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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 − 5
= −2 −5
= −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 3x3 − 2x2 − 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 (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 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 { (−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 3x3 − 2x2 + 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 3x3 − 2x2 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) x2 − 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 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) x2 − x + 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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