The availability of OP Malhotra Class 10 Solutions Chapter 7 Factor Theorem – Factorization Ex 7 encourages students to tackle difficult exercises.
S Chand Class 10 ICSE Maths Solutions Chapter 7 Factor Theorem – Factorization Ex 7
Question 1.
Find the remainder when the expression
(i) 3x³ + 8x² – 6x + 1 is divided by x + 3.
(ii) 5x³ – 8x² + 3x – 4 is divided by x – 1.
(iii) x³ + 3x² – 1 is divided by 3x + 2.
(iv) 4x³ – 12x² + 14x – 3 is divided by 2x – 1.
Solution:
(i) Let f(x) = 3x³ + 8x² – 6x + 1 and x + 3 = 0 ⇒ x = – 3
∴ Remainder = f(- 3) = 3 (- 3)³ + 8 (- 3)² – 6 (- 3) + 1
= – 81 + 72 + 18 + 1 = – 81 + 91 = 10
(ii) Let f (x) = 5x³ – 8x² + 3x – 4 and x – 1 = 0 ⇒ x = 1
∴ Remainder = f(1) = 5 (1)³ – 8(1)² + 3 x 1 – 4
= 5 – 8 + 3 – 4 = 8 – 12 = – 4
(iii) Let f(x) = x³ + 3x² – 1
and 3x + 2 = 0 ⇒ 3x = – 2 ⇒ x = \(\frac { -2 }{ 3 }\)
∴ Remainder = f(\(\frac { -2 }{ 3 }\))
= \(\left(\frac{-2}{3}\right)^3+3\left(\frac{-2}{3}\right)^2\) – 1
= \(\frac{-8}{27}+3 \times \frac{4}{9}\) – 1
= \(\frac { -8 }{ 27 }\) + \(\frac { 4 }{ 3 }\) – 1
= \(\frac{-8+36-27}{27}=\frac{36-35}{27}=\frac{1}{27}\)
(iv) Let f(x) = 4x³ – 12x² + 14x – 3
and 2x – 1 = 0 ⇒ 2x = 1 ⇒ x = \(\frac { 1 }{ 2 }\)
∴ Remainder = f\(\frac { 1 }{ 2 }\)
= 4 x (\(\frac { 1 }{ 2 }\))³ – 12(\(\frac { 1 }{ 2 }\))² + 14(\(\frac { 1 }{ 2 }\)) – 3
= 4 x \(\frac { 1 }{ 8 }\) – 12 x \(\frac { 1 }{ 4 }\) + 14 x \(\frac { 1 }{ 2 }\) – 3
= \(\frac { 1 }{ 2 }\) – 3 + 7 – 3
= 7\(\frac { 1 }{ 2 }\) – 6 = 1\(\frac { 1 }{ 2 }\) = \(\frac { 3 }{ 2 }\)
Question 2.
When x³ + 3x² – kx + 4 is divided by x – 2, the remainder is k. Find the value of the constant k.
Solution:
Let f(x) = x³ + 3x² – kx + 4
and x – 2 = 0, ⇒ x = 2
∴ Remainder =/(2)
= (2)³ + 3 (2)² – k (2) + 4
= 8 + 12 – 2A + 4 = 24 – 2k
∴ Remainder = k
∴ 24 – 2k = k ⇒ 2k = 24
∴ k = 8
Question 3.
Find the value of a if the division of ax³ + 9x² + 4x – 10 by x + 3 leaves a remainder 5.
Solution:
Let f (x) = ax³ + 9x² + 4x – 10 and x + 3 = 0 ⇒ x = – 3
∴ Remainder = f (- 3)
= a (-3)³ + 9 (-3)² + 4 (-3) – 10
= – 27a + 81 – 12 – 10
= – 27a + 81 – 22
= – 27a + 59
But remainder = 5
– 27a + 59 = 5 ⇒ – 27a = 5 – 59 = – 54
⇒ a = \(\frac { 1 }{ 2 }\)
∴ a = 2
Question 4.
If the polynomials ax³ + 4x² + 3x – 4 and x³ – 4x – a leave the same remainder when divided by x – 2, find the value of a.
Solution:
Let f(x) = ax³ + 4x² + 3x – 4
and q (x) = x³ – 4x – a
and x – 2 = 0 ⇒ x = 2
∴ Remainder f (2) = q (2)
Now f(2) = a (2)³ + 4 (2)² + 3x² – 4
= 8 a + 16 + 6 – 4
= 8o + 18
and q (2) = (2)³ – 4x² – a
= 8 – 8 – a = – a
∴ 8a + 18 = – a
⇒ 8a + a = – 18
⇒ 9a = – 18
⇒ a = \(\frac { -18 }{ 9 }\) = – 2
∴ a = – 2
Question 5.
