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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, *x*^{3} − 1

(ii)* x *− 1, *x*^{3} + 1

(iii) *x* + 1, *x*^{3} + 1

(iv) *x*^{2} − 1, *x*^{4} − 1

(v) *x* − 1, *x*^{4} − 1

(vi) *x* + 1, *x*^{4} − 1

(vii) *x* − 2, *x*^{2} − 5*x* + 1

(viii) *x* + 2, *x*^{2} + 5*x* + 6

(ix)

(x) 1.3*x* − 2.6, *x*^{2} − 5*x* + 6

#### Answer:

(i)

To find whether (*x* + 1) is a factor of (*x*^{3}^{ }− 1), we have to express (*x*^{3}^{ }− 1) in terms of (*x *+ 1).

It can be done as follows:

(*x*^{3}^{ }− 1) = (*x *+ 1)(*ax*^{2} + *bx* + *c*) + *d*

⇒(*x*^{3}^{ }− 1) = *ax*^{3} + *bx*^{2} + *cx* + *ax*^{2} + *bx* + *c* + *d*

⇒(*x*^{3 }− 1) =* ax*^{3} + (*b *+ *a*)*x*^{2} + (*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 (*x*^{3 }− 1) is divided by (*x* + 1), the remainder is −2.

∴ (*x* + 1) is not a factor of (*x*^{3 }− 1).

(ii)

To find whether (*x* − 1) is a factor of (*x*^{3 }+ 1), we have to express (*x*^{3 }+ 1) in terms of (*x *− 1).

It can be done as follows:

(*x*^{3 }+ 1) = (*x *− 1)(*ax*^{2} + *bx* + *c*) + *d*

⇒ (*x*^{3 }+ 1) = *ax*^{3} + *bx*^{2} + *cx* − *ax*^{2} − *bx* − *c* + *d*

⇒ (*x*^{3 }+ 1) =* ax*^{3} + (*b *− *a*)*x*^{2} + (*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 (*x*^{3 }+ 1) is divided by (*x* −1), the remainder is 2.

∴ (*x* −1) is not a factor of (*x*^{3 }+ 1).

(iii)

To find whether (*x* + 1) is a factor of (*x*^{3 }+ 1), we have to express (*x*^{3 }+ 1) in terms of (*x *+ 1).

It can be done as follows:

(*x*^{3 }+ 1) = (*x *+ 1)(*ax*^{2} + *bx* + *c*) + *d*

⇒ (*x*^{3 }+ 1) = *ax*^{3} + *bx*^{2} + *cx* + *ax*^{2} + *bx* + *c* + *d*

⇒ (*x*^{3 }+ 1) =* ax*^{3} + (*b *+ *a*)*x*^{2} + (*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 (*x*^{3 }+ 1) is divided by (*x* +1), the remainder is 0.

∴ (*x* + 1) is a factor of (*x*^{3 }+ 1).

(iv)

To find whether (*x*^{2} − 1) is a factor of (*x*^{4 }− 1), we have to express (*x*^{4 }− 1) in terms of (*x*^{2}* *− 1).

It can be done as follows:

(*x*^{4 }− 1) = (*x*^{2}* *− 1)(*ax*^{2} + *bx* + *c*) + *d*

⇒(*x*^{4 }− 1) = *ax*^{4} + *bx*^{3} + *cx*^{2} − *ax*^{2} − *bx* − *c* + *d*

⇒ (*x*^{4 }− 1) =* ax*^{4} + *bx*^{3} +* *(*c *− *a*)*x*^{2} * *− *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 (*x*^{4 }− 1) is divided by (*x*^{2} − 1), the remainder is 0.

∴ (*x*^{2 }− 1) is a factor of (*x*^{4 }− 1).

(v)

To find whether (*x* − 1) is a factor of (*x*^{4 }− 1), we have to express (*x*^{4 }− 1) in terms of (*x *− 1).

