Factor Theorem Factorization Class 10 OP Malhotra Exe-7 ICSE Maths Solutions of Ch-7 questions as latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Factor Theorem Factorization Class 10 OP Malhotra Exe-7 ICSE Maths Solutions of Ch-7
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 10th |
| Chapter-7 | Factor Theorem Factorization |
| Writer | OP Malhotra |
| Exe-7 | remainder and factor theorem |
| Edition | 2024-2025 |
Remainder and Factor Theorem
The remainder theorem tells us that for any polynomial f(x) , if you divide it by the binomial x−a , the remainder is equal to the value of f(a) . The factor theorem tells us that if a is a zero of a polynomial f(x) , then (x−a) is a factor of f(x) , and vice-versa
Exercise- 7
Factor Theorem Factorization Class 10 OP Malhotra Exe-7 ICSE Maths Solutions of Ch-7
Que-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.
Sol: (i) 3x³+8x²-6x+1
x+3 = 0
x = -3
3(-3)³+8(-3)²-6(-3)+1
3(-27)+8(9)+18+1
-81+72+19
= 10.
(ii) 5x³-8x²+3x-4
x-1 = 0
x = 1
5(1)³-8(1)²+3(1)-4
5-8+3-4
= -4.
(iii) x³+3x²-1
3x+2 = 0
x = -2/3
(-2/3)³+3(-2/3)²-1
(-8/27)+3(4/9)-1
(-8/27)+(4/3)-1
(-8+36-27)/27
= 1/27.
(iv) 4x³-12x²+14x-3
2x-1 = 0
x = 1/2
4(1/2)³-12(1/2)²+14(1/2)-3
4(1/8)-12(1/4)+7-3
(1/2)-3+4
(1-6+8)/2
= 3/2
Que-2: When x³+3x²-kx+4 is divided by x-2, the remainder is k. Find the value of constant k.
Sol: f(x) = x³+3x²-kx+4
Remainder = k
x-2 = 0
x = 2
f(2) = (2)³+3(2)²-2k+4 = k
8+3(4)-2k+4 = k
8+12-2k+4 = k
24-2k = k
24 = 3k
k = 8.
Que-3: Find the value of a if the division of ax³+9x²+4x-10 by x+3 leaves a remainder 5.
Sol: Let f(x) = ax3 + 9x2 + 4x – 10
x + 3 = 0 ⇒ x = –3
On dividing f(x) by x + 3, it leaves a remainder 5.
∴ f(–3) = 5
a(–3)3 + 9(–3)2 + 4(–3) – 10 = 5
–27a + 81 – 12 – 10 = 5
54 = 27a
a = 2
Que-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.
Sol: Let p(x) = ax3 + 4x2 + 3x – 4 and q(x) = x3 – 4x – a be the given polynomials.
When p(x) and q(x) are divided by (x – 2) the remainder are p(3) and q(3) respectively.
p(2) = q(2) given
a(2)3 + 4(2)2 + 3 x 2 – 4 = 23 – 4 x 2 – a
⇒ 8a + 16 + 6 – 4 = 8 – 8 – a
⇒ 8a + 18 = -a
⇒ 8a + a = -18
∴ 9a = -18
a = -18/9.
a = -2.
Que-5: Use factor theorem in each of the following to find whether g(x) is a factor of 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√8x-6 ; g(x) = x-√2
(iv) f(x) = 3x³+x²-20x+12 ; g(x) = 3x-2
Sol: (i) f(x) = x³-6x²+11x-6
g(x) = x-3, x = 3
f(3) = (3)³-6(3)²+11(3)-6
= 27-54+33-6
= 60-60
= 0.
As f(3) is zero therefore g(x), is the factor of polynomial f(x).
(ii) f(x) = 2x³-9x²+x+12
g(x) = x+1, x = -1
f(-1) = 2(-1)³-9(-1)²+(-1)+12
= -2-9-1+12
= 12-12
= 0.
As f(-1) is zero therefore g(x), is the factor of polynomial f(x).
(iii) f(x) = 7x²-2√8x-6
g(x) = x-√2, x = √2
f(√2) = 7(√2)²-2√8(√2)-6
= 14-8-6
= 14-14 = 0.
As f(√2) is zero therefore g(x), is the factor of polynomial f(x).
