2 POLYNOMIALS CHAPTER

CHAPTER
2
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POLYNOMIALS
Points to Remember :
1. A symbol having a fixed numerical value is called a constant. For e.g. 9 ,  7 , , 2 etc.
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2. A symbol which may take different numerical values is known as a variable. We usually denotes variable
by x, y, z etc.
3. A combination of constants and variables which are connected by basic mathematical operations, is
known as Algebraic Expression. For e.g. x2 – 7x + 2, xy2 – 3 etc.
4. An algebraic expression in which variable have only whole numbers as a power is called a Polynomial.
5. Highest power of the variable is called the degree of the polynomial. For e.g. 7x3 – 9x2 + 7x – 3 is a
polynomial in x of degree 3.
6. A polynomial of degree 1 is called a linear polynomial. For e.g. 7x + 3 is a linear polynomial in x.
7. A polynomial of degree 2 is called a Quadratic Polynomial. For e.g. 3y2 – 7y + 11 is a Quadratic polynomial in y.
8. A polynomial of degree 3 is called a Cubic Polynomial. For e.g. 3t3 – 7t2 + t – 3 is a cubic polynomial in
t.
9. According to number of terms, a polynomial having one non-zero term is a monomial, a polynomial
having two non-zero terms is a bionomial and a polynomial have three non-zero terms is a trinomial.
10. Remainder Theorem : Let f(x) be a polynomial of degree n  1 and let a be any real number. If f(x) is
divided by linear polynomial (x – a), then the remainder is f(a).
11. Factor Theorem : Let f(x) be a polynomial of degree n > 1 and a be any real number.
(i) If f(a) = 0, then x – a is a factor of f(x).
(ii) If (x – a) is a factor of f(x) then f(a) = 0.
12. Algebraic Identities :
(i) (x + y)2 = x2 + 2xy + y2
(ii) (x – y)2 = x2 – 2xy + y2
2
2
(iii) x – y = (x – y) (x + y)
(iv) (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz
3
3
3
(v) (x + y) = x + y + 3xy (x + y)
(vi) (x – y)3 = x3 – y3 – 3xy (x – y)
3
3
2
2
(vii) x – y = (x – y) (x + xy + y )
(viii) x3 + y3 = (x + y) (x2 – xy + y2)
3
3
3
2
(ix) x + y + z – 3xyz = (x + y + z) (x + y2 + z2 – xy – yz – xz)
13. If x + y + z = 0 then, x3 + y3 + z3 = 3xyz.
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ILLUSTRATIVE EXAMPLES
Example 1. Which of the following expressions are Polynomials? In case of a polynomial, give degree of
polynomial.
(i) x4 – 3x3 + 7x2 + 3
(iv) x 
14
1
x
(ii)
(v) 7
3 y3  7 y  6
(iii)
x 3
(vi) x 2 / 3 
POLYNOMIALS
2
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MATHEMATICS–IX
Solution.
(i) is a polynomial of degree 4.
(ii) is a polynomial of degree 3.
(v) is a polynomial of degree 0.
Example 2. Verify whether the following are zeros of the polynomial, indicated against them.
1
3
2
(iii) p(x) = x – 1, x = 1, –1
(ii) p ( x )  5 x   , x 
(v) p(x) = x2, x = 0
(vi) p(x) = lx + m, x 
(vii) p(x) = 3x2 – 1, x 
Solution.
1
,
3
(i) We have, p(x) = 3x + 1
At x 
2
3
1  1    1 
, p     3   1  1  1  0
3  3  3 
1
is a zero of p(x).
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(ii) We have, p(x) = 5x – 

4 4
4
, p   5     4    0
5 5
5
4
is not a zero of p(x).
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(iii) We have, p( x)  x 2  1
m
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(viii) p(x) = 2x + 1, x 

At x 
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5
(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
(i) p ( x )  3 x  1, x 
B
1
2
–NCERT
At x = 1, p(1) = (1)2 – 1 = 1 – 1 = 0
also, at x = – 1, p(–1) = (–1)2 – 1 = 1 – 1 = 0
 1, – 1 both are zeros of p(x).
(iv) We have, p(x) = (x + 1) (x – 2)
At x = – 1, p(–1) = (–1 + 1) (–1 –2) = 0 × (–3) = 0
also, at x = 2, p(2) = (2 + 1) (2 – 2) = 3 × 0 = 0
 –1, 2 both are zeros of p(x).
(v) We have, p(x) = x2
At x = 0, p(0) = 02 = 0
 0 is a zero of p(x).
(vi) We have, p(x) = lx + m
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At, x  
m m m
, p
  l
  m  m  m  0
l
 l   l 
m
is a zero of p(x).
l
(vii) We have, p(x) = 3x2 – 1

