2
Polynomial
CHAPTER
We are Starting from a Point but want to Make it a Circle of Infinite Radius
Constant: - A symbol having a fixed numerical value is called a constant.
5
7
Example: = 8,  6, ,  . etc.
Variables: A symbol which may be assigned different numerical values is known as variable.
Example: In equation 5x = 0, 5 is constant while x is variable.
Algebraic Expressions: A combination of constant and variable, connected by some or all of the operations,
-,  and  , is known as an algebraic expression.
+,
Terms of an algebraic Expressions:The several parts of an algebraic expression separated by + or – operations are called the terms of the
expression.
Example: x3 + 2x2y - 4xy2 + y3 + t, is an algebraic expression containing 5 terms, namely, x3, 2x2y,
xy2, y3 and 7.
Polynomials:An algebraic expression is which the variables involved have only non-negative integral powers is
called a polynomial.
Example: 3x3 – 2x2 + 4x – 3 a polynomial is one variable x.
Condition of an Algebraic Expression to be a
(i) Exponent should be a whole number; it should not be a negative integer.
(ii) Exponent should not be in fraction.
(iii) Variable should not be under radical sign.
Coefficients:
In polynomial 3x3 – 2x2 + 5x – 4, we say that the coefficient of x3, x2 and x are 3, – 2 and 5
respectively and also, – 4 is the constant term.
Degree of a polynomial: The highest power of the variable is called the degree of polynomial.
Example:
(i)
2x + 3, is a polynomial is x of degree 1.
(ii)
4 y2 
3
y is a polynomial is y of degree 2.
2
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Polynomial of Various Degrees-:
(i) Linear Polynomial: - A polynomial of degree 1 is called a linear polynomial.
Example: 3x + 5
2y – 9
(ii) Quadratic Polynomial:- A polynomial of degree 2 is called a quadratic polynomial.
Example: x 2  5 x
1
2
3y2  8 y  5
Cubic Polynomial: - A polynomial of degree 3 is called is a cubic polynomial
Example:
(i) 4x3 – 3x2 + 7x + 1
(ii) 8x3 – 2x2 – 5x – 8
Biquadratic Polynomial: - A polynomial of degree 4 is called a biquadratic polynomial
Example:
(i) x4 – 3x3 + 2x2 + 5x – 3
(ii) 4x4 – 7x3 + 6x2 + 9x – 7
Number of Terms:
(i) Monomial: - A polynomial containing one non zero term is called a monomial
Example: - 5, 3x,
1
xy are called monomial.
2
(ii) Binomial:- A polynomial containing two non-zero terms is called binomial.
Example: - 3 + 6x, 2x2 + 5x, are binomial.
(iii) Trinomial: - A polynomial containing three non-zero terms is called a trinomial.
Example: - 3x2 – 2x + 1, 2y2 – 1 + 2y are trinomial.
Constant polynomial: - A polynomial containing one term only, consisting of a constant is called a
constant is called a constant polynomial.
Examples: 3, - 5,
4
are all constant polynomial.
5
Note:
(i) Every real number is a constant polynomial.
(ii) The degree of a non-zero constant polynomial is zero.
Zero Polynomial: - A polynomial consisting of one term, namely zero only, is called a zero
polynomial.
(i) The degree of a zero polynomial is not defined.
(ii) A polynomial is one variable x of degree n is an expression of the form
an xn an1xn1..........a1x  a0 where a0, a1, a2 ……….. are constant.
Standard form of a polynomial:A polynomial is written in the descending powers in x is called standard form of a polynomial
Example: x3 – 3x2 + 2x + 1 is a polynomial is the standard form as the powers in descending order
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IX ACADEMIC QUESTIONS
Subjective
Assignment – 1
1.
Which of the following expression are
polynomials in one variable and which are
not? State reason for your answer
In case of a polynomial, write its degree.
(i)
4x2 – 3x + 7
1

1
1

x 1 2
(ii)
x2
(iii)
1
( x  1) ( x  2)
x
1 2
( x  4 x  6)
8
(iv)
(v)
y2  2 3
(vi)
3 x  2x
(vii) x  y  t
(viii) 1
12
(ix)
(x)
2.
1
3
x2 
3.
(ii)
x2 is
(vi)
x is
1 2
( x  4 x  6)
8
1
x
2