(iv) f(x) = 3x³+x²-20x+12
g(x) = 3x-2, x = 2/3
f(2/3) = 3(2/3)³+(2/3)²-20(2/3)+12
= 3(8/27)+(4/9)-(40/3)+12
= (8/9)+(4/9)-(40/3)+12
= (12/9)-{(40-48)/3}
= 4/3 – 4/3
= 0.
As f(2/3) is zero therefore g(x), is the factor of polynomial f(x).
Que-6: Find the value of a if x³+ax+2a-2 is exactly divisible by x+1.
Sol: f(x) = x³+ax+2a-2
x+1=0, x = -1
f(-1) = (-1)³+(-1)a+2a-2
= -1-a+2a-2
3 = a.
Que-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.
Sol: Let:
f(x) = x³+10x²+ax+b
Now,
x-1 = 0
⇒ x = 1
By the factor theorem, we can say:
f(x) will be exactly divisible by (x-1) if f(1)=0.
Thus, we have:
f(1) = 1³−10×(1)³+a×1+b
= 1+10+a+b
= 11+a+b
∴ f(1) = 0
⇒ a+b=-11 …(1)
Also,
x-2 = 0
⇒ x = 2
By the factor theorem, we can say:
f(x) will be exactly divisible by (x-2) if f(2)=0.
Thus, we have:
f(2) = 2³+10×(2)²+a×2+b
= 8+40+2a+b
= 48+2a+b
∴ f(2)=0
⇒ 2a+b = -48 …(2)
Subtracting (1) from (2), we get:
a = -37
Putting the value of a, we get the value of b, i.e., 26.
∴ a = -37 and b = 26.
Que-8: If (x-2) is a factor of x²+ax-6=0 and x²-9x+b=0, find the values of a and b.
Sol: f(x) = x²+ax-6 = 0
g(x) = x²-9x+b = 0
x-2 = 0, x = 2
f(2) = 2²+2a-6 = 0
= 4+2a-6 = 0
= 2a-2 = 0
2a = 2
a = 1
g(2) = 2²-9(2)+b = 0
= 4-18+b = 0
= b-14 = 0
b = 14.
Que-9: If both x-2 and x-(1/2) are factors of px²+5x+r, show that p=r.
Sol: Let f(x) = px² + 5x + r
Now, g(x) = x – 2
g(x) = 0
x – 2 = 0
x = 2
Put x = 2 in p(x),
p(2) = p(2)² + 5(2) + r
= 4p + 10 + r
At x = 2, p(x) = 0
So, 4p + 10 + r = 0 —————– (1)
Now, h(x) = x – 1/2
Put x = 1/2 in p(x),
p(1/2) = p(1/2)² + 5(1/2) + r
= p/4 + 5/2 + r
= (p + 2(5) + 4r)/4
= (p + 10 + 4r)/4
At x = 1/2, p(x) = 0
(p + 10 + 4r)/4 = 0
p + 10 + 4r = 0 ————— (2)
On comparing (1) and (2),
4p + 10 + r = p + 10 + 4r
By grouping,
4p – p + 10 – 10 = 4r – r
3p = 3r
p = r
Therefore, it is proved that p = r.
Que-10: If x³+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.
Sol: Let f(x) = x3 + ax2 + bx + 6
∴ x – 2 = 0 ⇒ x = 2
(2)3 + a(2)2 + b(2) + 6 = 0
8 + 4a + 2b + 6 = 0
4a + 2b + 14 = 0
2(2a + b + 7) = 0
2a + b + 7 = 02
2a + b + 7 = 0
2a + b = –7 …(i)
∴ x – 3 = 0 ⇒ x = 3
(3)3 + a(3)2 + b(3) + 6 = 3
27 + 9a + 3b + 6 = 3
9a + 3b + 33 = 3
9a + 3b = 3 – 33
9a + 3b = –30
3(3a + b) = –30
3a + b = -303
3a + b = –10 …(ii)
Subtracting (i) from (ii), we get,
2a + b = – 7
3a + b = – 10
– – +
– a = 3
∴ a = –3
Substituting the value of a = –3 in (i), we get,
2a + b = –7
2(–3) + b = –7
– 6 + b + 7 = 0
b = –7 + 6
∴ b = –1
Que-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.
