ML Aggarwal ICSE Solutions Class 10 Math 6th Chapter Factorisation
Class 10 Chapter 6 FactorisationExercise:
(1) Find the remainder (without division) on dividing f(x) by (x-2) where
(i) f(x) = 5x2 – 7x + 4
(ii) f (x) = 2x3 – 7x2 + 3
Solution:
Given that, x – 2 =0
x = 2
(ii) Put x = 2 in f (x) = 2x3 – 7x2 + 3
f (2) = 2 (2)3 – 7 (2)2 + 3
= 16 – 28 + 3
= 19 – 28
F (-2) = – 9 is the required remainder
(i) Put x = 2 in f (x) = 5x2 – 7x + 4
f (2) = 5 (2)2 – 7 (2) + 4
= 20 – 14 + 4
= 24 – 14
F (2) = 10 is the required remainder.
(2) Using the remainder theorem, find the remainder on dividing f (x) by (x+3) where.
(i) f (x) = 2x2-5x + 1
(ii) f (x) = 3x2 + 7x2 -5x + 1
Solution:
Given divider is x + 3 =0
x = – 3
(i) Put x = -3 in f (x) = 2x2 – 5x + 1
f (-3) = 2 (-3)2 – 5 (-3) + 1
= 18 + 15 + 1
F (-3) = 34 is the required remainder.
(ii) Put x = – 3 in f (x) = 3x3+ 7x2– 5x + 1
f (-3) = 3 (-3)3 + 7 (-3)2 – 5 (-3) + 1
= – 81+ 63 + 15 + 1
= – 81 + 79
F (-3) = -2 is the required remainder.
(3) Find the remainder (without division) on dividing f (x) by (2x+1) where
(i) f (x) = 4x2 + 5x + 3
(ii) f (x) = 3x3 – 7x2 + 4x + 11
Solution:
Given divider is 2x + 1 = 0
x= -1/2
(i) Put x = -1/2 in f(x) = 4x2 + 5x + 3
f (-1/2) = 4 (-1/2)2 + 5 (-1/2) + 3
= 1 -5/2 + 3 = 4 – 5/2 = 3/2
f (-1/2) = 3/2 is the required remainder.
(ii) Put x = -1/2 in f (x) = 3x3 – 7x2 + 4x + 11
f (1/2) = 3 (-1/2)3 – 7 (-1/2)2 + 4 (-1/2) + 11
=-3/8 – 7/4 – 2 + 11
= – 3/8 – 7/4 + 9
= -3-14/9 + 9 = -17/8 + 9
= -17+72/8 = 55/8
= f (-1/2) = 55/8 is the required remainder.
(4) Using remainder theorem, find the value, of K if on dividing (2x3 + 3x2 – kx + 5) by (x-2), leaves a remainder 7.
Solution:
Given polynomial is f (x) = 2x3 + 3x2-kx+5 and g (x) = x – 2
If g (x) = 0 = x – 2 = 0
=> x = 2
Put x = 2 in f (x) = 2x3 + 3x2 – kx + 5
f (2) = 2 (2)3+3 (2)2 – k(2) + 5
= 2 (8) + 3 (4) – 2k + 5
= 16 + 12 – 2k + 5
f (2) = 33 – 2K is the required remainder.
But, given that 7 is the required remainder.
= 33 – 2k – 7
33 – 7 = 2k
26 = 2k
K = 13 is the required value of k
(5) Using remainder theorem, find the value of ‘a’ if the division of (x3 + 5x2 – ax + 6) by (x-1) leaves the remainder 2a.
Solution:
Given polynomial is x3+ 5x2 = ax + 6 = f(x) and g (x) = x – 1
If g (x) = 0
=> x – 1 = 0 => x = 1
Put x = 1 in f (x) = x3 + 5x2 – ax + 6
f (1) = 13 + 512 –a + 6
= 1 + 5 – a + 6
f (1) = 12 – a is the required remainder.
But, given that 2a is the required remainder
= 12 – a = 2a
12 = 3a
a = 4 is the required value of a.
(6) What number must be subtracted from (2x2 – 5x) so that the resulting polynomial leaves the remainder 2. When divided by (2x + 1)?
Solution:
Let us consider, ‘a’ be subtracted from (2x2 – 5x).
Now, f (x) = 2x2– 5x and g (x) = 2x + 1
But given that remainder = 2
=> 3 – p = 2
P = 1 is the required number.
(7) (i) When divided by (x-3) the polynomials x3 – px2 + x + 6 & 2x3 – x2 – (p+3) x – 6 leaves the same remainder. Find P.