It can be done as follows:

(*x*^{4 }− 1) = (*x *− 1)(*ax*^{3} + *bx*^{2} + *cx *+ *d*) + *e*

⇒ (*x*^{4 }− 1) = *ax*^{4} + *bx*^{3} + *cx*^{2} + *dx *− *ax*^{3} − *bx*^{2} − *cx *− *d* + *e*

⇒ (*x*^{4 }− 1) =* ax*^{4} + (*b **−** a*)*x*^{3} + (*c **−** b*)*x*^{2} + (*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 (*x*^{4 }− 1) is divided by (*x *− 1), the remainder is 0.

∴ (*x *− 1) is a factor of (*x*^{4 }− 1).

(vi)

To find whether (*x* + 1) is a factor of (*x*^{4}− 1), we have to express (*x*^{4 }− 1) in terms of (*x *+ 1).

It can be done as follows:

(*x*^{4 }−1) = (*x *+ 1)(*ax*^{3} + *bx*^{2} + *cx *+ *d*) + *e*

⇒ (*x*^{4 }− 1) = *ax*^{4} + *bx*^{3} + *cx*^{2} + *dx *+ *ax*^{3} + *bx*^{2} + *cx *+ *d* + *e*

⇒ (*x*^{4 }− 1) = *ax*^{4} + (*b + a*)*x*^{3} + (*c + b*)*x*^{2} + (*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 (*x*^{4 }− 1) is divided by (*x* + 1), the remainder is 0.

∴ (*x *+ 1) is a factor of (*x*^{4 }− 1).

(vii)

To find whether (*x* − 2) is a factor of (*x*^{2 }− 5*x* + 1), we have to express (*x*^{2 }− 5*x* + 1) in terms of (*x *− 2).

It can be done as follows:

(*x*^{2 }− 5*x* + 1) = (*x *− 2)(*ax* + *b*) + *c*

⇒ (*x*^{2 }− 5*x* + 1) = *ax*^{2}* *+ *bx *− 2*ax* − 2*b* + *c*

⇒ (*x*^{2 }− 5*x* + 1) =* ax*^{2} + (*b **−** *2*a*)*x* − 2*b* + *c*

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

We have:

*a* = 1

*b* − 2*a *= −5

⇒ *b* = 2*a* − 5 = 2 − 5 = −3

−2*b *+ *c* = 1

⇒ *c* = 1 + 2*b* = 1 − 6 = −5

Thus, when (*x*^{2 }− 5*x* + 1) is divided by (*x* − 2), the remainder is −5.

∴ (*x* − 2) is not a factor of (*x*^{2 }− 5*x* + 1).

(viii)

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

It can be done as follows:

(*x*^{2 }+ 5*x* + 6) = (*x *+ 2)(*ax* + *b*) + *c*

⇒ (*x*^{2 }+ 5*x* + 6) = *ax*^{2}* *+ *bx *+ 2*ax* + 2*b* + *c*

⇒ (*x*^{2 }+ 5*x* + 6) =* ax*^{2} + (*b *+ 2*a*)*x* + 2*b* + *c*

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

We have:

*a* = 1

*b* + 2*a *= 5

⇒ *b* = −2*a* +5 = −2 + 5 = 3

2*b* + *c* = 6

⇒ *c* = 6 − 2*b* = 6 − 6 = 0

Thus, when (*x*^{2 }+ 5*x* + 6) is divided by (*x* + 2), the remainder is 0.

∴ (*x* + 2) is a factor of (*x*^{2 }+ 5*x* + 6).

(ix)

To find whether * * is a factor of (*x*^{2 }− 5*x* + 6), we have to express (*x*^{2 }− 5*x* + 6) in terms of * *.