1
x
1

1
2
( x 2  x  1) ( x  1)
(1  x)
2
(ix)
xy4 is 6y3 + 9y4 + 10x2 – 8y +
(x)
x3 is x2 (x – 1)
2
5
5.
Classify the following as linear, quadratic
and cubic polynomials: (i)
2x2 + 4x
(ii) P3
(iii) 2x – y – y2
(iv) x – x3
(v)
5t
(vi) – 7 + z
(vii) r2
(viii) y + y2 + 4
(ix) 1 + x
(x)
x3 + 3x
6.
Rewrite the following in standard form
2x  2
2x  5
2 x 1
(i)
3 y 3  4 y  11
Write the coefficient of: (i)
(v)
Give examples: (i)
Binomial of degree 35
(ii) Monomial of degree 100
(iii) Trinomial of degree 27
(iv) Binomial of degree 74
(v)
Monomial of degree 16.
2
y is y  3 y
2
x5 is 3x – 6x2 – 2x2 + 3x5
4.
50
5 x3  4 x 2  7 x
 2
(viii)
x x
2
(x)
(iv)
(viii) t is 3 + t
(vii)
(ix)
x 3 is
(vii) r2 is r2
Write the degree of each of the following
polynomials:
x9  x5  3x10  8
(i)
(ii) 3
(iii) 5 t
(iv) 12- x + 2x3
(v)
100x + 1
(vi)
2 x2  2x  1
(iii)
3
2
( x 2  x  1) ( x  1)
1 x
(ii)
(iii)
x – 7 + 8x2 + 9x3
– 5x2 + 6 – 3x3 + 4x
(iv)
– 4 + 6x3 – x + 7x4  2x 2
(v)
y2 + 5y3 – 11– 7y + 9y4
x is 2x + x
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Zero of a Polynomial
Let p (x) be a polynomial. If P (x) = 0, then we say that  is a zero of the polynomial p (x).
The value of the variable (s) is the equation which may satisfy the equation, is caused, the zero of the
polynomial.
Value of a polynomial:- The value of a polynomial f (x) at x =  is obtained by substituting x =  is the
given polynomial and is denoted by f (x).
Note: (i) The zero of the polynomial is also called the root of the polynomial.
(ii)
A non-zero constant polynomial has no zero.
(iii) Every real number is a zero of the zero polynomial.
Important Points
A zero of polynomial need to be zero.
0 may be a zero of a polynomial.
Every linear polynomial has one and only one zero.
A polynomial can have more than one zero.
Number of zeroes is same as the degree of polynomial.
Example: Find the value of the polynomial at 5x + 4x2 + 3 at
(i) x = 0
(ii) x = - 1
(iii) x = 2
Answer: (i) p(x) = 5x + 4x2 + 3 => p(0) = 5(0) + 4(0)2 + 3 = 3
(ii) p(x) = 5x + 4x2 + 3 => p(-1) = 5(-1) + 4(-1)2 + 3 => 5 - 4(1) + 3 = -6
(iii) p(x) = 5x + 4x2 + 3 => p(2) = 5(2) + 4(2)2 + 3 => 10 - 16 + 3 = -3
Example: Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x = -1/3
(ii) p(x) = 5x - π, x = 4/5
(iii) p(x) = x2 - 1, x = 1, -1
Answer: (i) If x = -1/3 is a zero of polynomial p(x) = 3x + 1 then p(-1/3) should be 0.
At, p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0
Therefore, x = -1/3 is a zero of polynomial p(x) = 3x + 1.
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(ii) If x = 4/5 is a zero of polynomial p(x) = 5x - π then p(4/5) should be 0.
At, p(4/5) = 5(4/5) - π = 4 - π
Therefore, x = 4/5 is not a zero of given polynomial p(x) = 5x - π.
(iii) If x = 1 and x = -1 are zeroes of polynomial p(x) = x2 - 1, then p(1) and p(-1) should be 0.
At, p(1) = (1)2 - 1 = 0 and
At, p(-1) = (-1)2 - 1 = 0
Hence, x = 1 and -1 are zeroes of the polynomial p(x) = x2 - 1.
Example: 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
Answer :(i) p(x) = x + 5 =>p(x) = 0 =>x + 5 = 0 => x = -5
Therefore, x = -5 is a zero of polynomial p(x) = x + 5
(ii) p(x) = x - 5 => p(x) = 0 => x - 5 = 0 => x = 5
Therefore, x = 5 is a zero of polynomial p(x) = x - 5.
(iii) p(x) = 2x + 5 => p(x) = 02x + 5 = 0 => 2x = -5 =>x = -5/2
Therefore, x = -5/2 is a zero of polynomial p(x) = 2x + 5
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IX ACADEMIC QUESTIONS
Subjective
Assignment - 2
1.
Find the value of the polynomial f (x) 2x3 – 13x2 + 17 x + 12 at:
(i) f (2)
2.
(ii) f (-3)
(iii) f = 0
Find P (0), p (1) and p (2) for the following polynomials:(i) P (x) = 4x2 + x – 5
(ii) P (z) = (z + 1) ( z – 1)
(iii) P (y) = 9y3 + 2y2 + y + 7
(iv) P (t) = t4 + t + 1
3.
If x = 0 and x = -1 are the roots of the polynomial f (x) = 2x3 – 3x2 + ax + b, find the value of a and b
4.
If x = 2 is a zero of the polynomial f (x) = 2x2 – 3x + 7 a, find the value of a.
5.
Find the zero (s) of the following polynomial: (i)
P (x) = ax + b, a  0
(ii)
h (x) = 6x – 1
(iii)
f (x) = 5x – 7
(iv)
b(x) = ( x + 1) ( x – 1 )
(v)
p (x) = x + 3
(vi)
p (t) = 2t – 3
(vii)
2x + 1
(viii) 2x + 0
6.
(ix)
x–5
(x)
24
x
13
Verify whether the indicate numbers are zeroes of the polynomial corresponding to them in
following cases: 1
2
(i)
p (x) = 2x + 1,
x=
(ii)
p (x) = x2,
x=0
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7.
4
5
(iii)
p (x) = 5 – 4x,
x=
(iv)
q (x) = 4x,
x=-4
(v)
q (x) = bx – b,
x=1
Verify that 2 and – 3 are the zeroes of the polynomial p (x) = x2 + x – 6.
Remainder Theorem:
Let f (x) be a polynomial of degree n  1 and let be any real number.
When f (x) is divided by (x – a), then the remainder is f (a).
If p (x) and g (x) are two polynomial such that degree of p (x)  degree of g (x) and
g (x)
 0, then we can find polynomials q (x) and r (x), such that: p (x) = g (x) q (x) + r (x) where r (x) = 0
or degree of r (x) < degree of g (x). p (x) is divided by 9 (x), gives q (x) as quotient and r (x) as
remainder.
Factor Theorem
Let p (x) be any polynomial of degree greater than or equal to 1 and ‘a’ be any real number, then
(i) (x – a) is a factor of p (x) if P (a) = 0
(ii) p (a) = 0 if (x – a) is a factor of p (x).
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IX ACADEMIC QUESTIONS
Subjective
Assignment - 3
(Remainder Theorem)
1.
Divide x + 3x2 – 1
2.
Find the remainder obtained on dividing p(x) = x3 + 1 by g (x) = x + 1
3.
Find the remainder when p (x) = 4x3 – 12x2 + 14x – 3 is dividing by g (x) =
4.
Divide the p (x) = x3 – ax2 + 6x – a by x – a
5.
If the polynomial ax3 + 4x2 + 3x – 4 and x3 – 4x + a leave the same remainder when divided by
(x – 3) find the value of a.
6.
Let R1 and R2 are the remainders when the polynomials x3 + 2x2 – 5ax - 7 and
x3
2
+ ax – 12x + 6 are divided by x + 1 and x – 2 respectively. If 2R1 + R2 = 6, find the value of a.
7.
Check whether the polynomial f (x) = 4x3 + 4x2 – x – 1 is a multiple of 2x + 1.
8.
The polynomial kx2 + 3x3 + 6 when divided by x – 2. leaves a remainder which is double the
remainder left by the polynomial 2x3 + 17x + k when divided by x – 2, find the value of k.
9.
Divide- 3x4 + 2x3 –
x 1
2
2
x2
2
by x 