Use factor theorem in each of the following to find whether g (x) is a factor f(x) or not :
(i) f(x) = x³ – 6x² + 11x – 6; g (x) = x – 3
(ii) f(x) = 2x³ – 9x² + x + 12 ; g (x) = x + 1
(iii) f(x) = 7x² – 2\(\sqrt{8}\) x – 6 ; g (x) = x – \(\sqrt{2}\)
(iv) f(x) = 3x³ + x² – 20x +12 ; g (x) = 3x – 2.
Solution:
(i) f(x) = x³ – 6x² + 11x – 6
Let x – 3 = 0, then x = 3
Remainder f(3) = (3)³ – 6 (3)² + 11 (3) – 6
= 27 – 54 + 33 – 6 = 60 – 60 = 0
∵ Remainder is zero
∴ x – 3 is a factor of f (x)
(ii) f (x) = 2x³ – 9x² + x + 12
Let x + 1 = 0 then x = – 1
∴ Remainder = f (- 1) = 2 (-1)³ – 9 (-1)² + (-1) + 12
= 2 x (-1) – 9 x 1 +(-1)+ 12
= – 2 – 9 – 1 + 12
= – 12 + 12 = 0
∵ Remainder = 0
∴ (x + 1) is a factor of f (x)
(iii) f(x) = 7x² – 2\(\sqrt{8}\) x – 6
Let x – \(\sqrt{2}\) = then x = \(\sqrt{2}\)
Remainder = f (\(\sqrt{2}\)) = 7 (\(\sqrt{2}\))² – 2\(\sqrt{8}\) x \(\sqrt{2}\) – 6
= 7 x 2 – 2 x \(\sqrt{16}\) – 6
= 14 – 2 x 4 – 6
= 14 – 8 – 6
= 14 – 14 = 0
∵ Remainder = 0
∴ x – \(\sqrt{2}\) is a factor of f(x)
(iv) f(x) = 3x³ + x² – 20x + 12
Let 3x – 2 = 0, then 3x = 2 ⇒ x = \(\frac { 2 }{ 3 }\)
Question 6.
Find the value of a if x³ + ax + 2a – 2 is exactly divisible by x + 1.
Solution:
Let f(x) = x³ + ax + 2a – 2
and x + 1 = 0 then x = – 1
∴ Remainder = f(-1) = (-1)³ + a x (-1) + 2a – 2
= – 1 – a + 2a – 2 = – 3 + a
∵ x + 1 is a factor of f (x)
∴ Remainder = 0
⇒ – 3 + a = 0 ⇒ a = 3
Question 7.
Find the values of a and b so that the expression x³ + 10x² + ax + b is exactly divisible by x – 1 as well as x – 2.
Solution:
Let f(x) = x³ + 10x² + ax + b
and let x – 1 = 0, then x = 1
∴ Remainder f(1) = (1)³ + 10 (1)² + a x 1 + b
= 1 + 10 + a + b
= 11 + a + b
∵ x – 1 is the factor of f (x)
∴ Remainder = 0
⇒ 11 + a + b = 0 ⇒ a + b = – 11 … (i)
Again x – 2 = 0, then x = 2
∴ f (2) = (2)³ + 10 (2)² + a x 2 + b
= 8 + 10 x 4 + 2a + b
= 8 + 40 + 2a + b
= 48 + 2 a + b
∵ x – 2 is a factor of f (x)
∴ Remainder = 0
∴ 48 + 2a + b = 0 ⇒ 2a + b = – 48 … (ii)
Subtracting (i) from (ii) a = – 37
But a + b = – 11 ⇒ – 37 + b = – 11
⇒ b = – 11 + 37 = 26
∴ a = – 37 and b = 26
Question 8.
If (x – 2) is a factor of x² + ax – 6 = 0 and x² – 9x + b = 0, find the values of a and b.
Solution:
Let f(x) = x² + ax – 6 and q (x) = x² – 9x + b
Now x – 2 = 0 ⇒ x = 2
∴ f(2) = (2)² + ax² – 6 = 4 + 2a – 6
= 2a – 2
∵ x – 2 is a factor of f (x)
∴ f(2) = 0 ⇒ 2a – 2 = 0
⇒ 2a = 2 ⇒ a = \(\frac { 1 }{ 2 }\) = 1
∴ a = 1
∴ Again ∵ x – 2 = 0, then x = 2
∴ q (2) = (2)² – 9 x 2 + b
= 4 – 18 + b = – 14 + b
∵ (x – 2) is a factor of q (x)
∴ – 14 + b = 0 ⇒ b = 14 (∵ Remainder = 0)
Hence a = 1, b = 14
Question 9.