MATHEMATICS–IX
POLYNOMIALS
15
At, x 
and at, x 

1
2
 1 
 1 
1
 3
  1  3   1  1  1  0
, p



3  3
 3
 3
1
is a zero of p(x), but
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(viii) We have p(x) = 2x + 1
At x 
2
is not a zero of p(x).
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1 1
1
, p   2   1  1  1  2  0
2 2
2
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is not a zero of p(x).
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Example 3. Find the zero of the polynomial in each of the following cases :
(i) p(x) = x + 5
(ii) p(x) = x – 5
(iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2
(v) p(x) = 3x
(vi) p(x) = ax ; a  0
(vii) p(x) = cx + d, c  0, c, d are real numbers.
Solution.
We know, finding a zero of p(x), is the same as solving the equation p(x) = 0.
(i) p(x) = 0  x + 5 = 0  x = – 5
 –5 is a zero of p(x).
(ii) p(x) = 0  x – 5 = 0  x = 5
 5 is a zero of p(x).

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(iii) p(x) = 0  2x + 5 = 0  2x = – 5  x 

5
is a zero of p(x).
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(iv) p(x) = 0  3x – 2 = 0  3x = 2  x 

5
2
2
3
( Given a  0)
(vii) p(x) = 0  cx + d = 0  cx = – d  x  
16
—NCERT
2
is a zero of p(x).
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(v) p(x) = 0  3x = 0  x = 0
 0 is a zero of p(x).
(vi) p(x) = 0  ax = 0  x = 0
 0 is a zero of p(x).

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 2   2 
4
  3
  1  3   1  4  1  3  0
, p
3  3  3
3
2
d
c
d
is a zero of p(x).
c
POLYNOMIALS
MATHEMATICS–IX
Example 4. Using remainder theorem, find the remainder when p(x) = 2x3 – 5x2 + 9x – 8 is divided by (x – 3).
Solution.
Remainder = p(3)
= 2(3)3 – 5(3)2 + 9 (3) – 8
= 54 – 45 + 27 – 8 = 81 – 53 = 28 Ans.
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Example 5. Find the remainder when x 3  3x 2  3x  1 is divided by :
(ii) x 
(i) x + 1
Solution.
1
2
(iv) x + 
(iii) x
(i) x + 1 = 0  x = – 1
 by remainder theorem, the required remainder is p (–1).
Now,
p(x) = x3 + 3x2 + 3x + 1

p(–1) = (–1)3 + 3(–1)2 + 3(–1) + 1
= –1 + 3 – 3 + 1 = 0
 Remainder = 0
(ii) x 
1
1
0x .
2
2
Now,
2
1 1
1
1
p      3   3   1
2 2
2
2

B
1 3 3
1  6  12  8
27
  1 

8 4 2
8
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—NCERT
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1
 by remainder theorem, the required remainder is p  .
2
3
(v) 5 + 2x
 Remainder  27
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(iii) x = 0 by remainder theorem, the required remainder is p(0).
Now,
p(0) = (0)3 + 3(0)2 + 3(0) + 1 = 1
 Remainder = 1
(iv) x +  = 0  x = – 
 by remainder theorem, the required remainder is p(–).
Now, p(–) = (–)3 + 3(–)2 + 3 (–) + 1
= –3 + 32 – 3 + 1
 Remainder = –3 + 32 – 3 + 1.
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(v) 5 + 2x = 0  2x = – 5  x =
5
2
 5
.
 By remainder theorem, the required remainder is p 
 2 
3
2
 5  5
 5
 5 
Now, p

  3
  3
 1
 2   2 
 2 
 2 

MATHEMATICS–IX
125 75 15

 1
8
4
2
POLYNOMIALS
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
125  150  60  8 27

