Sponsor Area
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
4x2 - 3x + 7
4x2 - 3x + 7
This expression is a polynomial in one variable x because in the expression there is only one variable (x) and all the indices of x are whole numbers.
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
y2 + 
y2 + 
This expression is a polynomial in one variable y because in the expression there is only one variable (y) and all the indices of y are whole numbers.
the exponent of t is 
which is not a whole number.
the exponent of y is (- 1) which is not a whole number.
x10+ y3 + t50
This expression is not a polynomial in one variable because in the expression, three variables (x, y and t) occur.
2 + x2 + x
Coefficient of x2 = 1.
2 - x2 + x3
Coefficient of x2 = - 1.
Sponsor Area
One example of a binomial of degree 35 is 3x35- 4.
One example of a monomial of degree 100 is 
5x3 + 4x2 + 7x
Term with the highest power of x = 5x3
Exponent of x in this term = 3
∴ Degree of this polynomial = 3.
4 - y2
Term with the highest power of y = - y2
Exponent of y in this term = 2
∴ Degree of this polynomial = 2.
3
It is a non-zero constant. So the degree of this polynomial is zero.
Write the various coefficients in each of the following polynomials:
(i) x7 - 3x5 + 4 (ii) 5y2 - 3y + 2
(iii) x4 - 3x2 + 1 (iv) 1 - 2y + 3y6
(v) 7.
(i) 1, - 3, 4
(ii) 5, - 3, 2 (iii) 1, - 3, 1
(iv) 1, - 2, 3 (v) 7.
How many terms are there in each of the following polynomials? Write them for each polynomial.
(i) 3x2 - 5x + 7 (ii) t - 7t2 + 5 - t3
(iii) x4 - 3x2 + 2 (iv) 5 - 3y + 2y6
(v) 7 (vi) 3 - 2x.
(i) 3x2 - 5x + 7
Number of terms = 3
Terms: 3x2, - 5x, 7.
(ii) t - 7t + 5 - t3
Number of terms = 4
Terms: t, - 7t2, 5, - t3.
(iii) x4 - 3x2 + 2
Number of terms = 3
Terms: x4, - 3x2, 2.
(iv) 5 - 3y + 2y6
Number of terms = 3
Terms: 5, - 3y, 2y6.
(v) 7
Number of term = 1
Term: 7 (only).
(vi) 3 - 2x
Number of terms = 2
Terms: 3, - 2x.
Solution not provided.
Ans. 1
Solution not provided.
Ans. - 1
Solution not provided.
Ans. 2
Solution not provided.
Ans. 
Find the degree of each of the polynomials given below:
(i) x5 - x4 + 3 (ii) 2 - y2 - y3 + 2y8 (iii) 2.
Solution not provided.
Ans. (i) 5 (ii) 8 (iii) 0
Point out which of the following polynomials are monomials, binomials or trinomials?
(i) x3 + 2x - 3 (ii) x3 + 8
(iii) 4x + 3x (iv) 8
(v) 5x3 + 2x3 (vi) 7x2 - 5.
(i) trinomial (ii) binomial
(iii) binomial (iv) monomial
(v) monomial (vi) binomial.
Let f(x) = 5x - 4x2 + 3
(i) Value of f(x) at x = 0
= f(0) = 5(0) - 4(0)2 + 3
= 3
Sponsor Area
Value of f(x) at x = - 1
= f(- 1) = 5(- 1) - 4(- 1)2 + 3
= - 5 - 4 + 3 = - 6
Value of f(x) at x = 2
= f(2) = 5(2) - 4(2)2 + 3
= 10 - 16 + 3 = - 3.
p(y) = y2 - y + 1
∴ p(0) = (0)2 - (0) + 1 = 1,
p(1) = (1)2 - (1) + 1 = 1,
and, p(2) = (2)2 - (2) + 1 = 4 - 2 + 1 = 3.
p(t) = 2 + t + 2t2 - t3
∴ p(0) = 2 + 0 + 2(0)2 - (0)3 = 2,
p(1) = 2 + 1 + 2(1)2 - (1)3
= 2 + 1 + 2 - 1 = 4,
and, p(2) = 2 + 2 + 2(2)2 - (2)3
= 2 + 2 + 8 - 8 = 4.
p(x) = x3
∴ p(0) = (0)3 = 0,
p(1) = (1)3= 1,
and, p( 2) = (2)3 = 8.
p(x) = (x - 1)(x + 1)
∴ p(0) = (0 - 1)(0 + 1) = (- 1)(1) = - 1,
p(1) = (1 - 1)(1 + 1) = (0)(2) = 0,
and, p(2) = (2 - 1)(2 + 1) = (1)(3) = 3
p(x) = 3x + 1, x = 
is a zero of p(x).