If both x – 2 and x – \(\frac { 1 }{ 2 }\) are factors of px² + 5x + r, show that p = r.
Solution:
Let f (x) = px² + 5x + r
and let x – 2 = 0 then x = 2
∴ f(x) = p (2)² + 5 x 2 + r = 4p + 10 + r = 4p + r + 10
∴ x – 2 is its factor
∴ f(2) = 0 ⇒ 4p + r + 10 = 0
⇒ 4p + r = – 10 … (i)
Again let x – \(\frac { 1 }{ 2 }\) = 0 then x = \(\frac { 1 }{ 2 }\)
∴ f(\(\frac { 1 }{ 2 }\)) = p(\(\frac { 1 }{ 2 }\))² + 5 x \(\frac { 1 }{ 2 }\) + r
= \(\frac { p }{ 4 }\) + \(\frac { 5 }{ 2 }\) + r
∵ x – \(\frac { 1 }{ 2 }\) is its factor
∴ f(\(\frac { 1 }{ 2 }\)) = 0 ⇒ \(\frac { p }{ 4 }\) + \(\frac { 5 }{ 2 }\) + r = 0 … (ii)
⇒ p + 4r + 10 = 0 ⇒ p + 4r = – 10 … (ii)
Adding (i) and (ii)
5p + 5r = – 20 ⇒ p + r = – 4 … (iii)
and subtracting
3p – 3r = 0 ⇒ p – r = 0 … (iv)
Again adding (iii) and (iv)
2p = – 4 ⇒ p = – 2
and subtracting
2r = – 4 ⇒ r = \(\frac {-4}{2}\) = – 2
∴ p = – 2, r = – 2
∴ p = r
Hence proved.
Question 10.
If x + ax² + bx + 6 has (x – 2) as a factor and leaves a remainder 3 when divided by (x – 3), find the values of a and b.
Solution:
Let f (x) = x³ + ax² + bx + 6
Let x – 2 = 0, then x = 2
∴ f(x) = (2)³ + a (2)² + b(2) + 6
= 8 + 4a + 2b + 6
= 4a + 2 b + 14
∵ x – 2 is its factor
∴ Remainder = 0
⇒ 4a + 2b + 14 = 0
⇒ 4a + 2b = – 14
⇒ 2a + b = – 7 (Dividing by 2) … (i)
Again let x – 3 = 0, then x = 3
∴ f(3) = (3)³ + a (3)² + b (3) + 6
= 27 + 9a + 3b + 6 = 9a + 3b + 33
∵ Remainder = 3,
∴ 9a + 3b + 33 = 3 ⇒ 9a + 3b = 3 – 33 = – 30
⇒ 3a + b = – 10 (Dividing by 3) … (ii)
Subtracting (i) from (ii)
a = – 10 + 7 = – 3
Substituting the value of a in (i)
2 x (- 3) + b = – 7
– 6 + b = – 7 ⇒ b = – 7 + 6 = – 1
∴ a = – 3, b = – 1
Question 11.
Factorise :
(i) x³ + 13x² + 32x + 20, if it is given that x + 2 is its factor.
(ii) 4x³ + 20x² + 33x + 18, if it is given that 2x + 3 is its factor.
Solution:
(i) Let f(x) = x³ + 13x² + 32x + 20
Let x + 2 = 0, then x = – 2
∴ f (- 2) = (-2)³ + 13 (-2)² + 32 (-2) + 20
= – 8 + 52 – 64 + 20 = 72 – 72 = 0
∵ Remainder = 0
∴ x + 2 is its factor
Now dividing f(x) by (x + 2), we get
(ii) 4x³ + 20x² + 33x + 18
and 2x + 3 = 0 then 2x = – 3 ⇒ x = \(\frac { -3 }{ 2 }\)
∵ Remainder = 0
∴ 2x + 3 is its factor
Now dividing f(x) by 2x + 3, we get
Question 12.
Show that :
(i) (x – 10) is a factor of x³ – 23x² + 142x – 120 and hence factorise it completely.
(ii) (3z + 10) is factor of 9z³ – 27z² – 100z + 300 and factorise it completely.
Solution:
(i) Let f(x) = x³ – 23x² + 142x – 120
and x – 10 = 0 then x = 10
∴ f(10) = (10)³ – 23 (10)² + 142 x 10 – 120
= 1000 – 23 x 100 + 142 x 10 – 120
= 1000 – 2300 + 1420 – 120
= 2420 – 2420 = 0
∵ Remainder = 0
∴ x – 10 is the factor of f (x)
Now dividing f(x) by (x – 10), we get
Question 13.