8
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 Remainder   27 .
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Example 6. Using Factor theorem, show that (x + 2) is a factor of x4 – x2 – 12.
Solution.
Let p(x) = x4 – x2 – 12.
Now, x + 2 = 0  x = – 2.
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(x + 2) is a factor of p(x)  p(–2) = 0
Now, p(–2) = (–2)4 – (–2)2 – 12
= 16 – 4 – 12 = 0,
which shows that (x + 2) is a factor of p(x).
Example 7. Use the factor theorem to determine whether g(x) is a factor of p(x) is each of the following cases:
(i) p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1
(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
(iii) p(x) = x3 – 4x2 + x + 6, g(x) = x – 3
—NCERT
Solution.
(i) In order to prove that g(x) = x + 1 is a factor of p(x) = 2x3 + x2 – 2x – 1, it is sufficient to show that
p(–1) = 0.
Now, p(–1) = 2 (–1)3 + (–1)2 – 2(–1) – 1
=–2+1+2–1=0
 g(x) is a factor of p(x).
(ii) In order to prove that g(x) = x + 2 is a factor of p(x) = x3 + 3x2 + 3x + 1, it is sufficient to show
that p(–2) = 0.
Now, p (–2) = (–2)3 + 3(–2)2 + 3(–2) + 1
= – 8 + 12 – 6 + 1 = – 1  0.
 g(x) is not a factor of p(x).
(iii) In order to prove that g(x) = x – 3 is a factor of p(x) = x3 – 4x2 + x – 6, it is sufficient to show
that p(3) = 0
Now, p(3) = (3)3 – 4(3)2 + 3 – 6 = 27 – 36 + 3 – 6 = – 12  0.
 g(x) is not a factor of p(x).
Example 8. If x3 + ax2 + bx + 6 has x – 2 as a factor and leaves a remainder 3, when divided by x – 3, find the
value of a and b.
Solution.
Let p(x) = x3 + ax2 + bx + 6
Since, x–2 is a factor of p(x)  p(2) = 0
( factor theorem)
 (2)3 + a(2)2 + b(2) + 6 = 0  2a + b = –7
...(1)
Also, p(x) leaves remainder 3, when divided by x – 3.

p(3) = 3
( Remainder theorem)
3
2
 (3) + a(3) + b(3) + 6 = 3
 3a + b = – 10
...(2)
Solving (1) and (2), we get a = – 3, b = – 1 Ans.
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Example 9. Factorize the following:
(i) 6x2 – 18xy
(ii) x3 + 7x2 – x – 7
(iii) 16a2 – 81b2
2
2
(iv) x + 5x – 24
(v) 9x – 22x + 8
Solution.
(i) 6x (x – 3y)
(ii) x2 (x + 7) – 1 (x + 7) = (x2 – 1) (x + 7) = (x – 1) (x + 1) (x + 7)
(iii) (4a)2 – (9b)2 = (4a + 9b) (4a – 9b)
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(iv) x 2  8 x  3x  24  x( x  8)  3( x  8) = (x – 3) (x + 8)
(v) 9x2 – 18x – 4x + 8 = 9x (x – 2) – 4(x – 2) = (x – 2) ( 9x – 4)
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Example 10. Factorise :
(i) 12x2 – 7x + 1
(ii) 2x2 + 7x + 3
(iii) 6x2 + 5x – 6
(iv) 3x2 – x – 4 –NCERT
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Solution.
(i) 12x – 7x + 1
here, p + q = coefficient of x = – 7
pq = coefficient of x2 × constant term = 12 × 1 =12

p + q = – 7 = – 4 – 3, pq = 12 = (–4) × (–3)
 12x2 – 7x + 1 = 12x2 – 4x – 3x + 1 = 4x (3x – 1) – 1(3x – 1) = (4x – 1) (3x – 1) Ans.
(ii) 2x2 + 7x + 3 = 2x2 + x + 6x + 3 = x (2x + 1) + 3 (2x + 1) = (x + 3) (2x + 1)
(iii) 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 = 3x (2x + 3) – 2(2x + 3) = (3x – 2) (2x + 3)
(iv) 3x2 – x – 4 = 3x2 + 3x – 4x – 4 = 3x(x + 1) – 4(x + 1) = (3x – 4) (x + 1)
Example 11. Factorize x3 – 3x2 – 9x – 5 using factor theorem.
Solution.
Let p(x) = x3 – 3x2 – 9x – 5.
factors of constant term are ±1 and ±5.
Now, p(1) = 13 – 3(1)2 – 9(1) – 5  0
p(–1) = (–1)3 – 3(–1)2 – 9(–1) – 5  0.
 (x + 1) is a factor of p(x).