, x = 
p(x) = 5x - 
, x = 
is not a zero of p(x).
p(x) = x2 - 1, x = 1, - 1
p(1) = (1)2 - 1 = 1 - 1 = 0
p(-1) = (-1)2 - 1 = 1 - 1 = 0
∴ 1, - 1 are zeroes of p(x).
p(x) = (x + 1)(x - 2), x = - 1, 2
p(- 1) = (-1 + 1)(- 1 - 2)
= (0)(- 3)
= 0
p(2) = (2 + 1)(2 - 2) = (3)(0) = 0
∴ -1, 2 are zeros of p(x).
p(x) = x2, x = 0
p(0) = (0)2 = 0
∴ 0 is a zero of p(x).
p(x) =lx +m, 
is a zero of p(x).
p(x) = 3x2 - 1, 
is a zero of p(x) but 
is not a zero of p(x).
p(x) = 2x + 1, 
is not a zero of p(x).
p(x) = x + 5
p(x) = 0
⇒ x + 5 = 0 ⇒ x = - 5
∴ - 5 is a zero of the polynomial p(x).
p(x) = x - 5
p(x) = 0
⇒ x - 5 = 0 ⇒ x = 5
∴ 5 is a zero of the polynomial p(x).
p(x) = 2x + 5
p(x) = 0
is a zero of the polynomial p(x).
p(x) = 3x - 2
p(x) = 0
is a zero of the polynomial p(x).
p(x) = 3x
p(x) = 0
⇒ 3x = 0 ⇒ x = 0
∴ 0 is a zero of the polynomial p(x).
p(x) = ax, a ≠ 0
p(x) = 0
⇒ ax = 0 ⇒ x = 0 | ∵ a ≠ 0
∴ 0 is a zero of the polynomial p(x).
, c, d are real numbers
p(x)=cx + d, 
, c, d are real numbers
p(x) = 0
is a zero of the polynomial p(x).
px + q + r = 0
px = - (q + r)
is the required zero.
p(x) = 5x3 - 2x2 + 3x - 2
∴ p(1) = 5(1)3 - 2(1)2 + 3(1) - 2
= 5 - 2 + 3 - 2 = 4
and p(0) = 5(0)3 - 2(0)2 + 3(0) - 2 = - 2
Also, p(- 1) = 5(- 1)3 - 2(- 1)2 + 3(- 1) - 2
= - 5 - 2 - 3 - 2 = - 12
We have
g(x) = 3x2 - 2
is not a zero of 8(x)
is not a zero of 8(x).
We have
8(x) = 3x2 - 2
f(x) = x3 - 6x2
f(1) = (1)3 - 6(1)2 + 11 (1) - 6 = 0
x = 1 is a zero f f(x)
f(3) = (3)3 - 6(3)2 + 11(3) - 6
= 27 - 54 + 33 - 6 = 0
x = 3 id s zero of f(x)
and is also the zero of the polynomial 
then find the value of k.
p = 2
Solution not provided.
Ans = 9
at y = 2Solution not provided.
Ans. 
Solution not provided.
Ans. 4a4 + 5a3 - a2 + 6.
Solution not provided.
Ans. - 2 is a zero but 2 is not.
Solution not provided.
Ans. 2 and 0, both are zeros of the polynomial.
Solution not provided.
Ans. 2
is a zero of the polynomial p(x) = 27x3 - ax2 - x + 3, then find the value of a.
Solution not provided.
Ans. 21
Solution not provided.
Ans. 4; 
Let p(x) = x3 + 3x2 + 3x + 1
x + 1
x + 1 = 0 ⇒ x = - 1
∴ Remainder
= p(- 1) = (- 1)3 + 3(- 1)2 + 3(- 1) + 1
= - 1 + 3 - 3 + 1 = 0.
x
Remainder
= (0)3 + 3(0)2 + 3(0) + 1 = 1.
Sponsor Area
x + 
x + 
= 0 ⇒ x = - 
∴ Remainder = (- 
)3 + 3(- 
)2 + 3(- 
) + 1
= - 
3 + 3
2 - 3
+ 1.
5 + 2x
5 +2x = 0 
2x = - 5 
x = 
Remainder
Let p(x) = x3 - ax2 + 6x - a
x - a = 0
⇒ x = a
∴ Remainder = (a)3 - a(a)2 + 6(a) - a
= a3 - a3 + 6a - a
= 5a.
3x = - 7
x = 
Remainder = 
7 + 3x is not a factor of 3x3 + 7x
Divisor = x + 3
x + 3 = 0 ⇒ x = - 3
∴ Zero of x + 3 is - 3
Remainder = -20 | Given
⇒ p(- 3) = -20
⇒ k(- 3)3 + 9(- 3)2 + 4(- 3) - 8 = - 20
⇒ - 27k + 81 - 12 - 8 = - 20
⇒ 27k = 81
⇒ k = 3.
p(x) = x4 - 2x3 + 3x2 - ax + 3a - 7
By remainder theorem.
p(- 1) = 19 | x + 1 = 0 
x = -1
(- 1)4 - 2 (- 1)3 + 3 (- 1)2 - a(- 1) + 3a - 7 = 19
4a = 20
Also, the remainder when p(x) is divided by x + 2
= p(-2)
= (-2)4 - 2 (-2)3 + 3(-2)2
- a(-2) + 3a - 7
= 16 + 16 + 12 + 2a + 3a - 7
= 5a + 37
= 5(5) + 37
= 25 + 37
= 62
Let f(x) = ax3 - 3x2 + 4
and g(x) = 2x3 - 5x + a
By remainder theorem,
f(2) = p ...(1) | x - 2 = 0 ⇒ x = 2
and g(2) = q ...(2) | x - 2 = 0 ⇒ x = 2
(1) gives
a(2)3 - 3(2)2 + 4 = p
⇒ 8a - 8 = p ...(3)
(2) gives
2(2)3 - 5(2) + a = q
⇒ 6 + a = q ...(4)
According to the question,
p - 2q = 4
⇒ (8a - 8) - 2 (6 + a) = 4
⇒ 8a - 8 - 12 - 2a = 4
⇒ 6a = 24
⇒ a = 4
Let f(x) = ax3 + 3x2 - 3
and g(x) = 2x3 - 5x + a
By remainder theorem,
f(4) = p ...(1) | x - 4 = 0 ⇒ x = 4
and g (4) = q ...(2) | x - 4 = 0 ⇒ x = 4
(1) gives
a(4)3 + 3(4)2 - 3 = p
⇒ 64a + 48 - 3 = p
⇒ 64a + 45 = p ...(3)
(2) gives
2(4)3 - 5(4) + a = q
⇒ 128 - 20 + a = q
⇒ 108 + a = q ...(4)
According to the question,
2p = q
2(64a + 45) = 108 + a
128a + 90 = 108 + a
128a - a = 108 - 90
127a = 18
Solution not provided.