It can be done as follows:

(*x*^{2 }− 5*x* + 6) = * *(*ax* + *b*) + *c*

⇒ (*x*^{2 }− 5*x* + 6) =

⇒ (*x*^{2 }− 5*x* + 6)

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

We have:

⇒ *a* = 3

= 5

⇒ *b* = 2*a* + 15 = 6 + 15 = 21

= 6

⇒ 3*c* = 18 + 2*b* = 18 + 42 = 60

Thus, when (*x*^{2 }− 5*x* + 6) is divided by * * the remainder is 60.

∴ is not a factor of (*x*^{2 }− 5*x* + 6).

(x)

To find whether (1.3*x* − 2.6) is a factor of (*x*^{2 }− 5*x* + 6), we have to express (*x*^{2 }−5*x* + 6) in terms of (1.3*x *+ 2.6).

It can be done as follows:

(*x*^{2 }− 5*x* + 6) = (1.3*x *+ 2.6)(*ax* + *b*) + *c*

⇒ (*x*^{2 }− 5*x* + 6) = 1.3*ax*^{2}* *+ 1.3*bx *+ 2.6*ax* + 2.6*b* + *c*

⇒ (*x*^{2 }− 5*x* + 6) =* *1.3*ax*^{2}* *+ (1.3*b *+ 2.6*a*)*x* + 2.6*b* + *c*

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

We have:

1.3*a* = 1

⇒ *a* =

1.3*b *+ 2.6*a *= −5

⇒ 1.3*b* = −2.6*a* − 5

= −2 −5

= −7

⇒ *b* =

2.6*b* + *c* = 6

⇒ *c* = 6 − 2.6*b* = 6 + 14 = 20

Thus, when (*x*^{2 }− 5*x* + 6) is divided by (1.3*x* − 2.6), the remainder is 20.

∴ (1.3*x* − 2.6) is not a factor of (*x*^{2 }− 5*x* + 6).

#### Page No 170:

#### Question 1:

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

(i) *x* − 1

(ii) 3*x* − 2

(iii) 2*x* − 3

(iv) *x* + 1

(v) 3*x* + 2

(vi) 2*x* + 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** **3*x*^{3}^{ }− 2*x*^{2}^{ }− 3*x** *+ 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 3*x*^{3}^{ }− 2*x*^{2}^{ }− 3*x** *+ 2.

(ii)

Divisor = (3*x* − 2)

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

(3*x* −2) = 3

Putting *x* = in the given polynomial:

3^{ }− 2^{ }− 3* *+ 2

Thus, _{ }is a factor of the given polynomial.

∴(3*x* −2) is a factor of the polynomial 3*x*^{3 }− 2*x*^{2}^{ }− 3*x** *+ 2.

(iii)

Divisor = (2*x* − 3)

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

(2*x* − 3) = 2

Putting *x* = in the given polynomial:

3^{ }− 2^{ }− 3* *+ 2

=

=

=

Thus, _{ }is a factor of the given polynomial.

∴(2*x* − 3) is not a factor of the polynomial 3*x*^{3}^{ }− 2*x*^{2}^{ }− 3*x** *+ 2.

(iv)

Divisor = (*x* + 1)

To check whether (*x* + 1) is a factor of 3*x*^{3 }− 2*x*^{2 }− 3*x *+ 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 3*x*^{3 }− 2*x*^{2 }− 3*x *+ 2.

(v)

Divisor = (3*x* + 2)

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

(3*x* + 2) = 3

= 3

Putting *x* = in the given polynomial:

3^{ }− 2^{ }− 3* *+ 2

=

=

=

_{Thus, } _{ is a factor of the given polynomial.}

∴(3*x* + 2) is not a factor of the expression 3*x*^{3 }− 2*x*^{2 }− 3*x *+ 2.

(vi)

Divisor = (2*x* + 3)

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

(2*x* + 3) = 2

= 2

Putting *x* = in the given polynomial:

3^{ }− 2^{ }− 3* *+ 2

=

=

=

_{Thus, } _{is not a factor of the given polynomial.}

∴(2*x* + 3) is not a factor of the polynomial 3*x*^{3 }− 2*x*^{2 }− 3*x *+ 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 (*x* − *a*), 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 *x*^{100} − 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 *x*^{100}^{ }− 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 *x*^{100}^{ }− 1.