Solution:
Given that, f(x) = x3 – px2 + x + 6 and g(x) = x – 3
If g(x) = 0 => x – 3 = 0
x = 3
Put x = 3 in f(x)
=> f(3) = (3)3 – P (3)2 + 3 + 6
= 27 – 9p + 9
f (3) = 36 – 9p is the required remainder.
f (x) = 2x3 – x2 – (p+3) x – 6 and g (x) = x – 3
If g (x) = 0 => x – 3 = 0
x = 3
Put, x = 3 in f(x) = 2x3 – x2 – (p + 3) x – 6
f (3) = 2 (3)3 – (3)2 – (p + 3) 3 – 6
= 54 – 9 – 3p – 9 – 6
= 54 – 24 – 3p
f (3) = 30 – 3p is the required remainder.
But, from given condition,
36 – 9p = 30 – 3p
36 – 30 = 9p – 3p
6 = 6p
P = 1 is the required value of p.
(ii) Find ‘a’ if the two polynomials ax3 + 3x2 -9 and 2x3 + 4x + a, leaves the same remainder when divided by (x+3).
Solution:
Given that,
f(x) = ax3 + 3x2 – 9
g (x) = x +
If g(x) = 0 => x + 3 = 0
x = -3
Put, x = -3 in f(x)
=> f (-3) = a (-3)3 + 3 (-3)2 – 9
= 27a + 27 – 9
f (-3) = – 27a + 18 is the required remainder.
Given that,
f (x) = 2x3 + 4x + a
g(x) = x + 3
If g(x) = 0 => x + 3 = 0
x = -3
Put x = -3 in f(x)
=> f(-3) = 2 (-3)3 + 4 (-3) + a
= -54 – 12 + a
f (-3) = -66 + a is the required remainder.
But, from given condition,
-27a +18 = – 66 + a
18 + 66 = 28a
84 = 28a
a = 12/4 = 3
a = 3 is the required value of ‘a’.
(8) Using remainder theorem find the remainder obtained. When x3 + (kx+8) x + k, is divided by (x+1) and (x-2). Hence, find k if the sum of the two remainders is 1.
Solution:
Given polynomial is f(x) = x3 + (kx + 8) x + k and g (x) = x + 1
If g (x) = 0 = x + 1= 0 ⇒ x = -1
Put x = – 1 in f (x)
=> f (-1) = (-1)3 + (-k+8) (-1) + K
= – 1 + K – 8 + K
f (-1) = – 9 + 2K is the required remainder.
If g (x) = x – 2 = 0
=> x = 2 put in f (x)
f (2) = 23+ (2k + 8) 2 + K
f (2) = 8 + 4k + 16 + K
f (2) = 24 + 5k is the required remainder.
But, from given condition, we can write
(-9+2k) + (24 + 5k) = 1
15 + 7k = 1
7 k = – 14
K= -2 is the required value of k.
(9) By factor theorem, show that (x+3) and (2x-1) are factor of (2x2 + 5x – 3).
Solution:
Let, Given that f (x) = 2x2 + 5x – 3
Let us consider x + 3 = 0 => x = – 3
Put x = – 3 in f (x) = f (-3) = 2 (-3)2 + 5 (-3)-3
= 18-15-3
= 18-18
f (-3) = 0 Remainder
Hence, (x+3) is the factor of f (x).
Now, 2x – 1 = x = 1/2
Put x = 1/2 in f (x) = f (1/2) = 2 (1⁄2)2 + 5 (1/2) – 3
= ½ + 5/2 – 3
= 3 – 3
= f (1/2) = 0 —-> Remainder
Thus, (2x – 1) is also the factor of f(x).
(10) Without actual division, prove that x4 + 2x3 – 2x2 + 2x – 3 is exactly divisible by (x2 + 2x – 3).
Solution:
Given that, f (x) = x4 + 2x3 – 2x2 + 2x – 3
And g (x) = x2 + 2x – 3
= X2+3X-X-3
= x(x+3)-1(x+3)
g (x) = (x-1) (x+3) ———- divider
When (x-1) = 0 => x = 1 put in f (x)
= f (1) = 14 + 2 (1)3 – 2(1)2 + 2 (1) -3
= 1 + 2 – 2 + 2 – 3
f (1) = 0 ———- Remainder
And When x + 3 = 0 => [x = – 3] put in f (x).
f (-3) = (-3)4 + 2 (-3)3 – 2 (-3)2 + 2 (-3) – 3
= 81 – 54 – 18 – 6-3
f (-3) = 0 ———- Remainder
Thus, (x-1) & (x+3) both are factors of f (x).