Ans. Quotient = 3x - 2
Remainder = 1
Solution not provided.
Ans. Quotient = 3x3 - x2 - x - 4
Remainder = - 5
Solution not provided.
Ans. Quotient = x2 - x + 1
Remainder = 0
Solution not provided.
Ans. 2
Solution not provided.
Ans. Yes
Solution not provided.
Ans. 10
Solution not provided.
Ans. 
Solution not provided.
Ans. a = 5, b = 8
Solution not provided.
Ans. 61, - 143
Solution not provided.
Ans. a = 1, b = 5
Solution not provided.
Ans. 0
Solution not provided.
Ans. 
Solution not provided.
Ans. 5
Solution not provided.
Ans. 1
x3 + x2 + x + 1
x4 + x3 + x2 + x + 1
Let p(x) = x4 + x3 + x2 + x + 1
The zero of x + 1 is - 1.
p(- 1) = (- 1)4 + (- 1)3 + (- 1)2 + (- 1) + 1
= 1 - 1 + 1 - 1 + 1 = 1 ≠ 0
∴ By factor theorem, x + 1 is not a factor of x4 + x3 + x2 + x + 1.
x4 + 3x3 + 3x2 + x + 1
Let p(x) = x4 + 3x3 + 3x2 + x + 1
The zero of x + 1 is - 1.
P(- 1) = (- 1)4 + 3(- 1)3 + 3(- 1)2 + (- 1) + 1
= 1 - 3 + 3 - 1 + 1 = 1 ≠ 0
∴ By factor theorem, x + 1 is not a factor of x4 + 3x3 + 3x2 + x + 1.
.
p(x) = 2x3 + x2 - 2x - 1, g(x) = x + 1
g(x) = 0
⇒ x + 1 = 0 ⇒ x = - 1
∴ Zero of g(x) is - 1.
Now, p(- 1)
= 2(- 1)3 + (- 1)2 - 2(- 1) - 1
= - 2 + 1 + 2 - 1 = 0
∴ By factor theorem, g(x) is a factor of p(x).
p(x) = x3 + 3x2 + 3x + 1,g(x) = x + 2
g(x) = 0
⇒ x + 2 = 0 ⇒ x = -2
∴ Zero of g(x) is - 2.
Now, p(- 2)
= (- 2)3 + 3(- 2)2 + 3(- 2) + 1
= - 8 + 12 - 6 + 1 = - 1 ≠ 0
∴ By factor theorem, g(x) is not a factor of p(x).
p(x) = x3 - 4x2 + x + 6, g(x) = x - 3
g(x) = 0
⇒ x - 3 = 0 ⇒ x = 3
∴ Zero of g(x) is 3.
Now, p( 3)
= (3)3 - 4(3)2 + 3 + 6
= 27 - 36 + 3 + 6 = 0
∴ By factor theorem, g(x) is a factor of p{x).
p(x) = x2 + x + k
If x - 1 is a factor of p(x), then p(1) = 0
| By Factor Theorem
(1)2 + (1) + k = 0
1 + 1 + k = 0
2 + k = 0
k = - 2.
p(x) = 2x2 + kx + 
If x - 1 is a factor of p(x), then p(1) = 0
| By Factor Theorem
p(x) = kx2 - 
If x - 1 is a factor of p(x), then p(1) = 0
| By Factor Theorem
k(1)2 - 
+ 1 = 0
p(x)=kx2 - 3x + k
If x - 1 is a factor of p(x) then p(1) = 0
| By Factor Theorem
k(1)2 - 3(1) + k = 0
k - 3 + k = 0
2k = 3
k = 
12x2 - 7x+ 1
12x2 - 7x + 1 = 12x2 - 4x - 3x + 1
= 4x(3x - 1) - 1(3x - 1)
= (3x - 1) (4x - 1)
2x2 + 7x + 3
2a2 + 7x + 3 = 2x2 + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (x + 3) (2x + 1)
6x2 + 5x - 6
6x2 + 5x - 6 = 6x2 + 9x - 4x - 6
= 3x(2x + 3) - 2(2x + 3)
= (2x + 3) (3x-2)
3x2 - x - 4
3x2 - x - 4 = 3x2 - 4x + 3x - 4
= x(3x - 4) + 1(3x - 4)
= (3x - 4)(x + 1)
x3 - 2x2 - x + 2
Let p(x) = x3 - x2 - x + 2
By trial, we find that
p(1) = (1)3 - 2(1)2 - (1) + 2
= 1 - 2 - 1 + 2 = 0
∴ By Factor Theorem, (x - 1) is a factor of p(x).
Now,
x3 - 2x2 - x + 2 = x2(x - 1) - x(x - 1) - 2(x - 1)
= (x - 1)(x2 - x - 2)
= (x - 1)(x2 - 2x + x - 2)
= (x - 1){x(x - 2) + 1(x - 2)}
= (x - 1)(x - 2)(x + 1).
x3 - 3x2 - 9x - 5
Let p(x) = x3 - 3x2 - 9x - 5
By trial, we find that
p(- 1) = (- 1)3 - 3(- 1)2- 9(- 1) -5
= - 1 - 3 + 9 - 5 = 0
∴ By Factor Theorem, x - (- 1), i.e., (x + 1) is a factor of p(x).