Given that (x – 2) and (x + 1) are factors of x³ + 3x² + ax + b, calculate the values of a and b, and hence find the remaining factor.
Solution:
Let f(x) = x³ + 3x² + ax + b
Let x – 2 = 0 then x = 2
∴ f(x) = (2)³ + 3 (2)² + a x 2 + b
= 8 + 12 + 2 a + b
= 2a + b + 20
∴ x – 2 is a factor of f(x)
∴ Remainder = 0
⇒ 2a + b + 20 = 0 ⇒ 2a + b = – 20 … (i)
Again x + 1 = 0 then x = – 1
∴ f( – 1) = (-1)³ + 3 (-1)² + a (-1) + b
= – 1 + 3 – a + b = – a + b + 2
∵ x + 1 is a factor, then
∴ Remainder = 0
∴ – a + b + 2 = 0
⇒ – a + b = – 2
⇒ a – b = 2 … (ii)
Adding (i) and (ii)
3 a = – 18 ⇒ a = \(\frac { -18 }{ 3 }\) = – 6
∴ a = – 6
∵ a – b – 2 ⇒ – 6 – b = 2
⇒ – b = 2 + 6 ⇒ – b = 8
⇒ b = – 8
∴ a = – 6, b = – 8
∴ f(x) = x³ + 3x² – 6x – 8
∵ (x – 2) is a factor of f(x)
∴ Dividing f(x) by x – 2, we get
f(x) = (x – 2)(x² + 5x + 4)
= (x – 2){x² + x + 4x + 4}
= (x – 2){x(x + 1) + 4(x + 1)}
= (x – 2)(x + 1)(x + 4)
∴ Remaining factor is (x + 4)
Question 14.
Given that (x + 2) and (x – 3) are factors of x³ + ax + b, calculate the values of a and b, and find the remaining factor.
Solution:
Let f (x) = x³ + ax + b
and x + 2 = 0,then x = – 2
∴ f(- 2) = (- 2)³ + a (- 2) + b
= – 8 – 2a + b ⇒ – 2a + b – 8 … (i)
∵ x + 2 is its factor
∴ Remainder = 0
⇒ – 2a + b – 8 = 0 ⇒ – 2a + b = 8 … (i)
Let x – 3 = 0, then x = 3
∴ f(3) = (3)³ + a (3) + b
= 27 + 3 a + b
∵ x – 3 is its factor
∴ Remainder = 0
⇒ 27 + 3a + b = 0 ⇒ 3a + b = – 27 … (ii)
Subtracting (i) and (ii)
5a = -35 ⇒ a = \(\frac { -35 }{ 5 }\) = – 7
Substituting the value of a in (i)
– 2 (- 7) + b = 8 ⇒ + 14 + b = 8
⇒ b = 8 – 14 = – 6
∴ a = – 7, b = – 6
∵ x + 2 is a factor of f(x) = x³ – 7x – 6
Dividing it by x + 2, we get
f(x) = (x + 2) (x² – 2x – 3)
= (x + 2) {x² – 3x + x – 3}
= (x + 2) {x (x – 3) + 1 (x – 3)}
= (x + 2) (x – 3) (x + 1)
∴ Remaining factor of f(x) is (x + 1)
Question 15.