Hence g (x) = (x-1) (x+3) is also the factor of f (x)
(12) Using the factor theorem show that (x-2) is a factor of (x3 + x2 – 4x – 4). Hence factorise the polynomial completely
Solution:
Given polynomial is f(x) = x3 + x2 – 4x – 4.
And g (x) = x – 2
If g (x) = 0 = x – 2 = 0 => x = 2
Put x = 2 in f(x) => f(2) = 23 +22 – 8 – 4
=8 + 4 – 8 -4
f (2) = 0 —- Remainder
Hence, (x – 2) is a factor of f (x).
Now, x3 + x2 – 4x + 4 = x2 (x+1) – 4 (x+1)
= (x+1) (x2-4)
f (x) = (x+1) (x+2) (x-2)
(13) Show that (2x + 7) is a factor of (2x2+ 5x2 – 11x – 14). Hence factorise the given expression completely, using factor theorem.
Solution:
Given polynomial is f (x) = 2x3 + 5x2– 11x – 14 and g (x) = 2x + 7
If g (x) = 2x + 7 = 2x + 7 = 0
=> x = -7/2 put in f (x)
f (-7/2) = 2(-7/2)3 + 5(-7/2)2 – 11 (-7/2) -14
= -343/4 +245/4 – 77/2 – 14
= (-343+245 +154-56)/4
= (-399+399)/4
f (-7/2)= 0 → Remainder
Hence, (2x+7) is the factor of f(x).
Thus, f (x) = 2x3+ 5x2 – 11x – 14
f (x)= (2x+7) (x2-x-2)
= (2x+7) (x2-2x+x-2)
= (2x+7) [x(x-2)+1(x-2)]
F(x) = (2x+7) (x-2) (x+1)
Thus, (2x+7), (x-2) & (x+1) are factors of f(x).
(14) Use factor theorem to factorise the following polynomial completely.
(i) x3 + 2x2 – 5x-6
Solution:
Given that f (x) = x3 + 2x2 – 5x – 6
Let, x = -1 put in f (x)
f (-1) = (-1)3+2 (-1)2 -5(1)-6
= -1+2+5-6
=7-7
f (-1) =0 => Remainder
Hence, (x+1) is the factor of f(x).
Now, by Actual division method,
Thus, f(x) = x3+ 2x2 – 5x – 6
f (x) = (x + 1) (x2 + x – 6)
= (x + 1) [x2 + 3x – 2x – 6]
= (x + 1) [x (x + 3)-2(x+3)]
= (x + 1) [(x – 2) (x + 3)
f (x) = (x + 1) (x – 2) (x + 3)
Thus, (x +1), (x – 2) & (x + 3) are the factors of f (x).
(15) Use the Remainder theorem to factorise the following expression
(i) 2x3+ x2 – 13x + 6
Solution:
Given that, f (x) = 2x3 + x2– 13x + 6
Let x = 2 put in f (x)
= f (2) = 2 (2)3 + (2)2-13 (2)+6
=16 +4-26+6
=26-26
f(2) = 0 => Remainder
Thus, (x-2) is the factor of f(x).
Now, by actual division method,
Thus, f(x) = 2x3+x2-13x+6
= (x-2) (2x2+5x-3)
= (x-2) [2x2 + 6x – x – 3]
= (x-2) [2x(x+3)-1(x+3)]
f (x) = (x-2) (x+3) (2x-1)
Thus, (x-2), (x+3) & (2x-1) are the factors of f (x)
(ii) 2x2+3x2-9x-1
Solution:
Given polynomial is f(x) = 2x3 + 3x2 – 3x – 10
Let x = 2 put in f (x)
f (2) = 2(2)3+3(2)2-9(2)-10
= 16 + 12 – 18 – 10
=28-28
f (2) = 0 => Remainder
Hence, (x-2) is the factor of f(x)
Now, by actual division method,
Thus, f(x) = 2x3 + 3x2 – 9x – 10
f (x) = (x-2) [2x2+7x+5]
= (x-2) [2x2+2x+5x+5]
= (x-2) [2x (x+1) + 5 (x+1)]
f (x) = (x-2) (2x+5) (x+1)
Hence, (x-2), (2x+5) & (x+1) are the factors of f(x).
(16) If (2x + 1) is a factor of (6x3 + 5x2 + ax – 2), find the value of a.
Solution:
Given polynomial is f (x) = 6x3 + 5x2 + ax – 2 and g(x) =2x + 1is a factor of f(x)
Then, g(x) = 0 => 2x + 1 = 0
x = -1/2 put in f (x)
f (-1/2) = 0
6 (-1/2)3 + 5 (-1/2)2 + a (-1/2) -2 =0
-6/8 + 5/4 -a/2 -2 = 0
-6 + 10-4a -16=0
10 – 22 = 4a
– 12 = 4a
= 0
a= -3 is the required value of a f(x)
(17) If (3x – 2) is a factor of 3x3-kx2 + 21x-10, find K.