Similarly, checking for (*x *+ 1).

To check whether (*x* + 1) is a factor of *x*^{100}^{ }− 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 *x*^{100}^{ }− 1.

#### Page No 171:

#### Question 3:

Is *x* − 1 a factor of *x*^{101} − 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 *x*^{101}^{ }− 1

Divisor = (*x* − 1)

Putting *x* = 1 in the given polynomial:

(1)^{101 }− 1

= 0

∴(* x* − 1) is a factor of the polynomial *x*^{101}^{ }− 1.

Similarly, checking for *x *+ 1.

To check whether (*x* + 1) is a factor of *x*^{101}^{ }− 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 *x*^{101}^{ }− 1.

#### Page No 171:

#### Question 4:

Prove that *x* − 1 is a factor of *x*^{n} − 1 for every natural number *n*.

#### Answer:

*p*(*x*) and for the number *a*, if we have *p*(*a*) = 0, then (*x *− *a*)* *is a factor of *p*(*x*).

Given: Polynomial *x*^{n}^{ }− 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 *x*^{n}^{ }− 1 for every natural number *n*.

#### Page No 171:

#### Question 5:

Prove that *x* + 1 is a factor of *x*^{n} − 1 for every even number *n*.

#### Answer:

*p*(*x*) and for the number *a*, if we have *p*(*a*) = 0, then (*x *− *a*)* *is a factor of *p*(*x*).

Given: Polynomial *x*^{n}^{ }− 1

Divisor = (*x* + 1)

To check whether (*x* + 1) is a factor of *x*^{n}^{ }− 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 even number *n*}

= 0

∴(* x* + 1) is a factor of the polynomial *x*^{n}^{ }− 1 for every even number *n*.

#### Page No 171:

#### Question 6:

Prove that *x* + 1 is a factor of *x*^{n} − 1 for every odd number *n*.

#### Answer:

*p*(*x*) and for the number *a*, if we have *p*(*a*) = 0, then (*x *− *a*)* *is a factor of *p*(*x*).

Given: Polynomial *x*^{n}^{ }− 1

Divisor = (*x* + 1)

To check whether (*x* + 1) is a factor of *x*^{n}^{ }− 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 *x*^{n}^{ }− 1 for every odd number *n*.

#### Page No 171:

#### Question 7:

What number added to 3*x*^{3} − 2*x*^{2} + 5*x* given a polynomial for which *x* − 1 is a factor?

#### Answer:

Let the number that has to be added to the polynomial 3*x*^{3}− 2*x*^{2} + 5*x* be *c*.

So, the polynomial is 3*x*^{3}− 2*x*^{2} + 5*x* + *c*.

Divisor = (*x* − 1)

*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 3*x*^{3}− 2*x*^{2} + 5*x* + *c*.

⇒ 3 − 2 + 5 + *c* = 0

⇒ *c* = −6

Thus, −6 must be added to 3*x*^{3}− 2*x*^{2} + 5*x* to obtain a polynomial which has (*x *− 1) as a factor.

#### Page No 171:

#### Question 8:

What first degree polynomial added to 3*x*^{3} − 2*x*^{2} 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 3*x*^{3}− 2*x*^{2} be *ax* + *b*.

So, the polynomial is 3*x*^{3}− 2*x*^{2} + *ax* + *b*.

Divisor = (*x* − 1)

*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 3*x*^{3}− 2*x*^{2} + *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 3*x*^{3}− 2*x*^{2} + *ax* + *b*.