Factorise, using remainder theorem :
(i) x³ – 19x – 30
(ii) x³ + 7x² – 21x – 27
(iii) x³ – 3x² – 9x – 5
(iv) 2x³ + 9x² + 7x – 6
Solution:
(i) Let f(x) = x³ – 19x – 30
Factor of 30 = +1, ±2, +3, +5, +6, +10, ±15, ±30
Now by trial and error method,
Let x = – 2, then
∴ f(-2) = (-2)³ – 19 (-2) – 30
= – 8 + 38 – 30 = 0
∴ (x + 2) is its factor
Let x = -3, then
f(-3) = (-3)³ – 19 (-3) – 30
= -27 + 57 – 30 = 0
∴ x + 3 is another factor of / (x)
Let x = 5, then f(5) = (5)³ – 19 x 5 – 30
= 125 – 95 – 30
= 125 – 125 = 0
∵ x – 5 is the third factor
∴ x³ + 9x – 30 = (x + 2) (x + 3) (x – 5)
(ii) Let f(x) = x³ + 7x² – 21x – 27
Factors of 27 = ±1, ±3, ±9, ±27
By trial and error method,
Let x = – 1
f(- 1) = (- 1)³ + 7(- 1)² – 21 (- 1) – 27
= – 1 + 7 + 21 – 27 = 0
∴ x + 1 is its factor
Now dividing f(x) by x + 1, we get
(iii) Let f (x) = x³ – 3x² – 9x – 5
Factors of 5 = ±1, ±5,
By trial and error method,
Let x = – 1
Then f (- 1) = (- 1)³ – 3 (-1)² – 9 (-1) – 5
= – 1 – 3 x 1 + 9 – 5
= – 1 – 3 + 9 – 5 = 0
∴ x + 1 is one factor of f(x)
Now dividing f(x) by x + 1, we get
f(x) = (x + 1) (x² – 4x – 5)
= (x + 1) {x² – 5x + x – 5}
= (x + 1) {x (x – 5) + 1 (x – 5)}
= (x + 1) (x – 5) (x + 1)
(iv) Let f(x) = 2x³ + 9x² + 7x – 6
Factors of 6 = ±1, ±2, ±3, ±6
By trial and error method,
Let x = – 2
Then f (- 2) = 2 (-2)³ + 9 (-2)² + 7 (-2) – 6
= 2 x (-8) + 9 x 4 + 7 x (-2) – 6
= – 16 + 36 – 14 – 6 = 0
∴ x + 2 is the factor of f (x)
Now dividing f(x) by x + 2, we get
f(x) = (x + 2) (2x² + 5x – 3)
= (x + 2) {2x² + 6x – x – 3)
= (x + 2) {2x (x + 3) – 1 (x + 3)}
= (x + 2)(x + 3) (2x – 1)
Question 16.
(x – 3) is the H.C.F. of x³ – 2x² + px + 6 and x² – 5x + q. Find 6p + 5q.
Solution:
Let p (x) = x³ – 2x² + px + 6
q (x) = x² – 5x + q
H.C.F. = x – 3
∴ H.C.F. is a factor of p (x) and q (x)
Let x – 3 = 0, then x = 3
∴ P (3) = (3)² – 2 (3)² + p (3) + 6
= 27 – 18 + 3p + 6
= 15 + 3p
∴ Remainder = 0
∴ 15 + 3p = 0
⇒ 3p = – 15
⇒ P = \(\frac { -15 }{ 3 }\)
p = – 5
and q (3) = (3)² – 5 x 3 + q
= 9 – 15 + q = – 6 + q
∴ Remainder = 0
∴ – 6 + q = 0 ⇒ q = 6
Hence p = – 5 and q = 6
Now 6p + 5q = 6 (- 5) + 5 x 6
= – 30 + 30 = 0
Question 17.
(i) What number must be subtracted from x³ – 6x² – 15x + 80, so that the result is exactly divisible by x + 4.
(ii) What number must be added to x³ – 3x² – 12x + 19, so that the result is exactly divisible by x- 2.
Solution:
(i) Let f(x) = x³ – 6x² – 15x + 80
and let p be subtracted from f (x) so that it may he exactly divisible by x + 4
∴ P (x) = x³ – 6x² – 15x + 80 – p
and let x + 4 = 0, then x = – 4
∴ P (- 4) = (-4)³ – 6 (-4)² – 15 (-4) + 80 – p
= – 64 – 96 + 60 + 80 – p
= 140 – 160 – p
= – 20 – p
∵ P (x) is divisible by x + 4
∴ Remainder = 0
∴ – 20 – p = 0 ⇒ p = – 20
(ii) Let q (x) = x³ – 3x² – 12x + 19
and let p be added to q (x)
Then (x) = x³ – 3x² – 12x + 19 + p
But it is divisible by x – 2
∴ x – 2 = 0 ⇒ x = 2
Q (2) = (2)³ – 3 (2)² – 12 (2) + 19 + p
= 8 – 12 – 24 + 19 + p
= 27 – 36 + p = – 9 + p
∵ Q (x) is divisible by x – 2
∴ Remainder = 0
∴ – 9 + p = 0 ⇒ p = 9
Self Evaluation And Revision (LATEST ICSE QUESTIONS)
Question 1.
Find the remainder when 2x³ – 3x² + 7x – 8 is divided by x – 1.
Solution:
Let f(x) = 2x³ – 3x² + lx – 8
and let x – 1 = 0, then x = 1
Remainder = f(1) = 2(1)³ – 3(1)² + 7 x 1 – 8
= 2 x 1 – 3 x 1 + 7 x 1 – 8
= 2 – 3 + 7 – 8
= 9 – 11
= – 2
Question 2.
Find the value of the constants a and b if (x – 2) and (x + 3) are both factors of the expression x³ + ax² + bx – 12.
