Mathematics
Class-X
2015-2016
Q.1. Show that (x – 1) is a factor of x3 – 7x2 + 14x – 8. Hence, factorize the above
polynomial completely.
Solution :
If (x – 1) is a factor of f(x) = x3 – 7x2 + 14x – 8, then f(1) = 0.
Now, f(1) = 13 – 7(1)2 + 14(1) – 8 = 1 – 7 + 14 – 8 = 0.
Hence, (x – 1) is a factor of f(x).
To find other factor, f(x) = x3 – 7x2 + 14x – 8 = x2(x – 1) – 6(x – 1) + 8(x – 1)
=(x – 1) (x2 – 6x +8) = (x – 1){x(x – 4) – 2(x – 4)}
= (x – 1)(x – 2)(x – 4).[Ans.]
Q.2. Show that 2x + 7 is a factor of 2x3 + 5x2 – 11x – 14. Hence factorize the given
expression completely using factor theorem.
Solution :
If 2x + 7 is a factor of 2x3 + 5x2 – 11x – 14 , then putting 2x + 7 = 0
Or x = – 7/2 we get,
f (– 7/2) = 2 (– 7/2)3 + 5 (– 7/2)2 – 11 (– 7/2) – 14
= – 343/4 + 245/4 + 77/4 – 14 = – 399/4 + 245/4 + 154/4
= (399 – 399)/4 = 0.
Hence, 2x + 7 is a factor.
Dividing 2x3+ 5x2 – 11x – 14 by 2x + 7 we get
2x3 + 5x2 – 11x – 14 = (2x + 7)(x2 – x – 2).
Now, x2 – x – 2 = x(x – 2) + (x – 2) = (x + 1)(x – 2).
Hence, 2x3 + 5x2 – 11x – 14 = (2x + 7)(x + 1)(x – 2). [Ans.]
Q.3. Find the remainder when f(x) = x3 – 6x2 + 9x + 7 is divided by
g(x) = x – 1.
Solution :
When f(x) is divided by g(x) = x – 1, then remainder, R = f(1), by remainder theorem. Hence, R
= f(1) = (1)3 – 6 (1)2 +9(1) + 7
= 1 – 6 + 9 + 7 = 11. [Ans.]
Q.4. Find the remainder when 2x3 – 3x2 + 7x – 8 is divided by x – 1.
Solution :
Do yourself. [Ans. = – 2.]
Q.5. For what value of ‘a’ , the polynomial g(x) = x – a is a factor of
f(x) = x3 – ax2 +x + 2.
Solution :
As, x – a is a factor of f(x), therefore, f(a) = 0
i.e. a3 – a × a2 + a + 2 = 0
or, a3 – a3 + a + 2 = 0
or, a + 2 = 0
or, a = – 2. [Ans.]
Q.6. Find the value of a, if x – a is a factor of x3 – a2x + x + 2.
Solution :
Do yourself [Ans. a = – 2 ]
Sudheer Gupta .
Be positive and constructive. 
Page 1
Mathematics
Class-X
2015-2016
Q.7. Find the value of p and q, if (x + 3) and (x – 4) are the factors of
x3 – px2 – qx + 24.
Solution :
Le f(x) = x3 – px2 – qx + 24.
As, x + 3 is a factor of f(x), f(– 3) = 0
i.e. (– 3)3 – p(– 3)2 – q(– 3) + 24 = 0
or, – 27 – 9p + 3q + 24 = 0
or, – 9p + 3q – 3 = 0 --------------------------- (i)
Also x – 4 is a factor of f(x), then f(4) = 0
i.e. (4)3 – p(4)2 – q(4) + 24 = 0
or, 64 – 16p – 4q + 24 = 0
or, – 16p – 4q + 88 = 0 ---------------------- (ii)
Multiplying eqn.(i) by 4 and eqn.(ii) by 3 and then adding we get
– 84p + 252 = 0 or, p = 3. [Ans.]
Putting value of p in eqn.(i) we get,
9×3 + 3q – 3 = 0
Or, 3q – 30 = 0
Or, q = 10. [Ans.]
Q.8. 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 :
Do yourself. [Ans. a = 3, b = – 4.]
Q.9. Find the value of q if the polynomial f(x) = x3 + 2x2 – 13x + q is divisible by g(x) = x – 2.
Hence find all the factors.
Solution :
As, f(x) = x3 + 2x2 – 13x + q is divisible by g(x) = x – 2,
Therefore, f(2) = 0
Or, (2)3 + 2(2)2 – 13(2) + q = 0
Or, 8 + 8 – 26 + q = 0
Or, q = 10. [Ans.]
Now, f(x) = x3 + 2x2 – 13x + 10
= x3 – 2x2 + 4x2 – 8x – 5x + 10
= x2(x – 2) + 4x(x – 2) – 5(x – 2)
= (x – 2)(x2 + 4x – 5)
= (x – 2)(x2 + 5x – x – 5)
= (x – 2){x(x + 5) – 1(x + 5)}
= (x – 2)(x – 1)(x + 5). [Ans.]
Q.10. (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 value of a and b.
Solution :
As (x – 2) is a factor of f(x) = x3 + ax2 + bx + 6,
Therefore, f(2) = 0
Or, (2)3 + a(2)2 + b(2) + 6 = 0
Sudheer Gupta .
Be positive and constructive. 
Page 2
Mathematics
Class-X
2015-2016
Or, 8 + 4a + 2b + 6 = 0
Or, 4a + 2b = – 14
Or, 2a + b = – 7 -------------------- (i)
On dividing f(x) by (x – 3) the remainder is 3
Therefore, f(3) = 3
Or, (3)3 + a(3)2 + b(3) + 6 = 3
Or, 27 + 9a + 3b + 6 = 3
Or, 9a + 3b = – 30
Or, 3a + b = – 10 -------------------- (ii)
Subtracting (ii) from (i) we get a = – 3
Substituting value of a = – 3 in (i) we get
2×(– 3) + b = – 7
Or, – 6 + b = – 7
Or, b = – 1.
Therefore, a = – 3, b = – 1. [Ans.]
Q.11. Use the factor theorem to factorize completely.
x3 + x2 – 4x – 4.
Solution :
We have, x3 + x2 – 4x – 4.
Let x + 1 is a factor, then x + 1 = 0 => x = – 1.
Putting x = – 1, we get
(– 1)3 +(– 1)2 – 4 (– 1) – 4 = – 1 + 1 + 4 – 4 = 0.
Therefore, x + 1 is a factor.
To factorize completely x3 + x2 – 4x – 4 = x2(x + 1) – 4(x + 1)
= (x + 1)(x2 – 4)
= (x + 1)(x + 2)(x – 2). [Ans.]
Q.12. Using Factor theorem, show that (x – 3) is a factor of x3– 7x2 + 15x – 9. Hence,
factorise the given expression completely.
Solution :
Do yourself. [Ans. = (x – 1)(x – 3)2]
Sudheer Gupta .
Be positive and constructive. 
Page 3