3
3 27
10. If f (x) = x4 – 2x3 + 3x2 – ax + b is a polynomial such that when it is divided by (x- 1) and (x + 1)
the remainders are respectively 5 and 19. Determine the values of a and b.
(Factor Theorem)
1.
Use factor theorem to determine whether x  2 is a factor of 2 2 x 2  5 x  2
2.
Find the value of k, if (x- 3) is factor of k2x2 – kx – 2.
3.
Show that:(i)
(x + 5) is a factor of (2x3 + 9x2 – 11x – 30)
(ii)
( 2 ) is a factor of (7x2 - 4 2 x – 6)
(iii) (x – 3) is a factor of (2x3 + 7x2 – 24x – 45)
(iv) (x2 + 2x – 3) is a factor of x3 – 3x2 – 13x + 15
4.
Show that 3x3 + x2 – 20x + 12 is exactly divisible by 3x – 2.
5.
For what value of m is the p (x) = 2x4 + 3x3 + 2mx2 + 3x + 6 exactly divisible by (x +2)?
Factors: A polynomial g (x) is called a factor of the polynomial p (x) if g (x) divides the p (x)
exactly.
Example: - (x – 2) is a factor of (x2 + 3x – 10)
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Factorization: To express a given polynomial as the product of polynomials, each of degreeless
than that of the given polynomial such that no such factor has a factor of lower degree, is called
factorization.
Example: (x2 – 16) can be expressed as (x – 4 ) (x + 4) also (x2 – 3x + 2) = (x – 2) (x – 1)
Methods of Factorization
1.
Factorization by taking out the common factor:-