Solution:
Let f (x) = x³ + ax² + bx – 12
∵ x – 2 is a factor of f(x)
Let x – 2 = 0 ⇒ x = 2
∴ f(2) = (2)³ + a (2)² + b (2) – 12
= 8 + 4 a + 2 b – 12
= 4a + 2b – 4
∵ Remainder = 0
∴ 4a + 26 – 4 = 0 ⇒ 4a + 26 = 4 … (i)
Again ∵ x + 3 is the factor of f(x)
Let x + 3 = 0, then x = – 3
∴ f(- 3) = (-3)³ + a (-3)² + b (-3) – 12
= – 27 + 9a – 36 – 12
= 9a – 36 – 39
∵ Remainder = 0
∴ 9a – 36 – 39 = 0 ⇒ 9a – 36 = 39
⇒ 3a – b = 13 … (ii)
Multiply (i) by 1 and (ii) by 2,
Substituting the value of a in (i)
4 x 3 + 26 = 4 ⇒ 12 + 26 = 4
⇒ 2b = 4 – 12 = – 8
b = \(\frac { -8 }{ 2 }\) = – 4
∴ a = 3, b = – 4
Question 3.
Using factor theorem, show that (x – 3) is a factor of x³ – 7x² + 15x – 9. Hence factorise the given expression completely.
Solution:
Let f(x) = x³ – 7x² + 15x – 9
Let x – 3 = 0. then x = 3
∴ f(3) = (3)³ – 7 (3)² + 15 x 3 – 9
= 27 – 63 + 45 – 9
= 72 – 72 = 0
∵ Remainder = 0
∴ x – 3 is a factor of f (x)
Now dividing f(x) by x – 3, we get
f(x) = (x – 3) (x² – 4x + 3)
= (x – 3) (x² – x – 3x + 3}
= (x – 3) {x(x- 1) – 3 (x- 1)}
= (x – 3)(x- 1) (x – 3)
= (x – 3)² (x – 1)
Question 4.
Find the value of a, if (x- a) is a factor of x³ – a²x + x + 2.
Solution:
Let f(x) = x³ – ax² + x + 2
and x – a = 0, then x = a
∴ f(a) = a³ – a.a² + a + 2
= a³ – a³ + a + 2 = a + 2
∵ x – a is the factor of f(x)
∴ Remainder = 0
∴ a + 2 = 0 ⇒ a = – 2
Question 5.
Use the factor theorem to factorise completely :
x³ + x² – 4x – 4.
Solution:
Let f(x) = x³ + x² – 4x – 4
Factors of 4 = +1, +2, ±4
By trial and error.
x = – 1
Then f (- 1) = (- 1)² + (- 1)² – 4 x (- 1) – 4
= – 1 + 1 + 4 – 4
= 5 – 5 = 0
∵ Remainder = 0
∴ x + 1 is its factor
Now dividing f(x) by x + 1, we get
f(x) = x³ + x² – 4x – 4
= (x + 1)(x² – 4)
= (x+ 1) {(x)² – (2)²}
= (x + 1) (x + 2) (x – 2)
Question 6.
(x – 2) is factor of the expression x³ + ax² + bx + 6. When this expression is divided by (x – 3), it leaves a remainder 3. Find the values of a and b.
Solution:
Let f(x) = x³ + ax² + bx + 6
∵ x – 2 is its factor
Let x – 2 = 0, then x = 2
∴ f (2) = (2)³ + a (2)² + b(2) + 6
= 8 + 4a + 2b + 6 = 4a + 2b + 14
∵ Remainder = 0
∴ 4a + 2b + 14 = 0
⇒ 2a + b + 7 = 0
⇒ 2a + b = -7 … (i)
Again when f(x) is divided by x – 3,
Let x – 3 = 0, then x = 3
∴ f(3) = (3)³ + a (3)² + b(3) = 6
= 27 + 9a + 3b + 6 = 9a + 3b – 33
∵ Remainder = 3
∴ 9a + 3b + 33 = 3
⇒ 9a + 36 = 3 – 33
⇒ 9a + 3b = – 30
⇒ 3a + b = – 10 ….(ii)
Subtracting (i) from (ii)
a = – 3
Substituting the value of a in (i)
2 x (- 3) + b = – 7
⇒ – 6 + b = – 7
⇒ b = – 7 + 6 = -1
∴ a = -3, b = – 1
Question 7.
Show that 2x + 7 is a factor of 2x³ + 5x² – 11x – 14. Hence factorise the given expression completely, using factor theorem.
