 1 in the given polynomial:

(1)n 1

= 1 1 { (1)n = 1, for all natural numbers n}

= 0 

( x 1) is a factor of the polynomial xn 1 for every natural number n.

Page No 171:

Question 5:

Prove that x + 1 is a factor of xn − 1 for every even number n.

Answer:

Factor theorem says that for the polynomial p(x) and for the number a, if we have p(a) = 0, then (x a) is a factor of p(x).

Given: Polynomial xn 1

Divisor = (x + 1)

To check whether (x + 1) is a factor of xn 1, we have to convert (x + 1) in a form that is suitable for the application of the Factor theorem. 

(x + 1) = {x − (1)}

Putting x = 1 in the given polynomial:

(1)n

= 1 − 1 { (1)n = 1, for every even number n}

= 0

( x + 1) is a factor of the polynomial xn 1 for every even number n.

Page No 171:

Question 6:

Prove that x + 1 is a factor of xn − 1 for every odd number n.

Answer:

Factor theorem says that for the polynomial p(x) and for the number a, if we have p(a) = 0, then (x a) is a factor of p(x).

Given: Polynomial xn 1

Divisor = (x + 1)

To check whether (x + 1) is a factor of xn 1, we have to convert (x + 1) in a form that is suitable for the application of the Factor theorem. 

(x + 1) = {(x − (1)}

Putting x = 1 in the given polynomial:

(1)n 1

= 1 − 1 ( (1)n = 1, for every odd number n)

= 2

( x + 1) is a factor of the polynomial xn 1 for every odd number n.

Page No 171:

Question 7:

What number added to 3x32x2 + 5x given a polynomial for which x − 1 is a factor?

Answer:

Let the number that has to be added to the polynomial 3x3 2x2 + 5x be c.

So, the polynomial is 3x3 2x2 + 5x + c.

Divisor = (x − 1) 

Factor theorem says that for the polynomial p(x) and for the number a, if we have p(a) = 0, then (x a) is a factor of p(x).

Thus, we must have:

3(1)3 − 2(1)2 + 5(1) + c = 0 for (x − 1) to be a factor of 3x3 2x2 + 5x + c.

3 − 2 + 5 + c = 0

c = 6

Thus, 6 must be added to 3x3 2x2 + 5x to obtain a polynomial which has (x 1) as a factor.

Page No 171:

Question 8:

What first degree polynomial added to 3x32x2 gives a polynomial for which both x − 1 and x + 1 are factors?

Answer:

Let the first degree polynomial that has to be added to the polynomial 3x3 2x2 be ax + b.

So, the polynomial is 3x3 2x2 + ax + b.

Divisor = (x 1)

Factor theorem says that for the polynomial p(x) and for the number a, if we have p(a) = 0, then (x a) is a factor of p(x).

Thus, we must have:

3(1)3 2(1)2 + a(1) + b = 0 for (x − 1) to be a factor of 3x3 2x2 + ax + b.

3 2 + a + b = 0

a + b = 1 …(1)

Similarly, checking for (x + 1).

(x + 1) = {x − (1)}

We must have: 

3(1)3 2(1)2 + a(1) + b = 0 for (x + 1) to be a factor of 3x3 2x2 + ax + b.

⇒ −3 2 a + b = 0

b a = 5 …(2)

Adding (1) and (2):

2b = 4

b = 2

a = 3 

Therefore, the first degree expression that has to be added to 3x3 2x2 is 3x + 2, so that the polynomial obtained has both (x + 1) and (x 1) as the factors.



Page No 174:

Question 1:

Write each of the polynomial listed below as a product of two first degree polynomials:

(i) 2x2 + 5x + 3

(ii) x2 + 2x − 1

(iii) x2 + 3x + 2

(iv) x2 − 2

(v) 4x2 + 20x + 25

(vi) x2x − 1

Answer:

According to the Factor theorem, to find the first degree factors of a polynomial, we need to find those numbers x, which make the given polynomial zero.

(i)

The first degree factors of the polynomial 2x2 + 5x + 3 can be obtained by finding out the solutions of the equation 2x2 + 5x + 3 = 0.