Sol: (i) x³+13x²+32x+20
= x³ + x²+ 12x² + 12x + 20x + 20
=x²(x+1) + 12x(x+1) +20(x+1)
=(x² + 12x + 20)(x+1)
=(x² + 10x + 2x + 20)(x + 1)
={x(x + 10) + 2(x + 10)}(x + 1)
= (x+1)(x+2)(x+10)
(ii) Let f(x) = 4x³+20x²+33x+18 be the given polynomial.
Therefore (2x + 3)is a factor of the polynomial f(x).
Now,
f(x) = 2x²(2x+3)+7x(2x+3)+6(2x+3)
= (2x+3){2x²+4x3x+6}
= (2x+3){2x²+4x+3x+6}
= (2x+3)(2x+3)(x+2)
Hence (x +2),(2x+3) and (2x + 3 ) are the factors of polynomial f(x).
Que-12: Show that :
(i) (x-10) is a factor of x³-23x²+142x-120 and factorise it completely.
(ii) (3z+10) is a factor of 9z³-27z²-100z+300 and factorise it completely.
Sol: (i) f(x) = x³-23x²+142x-120
x-10 = 0, x = 10
f(10) = (10)³-23(10)²+142(10)-120
= 1000-2300+1420-120
= 2420-2420
= 0.
Divide x³-23x²+142x-120 by (x-10)
we get,
(x²-13x+12)(x-10)
(x-1)(x-12)(x-10)
(ii) f(x) = 9z³-27z²-100z+300
3z+10 = 0, z = -10/3
f(-10/3) = 9(-10/3)³-27(-10/3)²-100(-10/3)+300
= 9(-1000/27)-27(100/9)+(1000/3)+300
= (-1000/3)-300+(1000/3)+300
= 0.
Divide 9z³-27z²-100z+300 by (3z+10)
we get,
(3z²-19z+30)(3z+10)
(3z+10)(3z-10)(z-3)
Que-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.
Sol: f(x)= x3 + 3x2 + ax + b
Since, (x – 2) is a factor of f(x), f(2) = 0
⇒ (2)3 + 3(2)2 + a(2) + b = 0
⇒ 8 + 12 + 2a + b = 0
⇒ 2a + b + 20 = 0 …(i)
Since, (x + 1) is a factor of f(x), f(–1) = 0
⇒ (–1)3 + 3(–1)2 + a(–1) + b = 0
⇒ –1 + 3 – a + b = 0
⇒ –a + b + 2 = 0 …(ii)
Subtracting (ii) from (i), we get,
3a + 18 = 0
⇒ a = – 6
Substituting the value of a in (ii), we get,
b = a – 2
= – 6 – 2
= – 8
∴ f(x) = x3 + 3x2 – 6x – 8
Now, for x = –1,
f(x) = f(–1)
= (–1)3 + 3(–1)2 – 6(–1) – 8
= –1 + 3 + 6 – 8
= 0
Hence, (x + 1) is a factor of f(x).
∴ x3 + 3x2 – 6x – 8 = (x + 1)(x2 + 2x – 8)
= (x + 1)(x2 + 4x – 2x – 8)
= (x + 1)[x(x + 4) – 2(x + 4)]
= (x + 1)(x + 4)(x – 2)
Que-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.
Sol: Let x + 2 = 0, then x = –2
Substituting the value of x in f(x),
f(x) = x3 + ax + b
f(–2) = (–2)2 + a(–2) + b
= –8 – 2a + b
∵ x + 2 is a factor
∴ Remainder is zero
∴ –8 – 2a + b = 0
⇒ –2a + b = 8
∴ 2a – b = –8 …(i)
Again let x – 3 = 0, then x = 3
Substituting the value of x in f(x),
f(x) = x3 + ax + b
f(3) = (3)3 + a(3) + b
= 27 + 3a + b
∵ x – 3 is a factor
∴ Remainder = 0
⇒ 27 + 3a + b = 0
⇒ 3a + b = –27 …(ii)
Adding (i) and (ii)
5a = –35
⇒ a = -35/5 = –7
Substituting value of a in (i)
2(–7) –b = –8
⇒ –14 – b = –8
–b = –8 + 14
⇒ –b = 6
∴ b = –6
Hence a = –7, b = –6
(x + 2) and ( x – 3) are the factors of
x3 + ax + b
⇒ x3 – 7x – 6
Now dividing x3 – 7x – 6 by (x + 2)
(x – 3) or x² – x – 6, we get
∴ Factors are (x + 2), (x – 3) and (x + 1).