Solution:
Given polynomial is 3x3 – kx2 + 21x – 10 = f(x)
And g (x) =3x-2 is the factor of f (x).
Then, g (x) = 0 => 3x – 2 = 0
x = 2/3 put in f (x)
f (2/3)=0
3 (2/3)3– k (2/3)2 + 21(2/3) – 10 = 0
8/9 – K (4/9) + 14 – 10 = 0
8/9 – 4k/9 + 4 = 0
8 – 4k + 36 = 0
– 4k = – 44
K=11 is the required value of k
(18) If (x – 2) is a factor of 2x3 – x2 + px – 2, then
(i) Find the value of p
(ii) With this value of p, factorize the above expression completely.
Solution:
Given polynomial is f (x) = 2x3 – x2 + px-2
g (x) = x-2 is the factor of f (x)
(i) Then g (x) = 0 => x – 2=0
=> x =2 put in f (x)
f (2) = 2(2)3 – 22+2p-2 = 0
16-4 +2p -2=0
8 + 2p =0
2p =-8
p= – 4 is the required value of p.
(ii) put p = – 4 in f(x) = 2x3 – x2 – 4x – 2
f(x) = (x-2) (2x2+3x+1)
= (x-2) (2x2+2x+x+1)
=(x-2) [2x(x+1)+1(x+1)]
f (x) = (x-2) (2x+1)(x +1)
Thus, (x-2), (2x+1) & (x+1) are the factors of f(x)
(19) What number should be subtracted from 2x3-5x2+5x so that the resulting polynomial has (2x-3) as a factor.
Solution:
Given polynomial is f (x) = 2x3-5x2+5x
Let p be the no. subtracted from f (x).
Given that, g(x) = 2x-3=0
x = 3/2 put in f (x)
f (3/2) = 2(3/2)3 -5(3/2)2 +5(3/2)
0 = 2(27/8) -5 (9/4) + 15/2
27/4 – 45/4 + 15/2 – p=0
27-45+30-4p =0
57-45-4p=0
12-4p=0
p = 3 is the required value of p.
(20) Find the value of the constants a and b, if (x-2) & (x+3) are both factors of the expression x3 + ax2 + bx = 12.
Solution:
Given polynomial is f (x) = x3+ ax2+bx-12
And x-2= 0 => x =2 put in f(x)
= f (2) = 23+ a (2)2 + b (2)-12
= 8+4a+2b-12
f (2) = 4a + 2b – 4 —- Remainder
But (x-2) is a factor of f (x).
= f (2) =0
= 4a +2b -4 = 0
= 4a +2b = 4
2a+b=2 ——— (i)
Now, x +3 = 0 => x = – 3 put in f(x)
f (-3) =(-3)3+ a (-3)2+ b(-3)-12
=-27+9a-3b-12
f (-3) = -39+9a- 3b → Remainder
But (x+3) is a factor of f(x)
= f (-3)=0
= -39 + 9a-3b=0
9a-3b=39
3a-b=13 —- (ii)
1 + 2 = 5a = 15
a = 3
6+b=2
b = – 4
a =3 & b = – 4 are the required values.
(22) (x-2) is a factor of the expression x3 + ax2 + bx + 6.
When this expression is divided by (x-3), it leaves the remainder 3. Find the values of a and b.
Solution:
Given polynomial is f (x) = x3 + ax2 + bx + 6
(x-2) is the factor of f (x)
x – 2 = 0 = x = 2 put in f(x)
f (2) = 23 + a22 + b2 + 6
=8 + 4a + 2b + 6
f (2) = 14 + 4a + 2b —– Remainder
f (2) = 0
14 + 4a + 2b = 0
4a+2b= -14
2a+b=-7 ———- (1)
When f(x) is divided by (x-3) remainder is 3.
= x – 3 = 0
=> x = 3 put in f(x)
f (3) = 33+ a (3)2 + b (3) +6
=27+9a+3b+6
f (3) = 33+9a +3b —— Remainder
3 = 33 + 9a + 3b
9a+3b= -30
3a+b=-10 ——- (2)
(1) – (2) => (2a+b)-(3a+b) = -7 +10
– a = 3
a = – 3 put in 1 = – 6 + b= – 7
b = – 1
Thus, a=-3 & b = -1 are the required values of a & b respectively.