On comparing with the general quadratic equation, ax2 + bx + c = 0, we have:

a = 2, b = 5 and c = 3

x =

= 1 or

So, by the Factor theorem, (x + 1) and are the factors of the given polynomial.


 

Now, (x + 1) = x2 + +

This is not the original polynomial.

We can write:

(x + 1) =

 

(ii)

The first degree factors of the polynomial x2 + 2x 1 can be obtained by finding out the solutions of the equation x2 + 2x 1 = 0.

On comparing with the general quadratic equation, ax2 + bx + c = 0, we have:

a = 1, b = 2 and c = 1

x =

= or  

So, by the Factor theorem, (x + ) and (x + ) are the factors of the given polynomial.

(x + ) (x + ) = x2 + 2x 1


 

(iii)

The first degree factors of the polynomial x2 + 3x + 2 can be obtained by finding out the solutions of the equation x2 + 3x + 2 = 0.

On comparing with the general quadratic equation, ax2 + bx + c = 0, we have:

a = 1, b = 3 and c = 2

x =

=

=

=

= 1 or 2

So, by the Factor theorem, (x + 1) and (x + 2) are the factors of the given polynomial.

(x + 1) (x + 2) = x2 + 3x + 2


 

(iv)

The first degree factors of the polynomial x2 2 can be obtained by finding out the solutions of the equation x2 2 = 0.

x2 2 = 0

(x)(x + ) = 0 

x = or

So, by the Factor theorem, (x + ) and (x ) are the factors of the given polynomial.

(x + )(x ) = x2 2


 

(v)

The first degree factors of the polynomial 4x2 + 20x + 25 can be obtained by finding out the solutions of the equation 4x2 + 20x + 25 = 0.

On comparing with the general quadratic equation, ax2 + bx + c = 0, we have:

a = 4, b = 20 and c = 25

x =

=

=

=

So, by the Factor theorem, and are the factors of the given polynomial.

Now,

This is not the original polynomial.

We can write:


 

(vi)

The first degree factors of the polynomial x2 x 1 can be obtained by finding out the solutions of the equation x2 x 1 = 0.

On comparing with the general quadratic equation, ax2 + bx + c = 0, we have:

a = 1, b = 1 and c = 1

x =

=

=

= or

So, by the Factor theorem, and are the factors of the given polynomial.

Page No 174:

Question 2:

Prove that none of the polynomials listed below has first degree factors:

(i) x2 + x + 1

(ii) x4 + 1

(iii) x2x + 1

(iv) x4 + x2 + 1

Answer:

According to the Factor theorem, to find the first degree factors of a polynomial, we need to find those numbers x, which make the given polynomial zero.

(i)

The first degree factors of the polynomial x2 + x + 1 can be obtained by finding out the solutions of the equation x2 + x + 1 = 0.

On comparing with the general quadratic equation, ax2 + bx + c = 0, we have:

a = 1, b = 1 and c = 1

x =

As we have obtained a negative value for the discriminant, the polynomial x2 + x + 1 has no first degree factors.


 

(ii)

The first degree factors of the polynomial x4 + 1 can be obtained by finding out the solutions of the equation x4 + 1 = 0.

Let x2 = p

So, the equation becomes p2 + 1 = 0. 

p2 = 1

p =

As the value of p is not real, the value of x is also not real.

Thus, the polynomial x4 +1 has no first degree factors.


 

(iii)

The first degree factors of the polynomial x2 x + 1 can be obtained by finding out the solutions of the equation x2 x + 1 = 0.

On comparing with the general quadratic equation, ax2 + bx + c = 0, we have:

a = 1, b = 1 and c = 1

x =

As we have obtained a negative value for the discriminant, the polynomial x2 x + 1 has no first degree factors.


 

(iv)

The first degree factors of the polynomial x4 + x2 + 1 can be obtained by finding out the solutions of the equation x4 + x2 + 1 = 0.

Let x2 = p

p2 + p + 1 = 0

On comparing with the general quadratic equation, ax2 + bx + c = 0, we have:

a = 1, b = 1 and c = 1

p =

As we have obtained a negative value for the discriminant, the polynomial x4 + x2 + 1 has no first degree factors.



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