Que-15: Factorise using remainder theorem:
(i) x³-19x-30 (ii) x³+7x²-21x-27 (iii) x³-3x²-9x-5 (iv) 2x³+9x²+7x-6
Sol: (i) f(x) = x3 – 19x – 30
Let x = –2, then
f(–2) = (–2)3 – 19(–2) – 30
= –8 + 38 – 30
= 38 – 38
= 0
∴ (x + 2) is a factor of f(x)
Now dividing f(x) by (x + 2), we get
f(x) = x3 – 19x – 30
= (x + 2)(x2 – 2x – 15)
= (x + 2){(x2 –5x + 3x – 15}
= (x + 2){x(x – 5) + 3(x – 5)}
= (x + 2)(x – 5)(x + 3)
(ii) x³ + 7x² – 21x – 27
➻ x³ – 27 + 7x² – 21x
➻ (x)³ – (3)³ + 7x² – 21x
➻ (x – 3) {(x)² + x × 3 + (3)² } + 7x(x – 3)
➻ (x – 3) (x² + 3x + 9) + 7x(x – 3)
➻ (x – 3) (x² + 3x + 9 + 7x)
➻ (x – 3) (x² + 3x + 7x + 9)
➻ (x – 3) (x² + 10x + 9)
➻ (x – 3) {x² +(9 + 1)x + 9}
➻ (x – 3) {(x² + 9x + x + 9)}
➻ (x – 3) {x(x + 9) + 1(x + 9)}
➻ (x – 3) (x + 9) (x + 1)
(iii) Substitute 5 for x in x3 – 3x2 – 9x – 5.
53 – 3(5)2 – 9(5) – 5 = 125 – 75 – 45 – 5 = 0
This means, by that (x – 5) is a factor of x3 – 3x2 – 9x – 5.
Divide x3 – 3x2 – 9x – 5 by x – 5 using long divisions of polynomial :
Here, the quotient is x2 + 2x + 1.
This quadratic expression can be factorized as follows:
x2 + 2x + 1 = x2 + x + x + 1
= x (x + 1) + 1 (x + 1)
= (x + 1)(x + 1)
= (x + 1)2
This means x3 – 3x2 – 9x – 5 = (x – 5) (x2 + 2x + 1) = (x – 5) (x + 1)2
(iv) For x = -2, the value of the given expression
= 2(-2)3 + 9(-2)2 + 7(-2) – 6
= 2 × -8 + 9 × 4 – 14 – 6
= -16 + 36 – 14 – 6
= -36 + 36
= 0.
⇒ x + 2 is a factor of 2x3 + 9x2 + 7x – 6.
Now dividing 2x3 + 9x2 + 7x – 6 by (x + 2),
we get quotient = 2x2 + 5x – 3
∴ 2x3 + 9x2 + 7x – 6 = (x + 2)(2x2 + 5x – 3)
= (x + 2)(2x2 + 6x – x – 3)
= (x + 2)[2x(x + 3) – 1(x + 3)]
= (x + 2)(2x – 1)(x + 3).
Que-16: (x-3) is the HCF of x³-2x²+px+6 and x²-5x+q. Find 6p+5q.
Sol: Since (x−3) is the HCF of both polynomials, (x−3) must divide both x³−2x²+px+6 and x²−5x+q.
Divide x²−5x+q by (x-3)
we get :
q-6 = 0
q = 6
Now, divide x³−2x²+px+6 by (x-3)
we get :
6+3(p+3) = 0
6+3p+9 = 0
3p = -15
p = -5
Now, 6p+5q is
6(-5)+5(6)
-30+30
= 0.
Que-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.
Sol: (i) f(x) = x³-6x²-15x+80
x+4 = 0, x = -4
f(-4) = (-4)³-6(-4)²-15(-4)+80
= -64-6(16)+60+80
= -64-96+140
= -160+140
= -20.
(ii) f(x) = x³-3x²-12x+19
x-2 = 0, x = 2
f(2) = (2)³-3(2)²-12(2)+19
= 8-12-24+19
= 27-36
= -9
The number added is 9.
–: End of Factor Theorem Factorization Class 10 OP Malhotra Exe-7 ICSE Maths Solutions of Ch-7 :–
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