Now,
x3 - 3x2 - 9x - 5
= x2(x + 1) - 4x(x + 1) - 5(x + 1)
= (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).
Sponsor Area
x3 + 13x2 + 32x + 20
Let p(x) = x3 + 13x2 + 32x + 20
By trial, we find that
p(- 1) = (- 1)3 + 13(- 1)2 + 32(- 1) + 20
= - 1 + 13 - 32 + 20 = 0
∴ By Factor Theorem, x - (- 1), i.e., (x + 1) is a factor of p(x).
Now,
x3 + 13x2 + 32x + 20
= x2(x + 1) + 12x(x + 1) + 20(x + 1)
= (x + 1)(x2+ 12x + 20)
= (x + 1)(x2 + 2x + 10x + 20)
= (x + 1) {x(x + 2) + 10(x + 2)}
= (x + 1)(x + 2)(x + 10).
2y3 + y2 - 2y - 1
Let p(y) = 2y3 + y2 - 2y - 1
By trial, we find that
p(1) = 2(1)3 + (1)2 - 2(1) - 1
= 2 + 1 - 2 - 1 = 0
∴ By Factor Theorem, (y - 1) is a factor of p(y).
Now,
2y3 + y2 - 2y - 1
= 2y2(y - 1) + 3y(y - 1) + 1(y - 1)
= (y - 1)(2y2 + 3y + 1)
= (y - 1)(2y2 + 2y + y+ 1)
= (y - 1){2y(y + 1) + 1(y + 1)}
= (y - 1)(y + 1) (2y + 1).
Let p(x) = xn + an
The zero of x + a is - a. | x + a = 0 ⇒ x = - a
Now,
p(- a) = (- a)n + an = (- 1)nan + an
= (- 1)an + an
n is an odd positive interger
(-1)n = - 1
= - an + an = 0
∴ By Factor Theorem, x + a is a factor of xn + an for any odd positive integer n.
Let p(x) = 5x3 - x2 + 4x + b
1 - 5x = 0
The zero of 1 - 5x is 
If p(x) is divisivle by 1 - 5x, then 
| By Factor Theorem
Let p (x) = x4 + ax3 - 3x2 + 2x + b
If (x + 1) and (x - 1) are factors of p(x), then by factor theorem,
p(-1) = 0 ...(1) | x + 1 = 0 ⇒ x = -1
and p(1) = 0 ...(2) | x - 1 = 0 ⇒ x = 1
Now,
P(-1) = 0
⇒ (-1)4 + a(-1)3 - 3(-1)2 + 2 (-1) + b = 0
⇒ 1 - a - 3 - 2 + b = 0
⇒ -a + b = 4 ...(3)
and p(1) = 0
⇒ (1)4 + a(1)3 - 3(1)2 + 2 (1) + b = 0
⇒ 1 + a - 3 + 2 + b = 0
⇒ a + b = 0 ...(4)
Solving (3) and (4), we get
a = -2,b = 2
Let p(x) = 2x3 + ax2 + 11x + a + 3
If p(x) is exactly divisible by 2x - 1, then by factor theorem,
Hence, the required polynomial is
2x3 - 7x2 + 11x - 7 + 3
or 2x3 - 7x2 + 11x - 4
x2 + 2x - 3
= x2 + 3x - x - 3
= x(x + 3) - 1 (x + 3)
= (x + 3) (x - 1)
Let p(x) = x4 + 2x3 - 2x2 + 2x - 3
We see that
p(-3) = (-3)4 + 2(-3)3 - 2(-3)2 + 2(-3) - 3
= 81 - 54 - 18 - 6 - 3
= 0
Hence by converse of factor theorem, (x + 3) is a factor of p(x).
Also, we see that
p(1) = (1)4 + 2(1)3 - 2(1)2 + 2(1) - 3
= 0
Hence by converse of factor theorem, (x - 1) is a factor of p(x).
From above, we see that
(x + 3) (x - 1), i.e., x2 + 2x - 3 is a factor of p(x)
⇒ p(x) is exactly divisible by (x2 + 2x - 3).
Let p(x) = 3x3 + x2 - 20x + 12
We see that
Hence by converse of factor theorem, (3x - 2) is a factor of p(x).
Let p(x) = x3 - 6x2 + 11x - 6
By trial, we find that
p(1) = (1)3 - 6(1)2 + 11(1) - 6 = 0
∴ By converse of factor theorem, (x - 1) is a factor of p(x).
Now, x3 - 6x2 + 11x - 6
= x2 (x - 1)- 5x (x - 1) + 6 (x - 1)
= (x - 1) (x2 - 5x + 6)
= (x - 1) {x2 - 2x - 3x + 6}
= (x - 1) {x(x - 2)-3 (x - 2)}
= (x - 1)(x - 2)(x - 3)
are factors of px2 + 5x + r, show that p = r.
Let f(x) = px2 + 5 x + r
If (x - 2) is a factor of f (x), then by factor theorem
f(2) = 0 | x - 2 = 0 ⇒ x = 2
⇒ p(2)2 + 5(2) + r = 0
⇒ 4p + r + 10 = 0 ...(1)
If 
is a factor of f (x), then by factor theorem,
Subtracting (2) from (1), we get
3p - 3r = 0
⇒ p = r
Let f(x) = x4 + x3 - 7x2 - x + 6
By trial, f(1) = 0 and f(2) = 0
So by factor theorem, (x - 1) and (x - 2) are factors of f(x).