⇒ −3 − 2 −* a *+ *b* = 0

⇒ *b *− *a* = 5 …(2)

Adding (1) and (2):

2*b* = 4

⇒ *b* = 2

*a* = −3

Therefore, the first degree expression that has to be added to 3*x*^{3}− 2*x*^{2} is −3*x* + 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) 2*x*^{2} + 5*x* + 3

(ii) *x*^{2} + 2*x* − 1

(iii) *x*^{2} + 3*x* + 2

(iv) *x*^{2} − 2

(v) 4*x*^{2} + 20*x* + 25

(vi) *x*^{2} − *x* − 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 2*x*^{2} + 5*x* + 3 can be obtained by finding out the solutions of the equation 2*x*^{2} + 5*x* + 3 = 0.

_{On comparing with the general quadratic equation, }_{ax}^{2}_{ + 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) = *x*^{2} + +

This is not the original polynomial.

We can write:

⇒ (*x *+ 1) =

⇒

⇒

(ii)

The first degree factors of the polynomial *x*^{2} + 2*x* − 1 can be obtained by finding out the solutions of the equation *x*^{2} + 2*x* − 1 = 0.

_{On comparing with the general quadratic equation, }_{ax}^{2}_{ + 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 *+ ) = *x*^{2} + 2*x* − 1

(iii)

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

_{On comparing with the general quadratic equation, }_{ax}^{2}_{ + 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) = *x*^{2} + 3*x* + 2

(iv)

The first degree factors of the polynomial *x*^{2} − 2 can be obtained by finding out the solutions of the equation *x*^{2} − 2 = 0.

*x*^{2} − 2 = 0

⇒ (*x* − )(*x* + ) = 0

⇒ *x* = or −

_{So, by the Factor theorem, }(*x *+ ) _{and }(*x *− ) _{are the factors of the given polynomial.}

⇒ (*x *+ )(*x *− ) = *x*^{2} − 2

(v)

The first degree factors of the polynomial 4*x*^{2} + 20*x* + 25 can be obtained by finding out the solutions of the equation 4*x*^{2} + 20*x* + 25 = 0.

_{On comparing with the general quadratic equation, }_{ax}^{2}_{ + 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 *x*^{2} − *x* − 1 can be obtained by finding out the solutions of the equation *x*^{2} − *x* − 1 = 0.

_{On comparing with the general quadratic equation, }_{ax}^{2}_{ + 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) *x*^{2} + *x* + 1

(ii) *x*^{4} + 1

(iii) *x*^{2} − *x* + 1

(iv) *x*^{4} + *x*^{2} + 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 *x*^{2} + *x* + 1 can be obtained by finding out the solutions of the equation *x*^{2} + *x* + 1 = 0.

_{On comparing with the general quadratic equation, }_{ax}^{2}_{ + 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 *x*^{2} + *x* + 1 has no first degree factors.

(ii)

The first degree factors of the polynomial *x*^{4} + 1 can be obtained by finding out the solutions of the equation *x*^{4} + 1 = 0.

Let *x*^{2}^{ }= *p*

So, the equation becomes *p*^{2} + 1 = 0.

⇒ *p*^{2} = −1

⇒ *p* =

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

Thus, the polynomial *x*^{4} +1 has no first degree factors.

(iii)

The first degree factors of the polynomial *x*^{2} − *x* + 1 can be obtained by finding out the solutions of the equation *x*^{2} − *x* + 1 = 0.

_{On comparing with the general quadratic equation, }_{ax}^{2}_{ + 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 *x*^{2} − *x* + 1 has no first degree factors.

(iv)

The first degree factors of the polynomial *x*^{4} + *x*^{2} + 1 can be obtained by finding out the solutions of the equation *x*^{4} + *x*^{2} + 1 = 0.

Let *x*^{2}^{ }= *p*

⇒ *p*^{2} + *p* + 1 = 0

_{On comparing with the general quadratic equation, }_{ax}^{2}_{ + 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 *x*^{4} + *x*^{2} + 1 has no first degree factors.

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