p(x) = (x + 1) (x2 – 4x – 5)
= (x + 1) [x2 – 5x + x – 5]
= (x + 1) [x(x – 5) + 1 (x – 5)]
= (x + 1) (x + 1) (x – 5) Ans.
Example 12. Factorise : x3 + 13x2 + 32x + 20
—NCERT
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Solution.
Let
p(x) = x + 13x + 32x + 20
Now, factors of constant term 20 are ± 1, ± 2, ± 5, ± 10 and ± 20.
Now, p(1) = (1)3 + 13(1)2 + 32(1) + 20 = 1 + 13 + 32 + 20 = 66  0
p(–1) = (–1)3 + 13(–1)2 + 32(–1) + 20 = –1 + 13 – 32 + 20 = – 33 + 33 = 0
 (x + 1) is a factor of x3 + 13x2 + 3x + 20.
Now, divide p(x) by x + 1.
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MATHEMATICS–IX
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p(x) = (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) Ans.
Example 13. Expand the following :
(i) (a – 2b + 3c)2
(ii) (2x + y)3
(iii) (3x – 2y)3
Solution.
(i) (a)2 + (–2b)2 + (3c)2 + 2(a) (–2b) + 2(–2b) (3c) + 2 (a) (3c)
= a2 + 4b2 + 9c2 – 4ab – 12bc + 6ac
(ii) (2x)3 + (y)3 + 3(2x)(y) [2x + y]
= 8x3 + y3 + 6xy (2x + y)
= 8x3 + y3 + 12x2y + 6xy2
(iii) (3x)3 – (2y)3 – 3(3x) (2y) [3x – 2y]
= 27x3 – 8y3 – 18xy (3x – 2y)
= 27x3 – 8y3 – 54x2y + 36xy2
Example 14. Factorize the following :
(i) x3 + 27y3
(ii) 27a3 – 64b3
(iii) a3 – 8b3 + 64c3 + 24abc
Solution.
(i) (x)3 + (3y)3 = (x + 3y) (x2 – 3xy + 9y2)
[  a3 + b3 = (a + b) (a2 – ab + b2)]
3
3
2
2
(ii) (3a) – (4b) = (3a – 4b) (9a + 12ab + 16b )
[ a3 – b3 = (a – b) (a2 + ab + b2)]
(iii) (a)3 + (–2b)3 + (4c)3 – 3 × a × (–2b) × (4c)
= (a – 2b + 4c) (a2 + 4b2 + 16c2 + 2ab + 8bc – 4ac)
[ a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)]

Consider,
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( x  y  z ).[( x  y ) 2  ( y  z ) 2  ( z  x) 2 ]
2
LHS = x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
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Example 15. Verify that : x3 + y3 + z3 – 3xyz 
Solution.
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
1
(x + y + z) . 2 (x2 + y2 + z2 – xy – yz – zx)
2

1
(x + y + z) (2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx)
2
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
1
(x + y + z) (x2 + x2 + y2 + y2 + z2 + z2 – 2xy – 2yz – 2zx)
2

1
(x + y + z) [(x2 + y2 – 2xy) + (y2 + z2 – 2yz) + (z2 +x2 – 2zx)]
2
1
(x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
2
= RHS. Hence verify.
Example 16. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
Solution.
We have x + y + z = 0
 x+y=–z
Cubing both sides, we get (x + y)3 = (–z)3
 x3 + y3 + 3xy (x + y) = –z3
 x3 + y3 + 3xy (–z) = – z3
( x + y = – z)
–NCERT