(x - 1) (x - 2) = x2 - 3x + 2
Now, f(x) = x2 (x2 - 3x + 2)+ 4x (x2 - 3x + 2) + 3 (x2 - 3x + 2)
= (x2 - 3x + 2) (x2 + 4x + 3)
= (x - 1) (x - 2) (x + 1) (x + 3)
Solution not provided.
Ans. Yes
Solution not provided.
Ans. - 3
Solution not provided.
Ans. (3x + 1) (2x + 5)
Solution not provided.
Ans. (y - 2) (y - 3)
Solution not provided.
Ans. 0
Solution not provided.
Ans. (x - 2) (x + 3) (2x - 5)
Solution not provided.
Ans. Yes
Solution not provided.
Ans. Yes, No
Solution not provided.
Ans. No
Solution not provided.
Ans. 
Use suitable identities to find the following products:
(i) (x + 4)(x + 10)
x + 4)(x + 10)
(x + 4)(x + 10) = x2 + (4 + 10)x + (4)(10)
| Using Identity IV
= x2 + 14x + 40.
Use suitable identities to find the following products:
(x + 8)(x - 10)
x + 8)(x - 10)
(x + 8)(x - 10) = (x + 8){x + (- 10)}
= x2 + {8 + (- 10)}x + (8)(- 10)
| Using Identity IV
= x2 - 2x - 80.
Use suitable identities to find the following products:
(3x + 4)(3x - 5)
(3x + 4)(3x - 5)
(3x + 4)(3x - 5) = (3x + 4) {3x + (- 5)}
= (3x)2 + (4 + (- 5)} (3x) + (4)(- 5)
| Using Identity IV
= 9x2 - 3x - 20.
Use suitable identities to find the following products:
(3 - 2x)(3 + 2x)
3 - 2x)(3 + 2x)
(3x - 2x)(3 + 2x) = (3)2 - (2x)2
| Using Identity III
= 9 - 4x2.
103 × 107
103 × 107 = (100 + 3) × (100 + 7)
= (100)2 + (3 + 7) (100 + (3)(7))
| Using Identity IV
= 10000 + 1000 + 21 = 11021.
Aliter
103 × 107 = (105 - 2) × (105 + 2)
= (105)2 - (2)2
| Using Identity III
= (100 + 5)2 - 4
= (100)2 + 2(100)(5) + (5)2 - 4
| Using Identity I
= 10000 + 1000 + 25 - 4 = 11021.
95 × 96
95 × 94 = (90 + 5) × (90 + 6)
= (90)2 + (5 + 6)(90) + (5)(6)
| Using Identity IV
= 8100 + 990 + 30 = 9120.
Aliter
95 × 96 = (100 - 5) × (100 - 4)
= {100 + (- 5)} (100 + (- 4)}
= (100)2 + {(- 5) + (- 4)}(100)
+ (- 5)(- 4) | Using Identity IV
= 10000 - 900 + 20
= 9120.
104 × 96
104 × 96 = (100 + 4) × (100 - 4)
= (100)2 - (4)2 | Using Identity III
= 10000 - 16 = 9984.
(999)3
= (1000 - 1)3
= (1000)3 - (1)3 - 3(1000)(1) (1000 - 1)
= 1000 000 000 - 1 - 3000 × 999
= 1000 000 000 - 1 - 2997000
= 997002999
(3x - 5y - 4) (9x2 + 25y2 + 15xy + 12x - 20y + 16)
= {(3x) + (-5y) + (-4)} {(3x)2 + (-5y)2 + (-4)2 - (3x) (-5y) - (-5y) (-4) - (-4) (3x)}
= (3x)3 + (-5y)3 + (-4)3 - 3 (3x) (-5y) (-4)
= 27x3 - 125y3 - 64 - 180cy.
15. Factorise:
a2 + b2 - 2 (ab - ac + bc)
a2 + b2 - 2(ab - ac + bc)
= a2 + b2 - 2ab + 2ac - 2bc
= (a - b)2 + 2c(a - b)
= (a - b) (a - b + 2c).
(ax + by)2 + (ay - bx)2
= a2x2 + b2y2 + 2abxy + a2y2 + b2x2 - 2abxy
= a2 (x2 + y2) + b2 (x2 + y2)
= (x2 + y2) (a2 + b2).