Solution:
Let f(x) = 2x³ + 5x² – 11x – 14
Let 2x + 7 = 0, then 2x = – 7 ⇒ y = \(\frac { -7 }{ 2 }\)
∵ Remainder = 0
∴ 2x + 7 is a factor of f(x)
Now dividing f(x) by 2x + 7, we get
f(x) = (2x + 7) (x² – x – 2)
= (2x + 7) {x² – 2x + x – 2}
= (2x + 7) {x (x – 2) + 1 (x – 2)}
= (2x + 7) (x – 2) (x + 1)
Question 8.
Show that (x – 1) is a factor of x³ – 7x² + 14x – 8. Hence completely factorise the above expression.
Solution:
Let f(x) = x³ – 7x² + 14x – 8
Let x – 1 = 0, then x = 1
∴ f(1) = (1)³ – 7(1)² + 14 x 1 – 8
= 1 – 7 + 14 – 8
= 15 – 15 = 0
∵ Remainder = 0.
∴ (x – 1) is a factor of f(x)
Now dividing f(x) by x – 1, we get
f(x) = (x – 1) (x² – 6x + 8)
= (x – 1) {x² – 4x – 2x + 8}
= (x – 1) {x(x – 4) – 2(x – 4)}
= (x – 1) (x – 4) (x – 2)
Question 9.
If (x – 2) is a factor of 2x³ – x² – px – 2.
(i) Find the value of p.
(ii) With the value of p, factorise the above expression completely.
Solution:
(i) Let f(x) = 2x³ – x² – px – 2
∵ x – 2 is its factor
Now x – 2 = 0, then x = 2
∴ f(2) = 2 (2)³ – (2)² – p x 2 – 2
= 2 x 8 – 4 – 2p – 2
= 16 – 4 – 2 – 2p
= 10 – 2p
v Remainder = 0
∵ 10 – 2p = 0 ⇒ 2p = 10
⇒ P = \(\frac { 10 }{ 2 }\) = 5
∴ p = 5
(iii) Now fix) = 2x³ – x² – 5x – 2
Now dividing, f(x) by x – 2, we get
f (x) = (x – 2) (2x² + 3x + 1)
= (x – 2) {2x² + 2x + x + 1}
= (x – 2) {2x (x + 1) + 1 (x + 1)}
= (x – 2) (x + 1) (2x + 1)
Question 10.
Given that x + 2 and x + 3 are factors of 2x³ + ax² + 7x – b. Determine the values of a and b.
Solution:
Given : (x + 2) and (x + 3) are factors of 2x³ + ax² + 7x – b.
∴ x + 2 = 0
⇒ x = – 2
2 (- 2)³ + a (- 2)² + 7 (- 2) – b = 0
– 16 + 4a – 14 – b = 0
4a – b – 30 = 0
⇒ 4a – b = 30 … (i)
Now, x + 3 = 0
⇒ x = – 3
2 (- 3)³ + a (-3)² + 7 (-3) – b = 0
– 54 + 9a – 21 – b = 0
9a – 6 – 75 = 0
⇒ 9a – 6 = 75 … (ii)
Subtracting (i) from (ii) we get
5a = 45
⇒ a = 9
Substituting the value of a in (i)
4 x 9 – b = 30
⇒ 3b – b = 30
⇒ b = 6
Question 11.
When divided by x – 3 the polynomials x³ – px² + x + 6 and 2x* – x² – (p + 3) x – 6 leave that same remainder. Find the value of ‘p’.
Solution:
When (x – 3) divides x³ – px² + x + 6, then
Remainder = p(3) = (3)³ – p(3)² + (3) + 6
= 27 – 9p + 9
= 36 – 9p
When (x – 3) divides 2x³ – x² – (p + 3) x – 6, then
Remainder = p(3) = 2(3)³ – (3)² – (p + 3) (3) – 6
= 54 – 9 – 3p – 9 – 6 = 30 – 3p
A.T.Q. both remainders are equal
⇒ 36 – 9p = 30 – 3p ⇒ 36 – 30 = -3p + 9p
⇒ 6 = 6p
∴ P = 1
Question 12.
Use the Remainder Theorem to factorise the following expression :
2x³ + x² – 13x + 6.
Solution:
By hit and trial, putting x = 2,
2 x (2)³ + (2)² – 13 x 2 + 6
⇒ 2 (8) + 4 – 26 + 6 = 0
⇒ 16 + 4 – 26 + 6 = 0
⇒ 26 – 26 = 0
(x – 2) is the factor of 2x³ + x² – 13x + 6
∴ 2x³ + x² – 13x + 6 = (x – 2) (2x² + 5x – 3)
= (x – 2) (2x² + 6x – x – 3)
= (x – 2) [2x (x + 3) – 1 (x + 3)]
2x³ + x² – 13x + 6 = (x – 2) (2x – 1) (x + 3)
Question 13.