20
POLYNOMIALS
—NCERT
MATHEMATICS–IX
 x3 + y3 + z3 – 3xyz = 0
 x3 + y3 + z3 = 3xyz. Hence shown.
Example 17. Give possible expressions for the length and breadth of each of the following rectangles, in which
their areas are given :
(i) Area : 25a2 – 35a + 12
(ii) 35y2 + 13y – 12
—NCERT
Solution.
(i) Given, Area of rectangle = 25a2 – 35a + 12
= 25a2 – 15a – 20a + 12
= 5a (5a – 3) – 4 (5a – 3)
= (5a – 3) (5a – 4)
 Possible length and breadth are (5a – 3) and (5a – 4) units.
(ii) Area of given rectangle = 35y2 + 13y – 12
= 35y2 + 28y – 15y – 12
= 7y (5y + 4) – 3 (5y + 4)
= (5y + 4) (7y – 3).
 Possible length and breadth are (5y + 4) and (7y – 3) units.
Example 18. What are the possible expressions for the dimensions of the cuboid whose volumes are given
below:
(i) Volume : 3x2 – 12x
(ii) Volume : 12ky2 + 8 ky – 20 k
—NCERT
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Solution.
(i) Volume : 3x – 12x = 3x (x – 4) = 3 × x × (x – 4)
 Possible dimensions of cuboid are 3, x and (x – 4) units.
(ii) Volume = 12ky2 + 8ky – 20k
= 4 k (3y2 + 2y – 5)
= 4k (3y2 – 3y + 5y – 5)
= 4k [3y (y – 1) + 5 (y – 1)]
= 4k (y – 1) (3y + 5)
 Possible dimensions of a cuboid are 4k, (y – 1) and (3y + 5) units.
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PRACTICE EXERCISE
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1. Which of the following expressions are polynomials in one variable?
(i) 3x2 – x3 + 7x + 1
4
x
(v) 2x4 – 7x3 + 3
(ii) x 
(iv) y 3 / 2  y 2  1
2. Write degree of each the following polynomial:
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(i) 7x4 – 9x3 + 2x + 4
(ii) 7 – y2 + y3
(iv) 10
(v) 3x2 – 7x + 4
3. Classify each of the following as linear, quadratic and cubic polynomial:
(i) 3x3 – 7x
(ii) 4 y2 + 3y – 1
2
(iii)
2 y2  7 y  2
(iii)
5t 3  7t 2  1
(iii) 7r
3
(iv) x  3x
(v) y  y
3
2
4. Find the value of p(x) = 5x – x + 3x + 4 at
(i) x = 0
(ii) x = 2
(iii) x = – 1
5. Find the value of p(0), p(2) and p(–3) where p(x) = x3 – x2 + x – 1.
6. Verify whether the following are zeros of the polynomial, indicated against them:
3
2
(iii) p(x) = x2 + x – 6; x = 3, –2
(i) p ( x )  2 x  3; x 
MATHEMATICS–IX
(ii) p(x) = (x + 3) (x – 4) ; x = – 3, 4
(iv) p(x) = x – x3 ; x = 0, 1, –1
POLYNOMIALS
21
7. Find the zero of the polynomial in each of the following :
(i) p(x) = x – 4
(ii) p(x) = 3x + 4
(iii) p(x) = 7x
(iv) p(x) = rx + s; r  0, r, s are real numbers.
8. Using Remainder Theorem, find the remainder when :
(i) 4x3 – 7x2 + 3x – 2 is divided by x – 1
(ii) x3 – 7x2 + 6x + 4 is divided by x – 3
(iii) x3 + 2x2 – x + 3 is divided by x + 3
(iv) 4x3 – 4x2 + x – 2 is divided by 2x + 1
(v) x3 – ax2 + 5x + a is divided by x – a
(vi) x3 + ax2 – 6x + 2a is divided by x + a
9. If 35 is the remainder when 2x2 + ax + 7 is divided by x – 4, find the value of a.
10. Without actual division, prove that 2x3 + 13x2 + x – 70 is exactly divisible by x – 2.
11. Show that x – 1 is a factor of x3 – 2x2 – 5x + 6.
12. Find the value of p for which the polynomial x4 – 2x3 + px2 + 2x + 8 is exactly divisible by x + 2.
13. Find values of a and b so that the polynomial x4 + 7x3 + 4x2 + ax + b is exactly divisible by x – 1
and x + 3.
14. The polynomials ax3 – 3x2 + 7 and 2x3 + 7x – 2a are divided by x + 3. If the remainder in each case is same,
find the value of a.
15. The polynomials ax3 + 4x2 – 3 and 4x3 + 4x – a when divided by x – 3, leaves the remainder R1 and R2
respectively. Find the value of a if R1 = 3R2.
16. Find the integral zeros of x3 – 3x2 – x + 3.
17. Find the integral zeros of x3 + 4x2 – x – 4.
18. Show that x – 2 is a factor of p(x) = x3 – 6x2 + 15x – 14.
19. Show that x + 4 is a factor of x4 + 3x3 – 4x2 + x + 4.
20. Find value of a for which x + a is a factor of f(x) = x3 + ax2 + 3x + a + 4.
21. Without actual division, prove that 2x4 + x3 – 2x2 + 5x – 6 is exactly divisible by x2 + x – 2.
22. For what value of a is the polynomial 2x3 – ax2 + 8x + a + 4 is exactly divisible by 2x + 1.
23. If x3 + ax2 + bx – 12 has x – 3 as a factor and leaves a remainder 10 when divided by x + 2, find a and b.
24. Factorise the following, using factor theorem :
(i) x3 – 6x2 + 11x – 6
(ii) x3 – 6x2 + 3x + 10
(iii) 2x3 – 5x2 – x + 6
(iv) x3 – 3x2 + 3x – 1
(v) x3 + 4x2 – 11x – 30
(vi) 4x3 + 9x2 – 19x – 30
3
2
25. Factorize 6x + 35x – 7x – 6, given x + 6 is one of its factor.
26. Factorize 2x3 + 5x2 – 124x – 63, given x + 9 is one of its factor.
27. Use suitable identity to find the following products :
(i) (x + 4) (x + 6)
(ii) (x – 3) (x + 8)
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(iii) (2x + 3) (3x – 2)
B
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 2 5 2 5
(iv)  t    t  
2
2