We have x = 2y + 6
⇒ x + (-2y) + (-6) = 0
∴ x3 + (-2y)3 + (-6)3 = 3x (-2y) (-6)
⇒ x3 - 8y3 - 216 = 36xy
∴ x3 - 8y3 - 36xy - 216 = 0
Factorise:
(x - 3y)3 + (3y - 7z + (7z - x)3
We have
(x - 3y) + (3y - 7z) + (7z - x) = 0
∴ (x - 3y)3 + (3y - 7z)3 + (7z - x)3
= 3(x - 3y) (3y - 7z) (7z - x)
We know that
(a + b + c)2 = a2 + b2 + c2 + 2 (ab +bc + ca)
⇒ (7)2 = a2 + b2 + c2 + 2(20)
⇒ a2 + b2 + c2 = 9
p = 2 - a
⇒ p + a - 2 = 0
∴ p3 + a3 + (- 2)3 = 3 (p) (a) (- 2)
| Using Identity VI
⇒ p3 + a3 - 8 = - 6ap
⇒ a3 + 6ap +p3 - 8 = 0
Simplify: (x + y)3 - (x - y)3 - 6y(x2 - y2)
Let x + y = a, x - y = b
Then, ab = (x + y) (x - y) = x2 - y2
and a - b = (x + y) - (x - y) = 2y
Now, (x + y)3 - (x - y)3 - 6y (x2 - y2)
= (x + y)3 - (x - y)3 - 3(2y) (x2 - y2)
= a3 - b3 - 3(a - b) ab
= (a - b)3 | Using Identity VII
= {(x + y) - (x - y)}3
= (2y)3 = 8y3
x2 + px + q = (x + a)(x + b)
⇒ x2+ px + q = x2 + ax + bx + ab
⇒ x2+ px + q = x2 + (a + b)x + ab
Comparing the coefficients, we get
a + b = p ...(1)
ab = q ...(2)
∴ x2 + pxy + qy2 = x2 + (a + b)xy + aby2
| Using (1) and (2)
= x2 + axy + bxy + aby2
= x(x + ay) + by(x + ay)
= (x + ay)(x + by).
L.H.S. = (a + b)3 + (b + c)3 + (c + a)3 - 3(a + b)(b + c)(c + a)
= {(a + b) + (b + c) + (c + a)} [(a + b)2 + (b + c)2 + (c + a)2 - (a + b)(b + c) - (b + c)(c + a) - (c + a)(a + b)]
| Using Identity VIII
= 2(a + b + c) [a2 + 2ab + b2 + b2 + 2bc + c2 + c2 + 2ca + c2 - ab - ac - b2 - bc - bc - ba - c2 - ca - ca - cb - a2 - ab]
| Using Identity I
= 2(a + b + c)(a2 + b2 + c2 - ab - be - ca)
= 2(a3 + b3 + c3 - 3abc).
| Using Identity VIII
(999)3
= (1000 - 1)3
= (1000)3 - (1)3 - 3(1000)(1) (1000 - 1)
= 1000 000 000 - 1 - 3000 × 999
= 1000 000 000 - 1 - 2997000
= 997002999
(3x - 5y - 4) (9x2 + 25y2 + 15xy + 12x - 20y + 16)
= {(3x) + (-5y) + (-4)} {(3x)2 + (-5y)2 + (-4)2 - (3x) (-5y) - (-5y) (-4) - (-4) (3x)}
= (3x)3 + (-5y)3 + (-4)3 - 3 (3x) (-5y) (-4)
= 27x3 - 125y3 - 64 - 180cy.
a(x - a) = 2ab - b(x - b)
⇒ ax - a2 = 2ab - bx + b2
⇒ ax + bx = a2 + 2ab + b2
∴ x(a + b) = (a + b)2 | Using Identity I
∴ x = a + b
| Cancelling (a + b) from both sides as (a + b) ≠ 0
Hence the given equation has a solution x = a + b.
2(a2 + b2) = (a + b)2
⇒ 2a2 + 2b2 = a2 + b2 + 2ab
⇒ a2 + b2 - 2ab = 0
⇒ (a - b)2 = 0
⇒ a - b = 0
⇒ a = b.
Factorise each of the following:
a4 - b4 (b)a4 - 16b4
a4 - b4
a4 - b4 = (a2)2 - (b2)2 = (a2 + b2) (a2 - b2)
| Using Identity III
= (a2 + b2)(a + b)(a - b)
| Using Identity III
Factorise each of the following:
a4 - 16b4
a4 - 16b4
a4 - 16b4 = (a2)2 - (4b2)2 = (a2 - 4b2)(a2 + 4b2)
| Using Identity III
= {(a)2 - (2b)2}(a2 + 4b2)
= (a + 2b)(a - 2b)(a2 + 4b2)
| Using Identity III
Factorise each of the following:
a2 - (b - c)2
a2 - (b - c)2
a2 - (b - c)2 = {a + (b - c)}{a - (b - c)}
| Using Identity III
= (a + b - c)(a - b + c)
x2 + 7xy + 12y2
x2 + 7xy + 12y2 = x2 + 3xy + 4xy + 12y2
= x(x + 3y) + 4y(x + 3y)
= (x + 3y)(x + 4y)
Factorise each of the following:
x2 + 2ax - b2 - 2ab
x2 + 2ax - b2 - 2ab
x2 + 2ax - b2 - 2ab = (x2 - b2) + (2ax - 2ab)
= (x- b)(x + b) + 2a(x - b)
| Using Identity III
= (x - b)(x + b + 2a)
Factorise each of the following:
(x2 + x)2 + 4(x2 + x) - 12
(x2 + x)2 + 4(x2 + x) - 12
(x2 + x)2 + 4(x2 + x) - 12 = y2 + 4y - 12
| where y = x2 + x
= y2 + 6y - 2y - 12
= y(y + 6) - 2(y + 6)
= (y + 6)(y - 2)
= (x2 + x + 6)(x2 + x - 2)
| ∵ = x2 + x
= (x2 + x + 6)(x2 + 2x - x - 2)
= (x2 + x + 6)[x(x + 2) - 1(x + 2)]
= (x2 + x + 6)(x + 2)(x - 1).