Find the vlaue of ‘k’ if (x – 2) is a factor of x³ + 2x² – kx + 10. Hence determine whether (x + 5) is also a factor.
Solution:
f (x) = x³ + 2x² – kx + 10
(x – 2) is a factor.
∴ x = 2
f (2) = (2)³ + 2 (2)² – k (2) + 10 = 0 ⇒ 8 + 8 – 2k + 10 = 0 ⇒ 26 = 2k
∴ k = \(\frac { 26 }{ 2 }\) = 13
If (x = 5) is a factor, then x = – 5
Now, f(- 5) = (-5)³ + 2(-5)² – 13(-5) + 10
= – 125 + 50 + 65 + 10 = 0
⇒ 125 – 125 = 0
Yes, (x + 5) is also a factor.
Question 14.
Using a Remainder Theorem factorise completely the following polynomial.
3x³ + 2x² – 19x + 6
Solution:
P(x) = 3x³ + 2x² – 19x + 6
P(1) = 3 + 2 – 19 + 6 = – 8 ≠ 0
P(- 1) = – 3 + 2 + 19 + 6 = – 24 ≠ 0
P(2) = 24 + 8 – 38 + 6 = 0
Hence, (x – 2) is a factor of P(x)
So, P(x) = 3x³ + 2x² – 19x + 6
= 3x³ – 6x² + 8x² – 16x – 3x + 6
= 3x² (x – 2) + 8x (x – 2) – 3 (x – 2)
= (x – 2) (3x² + 8x – 3)
= (x – 2) (3x² + 9x – x – 3)
= (x – 2) {3x (x + 3) – 1 (x + 3)
= (x – 2) (x + 3) (3x – 1)
Question 15.
If (x – 2) is a factor of the expression 2x³ + ax² + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b.
Solution:
f(x) = 2x³ + ax² + bx – 14
∵ f (x – 2) is factor of f(x)
∴ f (2) = 0
2(2)³ + a(2)² + b(2) – 14 = 0
16 + 4a + 2b – 14 = 0
4a + 2b = – 2
2a + b = – 1 … (i)
Also, (x – 3) it leaves remainder = 52
∴ f(3) = 52
2(3)³ + a(3)² + 6(3) – 14 = 52
54 + 9a + 3b – 14 = 52
9a + 3b = 52 – 40
9a + 3b = 12
3a + b = 4 … (ii)
From (i) and (ii)
∴ a = 5 put in (/)
∴ 2(5) + b = – 1
b = – 1 – 10
b = – 11
∴ a = 5, b = – 11
Question 16.
Using the Remainder and Factor theorem, factorise the following polynomial:
x³ + 10x² – 37x + 26.
Solution:
f(x) = x³ + 10x² – 37x + 26
f(1) = (1)³ + 10(1)² – 37(1) + 26
= 1 + 10 – 37 + 26 = 0
x = 1
x – 1 is factor of f (x)
∴ f (x) = (x – 1) (x² + 11x – 26)
= (x – 1) (x² + 13x – 2x – 26)
= (x – 1) [x(x + 13) – 2(x + 13)]
= (x – 1) [(x – 2) (x + 13)]
Question 17.
Find ‘a’ if the two polynomials ax³ + 3x² – 9 and 2x³ + 4x + a, leave the same remainder when divided by x + 3.
Solution:
The given polynomials are ax³ + 3x² – 9 and 2x³ + 4x + a
Let p(x) = ox³ + 3x² – 9
and q(x) = 2x³ + 4x + a
Given that p(x) and q(x) leave the same remainder when divided by (x + 3),
Thus by Remainder Theorem, we have
p(-3) = q(-3)
⇒ a(-3)³ + 3(-3)² – 9 = 2(-3)³ + 4(-3) + a
⇒ – 27a + 27 – 9 = – 54 – 12 + a
⇒ – 27a + 18 = – 66 + a
⇒ – 27a – a = – 66 – 18
⇒ – 28a = – 84
⇒ a = \(\frac { 84 }{ 28 }\)
∴ a = 3
Question 18.
Using remainder theorem, find the value of k if on dividing 2x² + 3x² – kx + 5 by x – 2, leaves a remainder 7.
Solution:
Let f(x) = 2x³ + 3x² – kx + 5
By the remainder theorem,
f(2) = 7
∴ 2(2)³ + 3(2)² – k(2) + 5 = 7
∴ 2(8) + 3(4) – k(2) + 5 = 7
∴ 16 + 12 – 2k + 5 = 7
∴ 2k = 16 + 12 + 5 – 7
∴ 2k = 26
∴ k = 13
The value of k is 13.