28. Evaluate the following products without multiplying directly:
(i) 87 × 93
(ii) 106 × 94
22
POLYNOMIALS
MATHEMATICS–IX
29. Factorize the following using appropriate identities :
y2
(ii) 9 x 2  6 x  1
144
(iii) x 4  y 4
(iv) 9 x 2  30 xy  25 y 2
30. Factorize by splitting the middle term :
(i) x2 – 21x + 108 (ii) 12y2 – y – 6
(iii) t2 – 11t – 42
(v) 8x2 – 2x – 15
(vi) 20x2 + 13x – 84
31. Expand each of the following using suitable identity :
(i) 9 x 2 
a

(ii)   b  3c 
3

(i) ( x  2 y  3z ) 2
y

(iv)  2 x  
2


3
(iii) (3x – 2y)
2
3
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32. Evaluate the following, using suitable identity:
(i) (101)3
(ii) (99)3
33. Factorize the following:
(i) 4x2 + y2 + 9z2 + 4xy – 6yz – 12xz
(ii) 64a3 – 144a2b + 108ab2 – 27b3
3
2
2
3
(iii) x + 6x y + 12xy + 8y
(iv) 8x3 + 125
(v) 27a3 – 64b3
(vi) 27a3 + 8b3 + c3 – 18abc
34. Without actually calculating the cubes, find the value of each of the following :
(i) (–14)3 + (8)3 + (6)3
(ii) (19)3 + (–11)3 + (–8)3
35. If x + y + z = 0, prove that x3 + y3 + z3 = 3xyz.
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(iv) 6z2 + 5z – 6
1
( x  y  z ) [( x – y ) 2  ( y – z ) 2  ( z – x) 2 ]
2
37. Using factor theorem, show that a – b, b – c and c – a are the factors of a(b2 – c2) + b (c2 – a2) + c (a2 – b2).
38. Factorize the following :
(i) 1 – 2ab – a2 – b2
(ii) x4 + 5x2 + 9
36. Prove that: x3 + y3 + z3 – 3xyz 
(iii) 5 5 x 2  30 x  8 5
(iv) 9( x  y) 2  24( x 2  y 2 )  16( x  y ) 2
(v) x6 – y6
(vi) x6 + y6
39. Factorize the following :
(i) x3 (y – z)3 + y3 (z – x)3 + z2(x – y)3
(ii) (a – 4b)3 + (4b – 3c)3 + (3c – a)3
3
3
3
40. Prove that : (a + b) + (b + c) + (c + a) – 3 (a + b) (b + c) (c + a) = 2 (a3 + b3 + c3 – 3abc)
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PRACTICE TEST
MM : 30
Time : 1 hour
General Instructions :
Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.
1.
2.
3.
4.
Find the remainder when 3x3 – 8x2 + 9x – 10 is divided by x – 3.
Find the value of a for which x3 + ax2 – 3x + 14 is exactly divisible by x + 2.
Factorize : 64a3 – 27b3 – 144a2b + 108ab2
Evaluate : (103)3 , using suitable identity.
MATHEMATICS–IX
POLYNOMIALS
23
5. Factorize : 3x2 + 13x – 10
x

(ii)   y 
3

2
6. Expand the following : (i) (x – 2y – 3z)
3
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7. Without actually calculating the cubes, evaluate the following :
(20)3 + (–15)3 + (–5)3
8. Factorize : x3 – 64y3 – 8z3 – 24xyz.
9. Factorize 6x3 + 25x2 + 21x – 10 using factor theorem.
10. Prove that : a3 + b3 + c3 – 3abc =