(i) (x - 2y - 3z)2
9x2 + 6xy + y2
9x2 + 6xy + y2 = (3x)2 + 2(3x)(y) + (y)2
= (3x + y)2 = (3x + y) (3x + y)
| Using Identity 1
4y2 - 4y + 1
4y2 - 4y + 1 = (2y)2 - 2(2y)(1)+(1)2
= (2y - 1)2 = (2y - 1) (2y - 1)
| Using Identity II
(x + 2y + 4z)2
(x + 2y + 4z)2 = (x)2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)
| Using Identity V
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8 zx
(2x - y + z)2
(2x - y + z)2 = {2x + (- y) + z}2
= (2x)2 + (- y)2 + (z)2 + 2(2x)(- y) + 2(- y)(z) + 2(z)(2x)
| Using Identity V
= 4x2 + y2 + z2- 4xy -2yz + 4 zx
(- 2x + 3y + 2z)2
(- 2x + 3y + 2 z)2 = {(- 2x) + 3y + 2z)}2
= (- 2x)2 + (3y)2 + (2z)2 + 2(- 2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)
| Using Identity V
= 4x2 + 9y2 + 4z2 - 12xy + 12yz - 8zx
(3a -7b - c)1
(3a - 7b - c)2 = {3a + (- 7b) + (- c)}2
= (3a)2 + (- 7b)2 + (- c)2 + 2(3a)(- 7b) + 2(- 7b)(- c) + 2(- c)(3a)
= 9a2 + 49b2 + c2 - 42ab + 14bc - 6ca
(- 2x + 5y - 3z)2 (-2x + 5y - 3z)2
={(-2x) + 5y + (- 3z)}2
= (- 2x)2 + (5y)2 + (- 3z)2 + 2(- 2x)(5y) + 2(5y)(- 3z) + 2(- 3z)(- 2x)
= 4x2 + 25y2 + 9z2 - 20xy - 30yz + 12zx
4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz
4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz
= (2x)2 + (3y)2 + (- 4z)2 + 2(2x) (3y) + (3y) (- 4z) + (- 4z) (- 2x)
= {2x + 3y) + (- 4z)}2 = (2x + 3y - 4z)2
= (2x + 3y - 4z) (2x + 3y - 4z)
(2x + 1)3
(2x + 1)3 = (2x)3 + (1)3 + 3(2x)(1)(2x + 1)
| Using Identity VI
= 8x3 + 1 + 6x(2x + 1)
= 8x3 + 1 + 12x2 + 6x
= 8x3+ 12x2 + 6x + 1
(2a - 3b)3
(2a - 3b)3 = (2a)3 - (3b)3 - 3(2a)(3b)(2a - 3b)
| Using Identity VII = 8a3 - 27b3 - 18ab(2a - 3b)
= 8a3 - 27b3 - 36a2b + 54ab2
(99)3
(99)3 = (100 - 1)3
= (100)3 - (1)3 - 3(100)(1)(100 - 1)
| Using Identity VII
= 1000000 - 1 - 300(100 - 1)
= 1000000 - 1 - 30000 + 300
= 970299
(102)3
(102)3 = (100 + 2)3
= (100)3 + (2)3 + 3(100)(2)(100 + 2)
| Using Identity VI
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
=1061208
(998)3
(998)3 = (1000 - 2)3
= (1000)3 - (2)2 - 3( 1000)(2)( 1000 - 2)
| Using Identity VII
= 1000000000 - 8 - 6000(1000 - 2)
= 1000000000 - 8 - 6000000 + 12000
= 994011992.
8a3 + b3+ 12a2b + 6ab2
8a3 + b3+ 12a2b + 6ab2
= (2a)3 + (b)3 + 3(2a)(b) (2a + b)
= (2a + b)3 | Using Identity VI
= (2a + b)(2a + b)(2a + b)
8a3 - b3 - 12a2b + 6ab2
8a3 - b3 - 12a2b + 6ab2
= (2a)3 - (b)3 - 3(2a)(b) (2a - b)
= (2a - b)3 | Using Identity VII
= (2a - b)(2a - b)(2a - b)
27 - 125a3 - 135a + 225a2
27 - 125a3 - 135a + 225a2
= (3)3 - (5a)3 - 3(3)(5a)(3 - 5a)
= (3 - 5a)3 | Using Identity VII
= (3 - 5a)(3 - 5a)(3 - 5a)
64a3 - 27b3 - 144a2* + 108ab2
64a3 - 27b3 - 144a2b + 108ab2
= (4a)3 - (3b)3 - 3(4a)(3b) (4a - 3b)
= (4a - 3b)3 | Using Identity VII
= (4a - 3b)(4a - 3b)(4a - 3b)
We know that
(x + y)3 = x3 + y3 + 3xy(x + y)
| Using Identity VI
⇒ x3 + y3 = (x + y)3 - 3xy(x + y)
⇒ x3 + y3 = (x + y){(x + y)2 - 3xy)
⇒ x3 + y3 = (x + y)(x2 + 2xy + y2 - 3xy)
| Using Identity I
⇒ x3 + y3 = (x + y)(x2 - xy + y2)
We know that
(x - y)3 = x3 - y3 - 3xy(x - y)
| Using Identity VII
⇒ x3 - y3 = (x - y)3 + 3xy(x - y)
x3 - y3 = (x - y){(x - y)2 + 3xy}
⇒ x3 - y3 = (x - y)(x2 - 2xy + y2 + 3ry)
| Using Identity IV
⇒ x3 - y3 = (x - y)(x2 + xy + y2).
27y3 + 125z3
27y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z){(3y)2 - (3y)(5z) + (5z)2}
= (3y + 5z)(9y2 - 15yz + 25z2)
64m3 - 343n3
64m3 - 343n3 = (4m)3 - (7n)3
= (4m - 7n){(4m)2 + (4m)(7n) + (7n)2}
= (4m - 7n)(16m2 + 28mn + 49n2).