(23) If (x-2) is a factor of the expression 2x3+ax2+bx-14 and when the expression is divided by (x-3), it leaves remainder 52 find the values of a and b.
Solution:
Given polynomial is f (x) = 2x3 + ax2+ bx – 14
(x-2) is the factor of f (x)
Then, x-2 = 0 => x = 2 put in f(x)
f (2) = 2(2)3+ a (2)2 + b (2) – 14
= 16 + 4a + 2b – 14
f (2) = 2+4a+2b → Remainder
= f (2)=0
4a +2b = -2
2a + b = – 1——– (1)
Now, when f (x) is divided by (x-3) remainder is 52.
= x – 3 = 0 => x = 3 put in f (x)
f (3) = 2(3)3 + a (3)2 + b (3) – 14
= 54 + 9a + 3b – 14
52= 40 +9a + 3b
9a + 3b = 52 – 40
9a + 3b = 12
3a + b = 4 —— (2)
(1) – (2) => (2a+b) – (3a+b) = – 1 – 4
– a = – 5
a = 5 put in (1)
10 + b = – 1
b=-11
Thus, a=5 & b=-11 are the values of a and b respectively.
(24) If ax3 + 3x2 + bx – 3 has a factor (2x+3) & leaves remainder (-3) when divided by (xyz), find the values of a & b. With these values of a & b, factorise the given expression.
Solution:
Given polynomial is f (x) = ax3 + 3x2 + bx – 3 (2x+3) is a factor of f (x)
Then, 2x + 3 = 0 => x = -3/2 put in f (x)
f (3/2) = a (-3/2)3 + 3 (-3/2)2 + b(-3/2)-3
=-27/8 a + 27/4 – 3b/2 – 3
0 = -27a/8 +27/4 – 3b/2 – 3
= -27a + 54 -12b – 24 = 0
-27a – 12b = -30
9a + 4b = 10 ——- (1)
When f (x) is divided by (x+2) the remainder is -3.
x +2 = 0 => x = -2 put in f(x)
f (-2) = a (-2)3 + 3 (-2)2+ b (-2)-3
= -8a + 12 – 2b – 3
f (-2) = – 8a – 2b + 9
– 3 = – 8a – 2b + 9
– 8a – 2b = – 12
4a +b= 6 ——- (2)
(1) = 9a + 4b = 10x
(2) × 4 16a + 4b = 24
_____________________
– 7a = – 14
a = 2 Put in (1)
9a + 4b = 10
8+4b=10
4b= – 8
b = -2
Thus, a = 2 & b = – 2 are the required values.
Now, f (x) = ax3 + 3x2+bx-3
f(x)2x3+3x2-2x-3
Given that, (2x+3) is also the factor of f (x).
Then, by actual division method,
Thus, f(x) = 2x3 + 3x2-2x-3
f(x) = (2x+3) (x2-1)
f(x) = (2x+3)(x-1) (x+1)
Thus, (x+1), (x-1) & (2x+3) are the factors of f(x).
(25) Given f(x) = ax2+ bx + 2 and g(x) = bx2 + ax+1. If (x-2) is A factor of f (x) but leaves the remainder – 15 when it divides g (x), find the values of a and b. With these values of a and b, factorise the expression f(x) + g (x) + 4x2 + 7x.
Solution:
Given polynomial is f(x) = ax2 + bx+2 and g(x) = bx2 + ax + 1
(x – 2) is the factor of f(x).
Then, x – 2 = 0 => x = 2 put in f(x)
f (2) = a (2)2+b (2) + 2 = 4a + 2b + 2
0 = 4a + 2b + 2
4a + 2b = -2
2a + b = -1 ——– (1)
When (x-2) divides g(x) remainder is -15.
=> x – 2 = 0 => x = 2 put in g (x)
g (2) = 4b + 2a + l
– 15 = 4b + 2a + 1
4b + 2a +1 = -15
4a +2a = – 16
a +2b = – 8 ——– (2)
2a +b = – 1 ——- (1)
(a + 2b) – (4a+2b) = – 8 + 2
– 3a = – 6
a = 2 put in (2)
2 + 2b = 8
2b = -10
b= – 5
Now, f (x) = ax2 + bx + 2 = 2x2 – 5x + 2
And g(x) = bx2+ ax + 1 = -5x2 + 2x + 1
Now, f(x) + g (x) + 4x2 + 7x
= 2x2 – 5x + 2 – 5x2 + 2x + 1 + 4x2 + 7x
= x2 + 4x + 3
= x2 + 3x + x + 3
= x (x + 3) + 1(x + 3)
=> (x + 1)(x+3) = f (x) + g (x) + 4x2 + 7x
This is the required answer.
Also See :