1
(a  b  c) (a  b) 2  (b  c )2  (c  a ) 2
2

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ANSWERS OF PRACTICE EXERCISE
1.
3.
4.
6.
(i), (iii) and (v)
2. (i) 4 (ii) 3 (iii) 3 (iv) 0 (v) 2
(i) Binomial (ii) Trinomial (iii) Monomial (iv) Bionomial (v) Bionomial
4, 46, –5
5. –1, 5, –40
(i) yes (ii) yes (iii) no (iv) yes
4
7. (i) 4
3
8. (i) – 2 (ii) 175
(ii) 
5
(iii) 0 (iv)
r
(iii) –3 (iv) – 4 (v) 6a
9. a = – 1
12. p = – 9
16. 1, –1, 3
17. 1, –1, –4
B
(vi) 8a
14. a 
13. a = – 21, b = 9
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24. (i) (x – 1) (x – 2) (x – 3)
(iv) (x – 1) (x – 1) (x – 1)
25. (x + 6) (3x + 1) (2x – 1)
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20. a = 2
(ii) (x + 1) (x – 2) (x – 5)
(v) (x + 2) (x – 3) (x + 5)
26. (x + 9) (x – 7) (2x – 1)
27. (i) x2 + 10 x + 24
(ii) x2 + 5x – 24
28. (i) 8091
(ii) 9964
11
5
15. a 
109
10
22. a  1
23. a = 1, b = – 8
3
(iii) (x + 1) (x – 2) (2x – 3)
(vi) (x – 2) (x + 3) (4x + 5)
(iii) 6x2 + 5x – 6
y 
y

29. (i)  3 x    3 x  
12  
12 

4
(iv) t 
25
4
(ii) (3x – 1) (3x – 1)
(iii) ( x  y)( x  y)( x 2  y 2 )
(iv) (3x + 5y) (3x + 5y)
30. (i) (x – 9) (x – 12) (ii) (3y – 2) (4y + 3)
(iii) (t + 3) (t – 14)
(iv) (2z + 3) (3z – 2) (iv) (2x – 3) (4x + 5)
(vi) (4x – 7) (5x + 12)
31. (i) x2 + 4y2 + 9z2 – 4xy – 12yz + 6xz
(iii) 27x3 – 8y3 – 54x2y + 36xy2
(ii)
a2
2
 b 2  9c 2  ab  6bc  2ac
9
3
(iv) 8 x 3 
y3
3
 6 x 2 y  xy 2
8
2
32. (i) 1030301 (ii) 970299
24
POLYNOMIALS
MATHEMATICS–IX
33.
(i) (2x + y – 3z) (2x + y – 3z) (ii) (4a – 3b) (4a – 3b) (4a – 3b)
(iv) (2x + 5) (4x2 – 10x + 25) (v) (3a – 4b) (9a2 + 12ab + 16b2)
(vi) (3a + 2b + c) (9a2 + 4b2 + c2 – 6ab – 2bc – 3ac)
34. (i) –2016 (ii) 5016
38.
(iii) (x + 2y) (x + 2y) (x + 2y)
(ii) (x2 + x + 3) (x2 – x + 3)
(i) (1 + a + b) (1 – a – b)
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(iii) ( 5 x  2) (5 x  4 5 )
(iv) (x + 7y) (x + 7y)
(v) (x – y) (x + y) (x2 – xy + y2) (x2 + xy + y2)
2
2
4
2 2
4
(vi) (x + y ) (x – x y + y )
39. (i) 3xyz (y – z) (z – x) (x – y) (ii) 3 (a – 4b) (4b – 3c) (3c – a)
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ANSWERS OF PRACTICE TEST
1. 26
4. 1092727
2. a = – 3
5. (3x – 2) (x + 5)
6. (i) x2 + 4y2 + 9z2 – 4xy + 12yz – 6xz
(ii)
x3
x2 y
 y3 
 xy 2
27
3
8. (x – 4y – 2z) (x2 + 16y2 + 4z2 + 4xy – 8yz + 2xz)
7. 4500
9. (x + 2) (2x + 5) (3x – 1)
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MATHEMATICS–IX
3. (4a – 3b) (4a – 3b) (4a – 3b)
B
POLYNOMIALS
25