27x3 + y3 + z3 - 9xyz
= (3x)3 + (y)3 + (z)3 - 3(3x)(y)(z)
= (3x + y + z){(3x)2 + (y)2 + (z)2 - (3x)(y) - (y)(z) - (z)(3x)}
| Using Identity VIII
= (3x + y + z)(9x2 + y2 + z2 - 3xy - yz - 3zx).
(x + y + z)[(x - y)2 + (y - z)2 + (z - x)2].
L.H.S. = x3 + y3 + z3 -3xyz
= (x + y + x) (x2 + y2 + z2 -xy - yz -zx)
| Using Identity VIII
(x+y+z){2x2 + 2y2 + z2 -xy -yz -zx)
(x+y+z){2x2 + 2y2 + 2z2 -2xy -2yz -2zx)
(x + y + z) {x2 - 2xy + y2 ) (y2 -2yz + z2) + (z2 - 2zx + x2 )}
| Using Identity II
We know that
x3 + y3 + z3 - 3xyz
= (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
| Using Identity VIII
= (0)(x2 + y2 + z2 - xy - yz - zx)
| ∵ x + y + z = 0
= 0
⇒ x3 + y3 + z3 = 3xyz.
(28)3 + (- 15)3 + (- 13)3.
(- 12)3 + (7)3 + (5)3
(- 12)3 + (7)3 + (5)3 = 3(- 12)(7)(5)
| ∵ (- 12) + (7) + (5) = 0
= - 1260
(28)3 + (- 15)3 + (- 13)3
(28)3 + (- 15)3 + (- 13)3
= 3(28)(- 15)(- 13)
| ∵ (28) + (- 15) + (- 13) = 0
= 16380.
25a2 - 35a + 12
= 25a2 - 20a - 15a + 12
= 5a(5a - 4) - 3(5a - 4)
= (5a - 4)(5a - 3)
∴ The possible expressions for the length and breadth of the rectangle are 5a - 3 and 5a - 4.
35y2 + 13y - 12
= 35y2 + 28y - 15y - 12
= 7y (5y + 4) - 3(5y + 4)
= (5y + 4)(7y - 3)
∴ The possible expressions for the length and breadth of the rectangle are 7y - 3 and 5y + 4.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
3x3 - 12x
= 3x(x - 4)
∴ The possible expressions for the dimensions of the cuboid are 3, x and x - 4.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
12ky2 + 8ky - 20k
= 4k(3y2 + 2y - 5)
= 4k(3y2 + 5y - 3y - 5)
= 4k{y(3y + 5) - 1(3y +5)}
= 4k(3y + 5)(y - 1)
∴ The possible expressions for the dimensions of the cuboid are 4k, 3y + 5 and y - 1.
Solution not provided.
Ans. 9a2 + 16b2 + 25c2 + 24ab + 40bc + 30ac
Solution not provided.
Ans. 27a3 + 64b3 + 108a2b + 144ab2
Solution not provided.
Ans. 125p3 - 27q3 - 225p2q + 135pq2
Solution not provided.
Ans. (i) 1124864 (ii) 997002999
Solution not provided.
Ans. (2x + y + 3z) (4x2 + y2 + 9z2 - 2xy - 3yz - 6xz)
Solution not provided.
Ans. (x - y) (x2 + xy + y2) (x + y) (x2 - xy + y2)
Solution not provided.
Ans. 1
Solution not provided.
Ans. 4
A.
zero polynomialA.
monomialA.
has one and only one zero1
-1
0
does not exist
D.
does not exist
0
abc
3abc
D.
3 (a - b) (b - c) (c - a)A.
(a + b + c) (a2 + b2 + c2 - ab - be - ca)C.
x2 + y2 + z2 + 2xy - 2yz - 2zx(x + 1)(x + 3)(x - 5)
A.
(x + 1)(x + 3)(x + 5)x2 + 4y2 + z2 + 4xy + 4yz + 2zx
A.
x2 + 4y2 + z2 + 4xy + 4yz + 2zx
A.
x2 + y2 + z2 - 2xy + 2yz - 2zxA.
(y - 2) (y - 3)A.
3x + 1) (2x + 5)A.
(2a + 3b + c)2C.
6(2a - b)(b - 2c) (c - a)Solution not provided.
Ans. (a - b + 1) (a2 + b2 + 1 + ab + b - a)
Solution not provided.
Ans. - 1; (x - 3) (x + 1)
Solution not provided.
Ans. (x - 3) (x + 1) (x - 4) (x + 2)
Prove that 2x3 + 2y3 + 2z3 - 6xyz
= (x + y + z) [(x - y)2 + (y - z)2 + (z - x)2].
Solution not provided.
Ans. 1624
If x2 - 3x + 2 is a factor of x4 - ax2 + b, then find a and b.
Solution not provided.
Ans. a = 5, b = 4
The volume of a cube is given by the polynomial
p(x) = 8x3 - 36x2 + 54x - 27
Solution not provided.
Ans. 2x - 3
Using factor theorem, factorise the polynomial:
x4 + 3x3 + 2x2 - 3x - 3.
Solution not provided.
Ans. (x - 1) (x + 1) (x - 4) (x + 2)
Solution not provided.
Ans. 
Solution not provided.
Ans. 3x3 - x2 -x -4 ; - 5
Solution not provided.
Ans. 20
Solution not provided.
Ans. 0
Sponsor Area
Sponsor Area
when z =3.
Remainder 
then find the value of 
is
then the value 
.
then the value 
.