When each term of an expression has a common factor, we divide each term by this factor and take it
out as a multiple.
Example:
(i) 5x2 – 20 xy
5 is common in each term of the given expression.
So,
5 (x2 – 4xy), now, also, x is common
so, the factorization is : 5x (x – 4)
so,
2.
5x2 – 20xy = 5x (x – 4y)
Factorization by grouping: Sometimes in a given expression it is not possible to take out a common factor directly. However,
the term of the given expression are grouped in such a manner that we may have a common factor.
This can now be factorize by the method of taking out a common factor.
Example: (i) 6 ab – b2 + 12ac – 2 bc
We can group the expression as follows: 6ab + 12ac – b2 – 2bc
now, by taking out a common factor as: 6a (b + 2c) – b (b + 2 c)
3.
so,
by taking (b + 2c) common, (b + 2c) (6a – b)
so,
p (x) can be factories as : p (x) = (b + 2c ) (6a – b)
Factorization of Quadratic Trinomials
Case 1: - Polynomial of the form x2 + bx + c. We find integers p and q such that
p + q = b and pq = c. then, x2 + bx + c = x2 + (p + q) x + pq
= x2 + px + qx + pq
= x (x + p) + q (x + p)
= (x + p) (x + q)
Example: - p (x) = x2 + 9x + 18
We try to split q into two parts whose sum is 9 and product 18.
so,
6 + 3 = 9 and 6  3 = 18

x2 + 9x +18 = x2 + 6x + 3x + 18
= x (x + 6) + 3 (x + 6)
= (x + 6) (x + 3)
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Hence, x2 + 9x + 18 = (x + 6) (x + 3)
Case 2 : Polynomial of the form ax2 + bx + c
In this case, we find integers p and q such that
p + q = b and pq = ac. Then,
ax2 + bx + c =ax2 + (p + q)x
pq
a
= a2x2 + apx + aqx + pq
= ax (ax + p) q (ax + p)
= (ax + p) (x + q)
Hence, (ax2 + bx + c) = (ax + p) (ax + q)
Example:
p (x) = 6x2 + 7x – 3
Here, 6  ( – 3) = – 18
so, we try to split 7 into two parts whose sum is 7 and product – 18.
Clearly, 9 + (-2) = 7 and 9  (– 2) = – 18
 6x + 7x – 3 = 6x + 9x – 2x – 3
2
2
= 3x (2x + 3) – (2x + 3)
= (3x – 1) (2x + 3)
Hence, (6x2 + 7x – 3) = (2x +3) (3x – 1)
 x + 9x + 18 = x + 6x + 3x + 18
2
2
= x (x + 6) + 3 (x + 6)
= (x + 6) (x + 3)
2
Hence, x + 9x + 18
= (x + 6) (x + 3)
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IX ACADEMIC QUESTIONS
Subjective
Assignment - 4
1.
Factorize the given expressions: (1)
a2 + b – ab – a
(11)
x (x – y)3 + 3x2y (x – y)
(2)
ab + bc + ax + cx
(12)
x3 – 23x2 + 14 2x – 120
(3)
6x2 + 5x – 6
(13)
x3 + 6x2 + 11x + 6
(4)
y2 – 5y + 6
(14)
2y3 – 5y2 – 19y + 42
(5)
6x2 + 17x + 5
(15)
x3 + 13x2 + 32x + 20
(6)
x2 + 3x + x + 3
(16)
4x3 + 4x2 – x – 1
(7)
6(2a + 3b)2 – 8 (2a + 3b)
(17)
x3 – 6x2 + 3x + 10
(8)
5a (b + c) – 7b (b + c)
(18)
6x3 + 11x2 – 3x – 2
(9)
x2 – 4x – 21
(19)
x3 – x2 – 9x + 9
(10)
6x2 + 7x – 3
(20)
2y3 + y2 – 2y – 1
2.
Without actual division prove that 2x4 – 6x3 + 3x2 + 3x – 2 is exactly divisible by x2–3x+2.
3.
Find the value of a, if x – a is a factor of x3 – x2x + x + 2.
4.
Factorise the following by the method of middle term splitting:
(i)
3 x 2  11x  6
3
(iv) 2x2 – 7x – 39
5.
(ii) 4 3 x 2  5 x  2 3
(v) 2 x 2 
(iii) 9x2 – 22x + 8
5x 1

6 12
Show that (x – 1) is a factor of x 10 – 1 and also of x11 – 1
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IDENTITIES FOR FACTORIZATION
(i)
(x + y)2 = x2 + 2xy + y2
2
Ex.
9x + 6xy + y
2
2
2
2
2
2
Ans: 9x + 6xy + y = (3x) + (2×3x×y) + y = (3x + y) = (3x + y) (3x + y)= (3x + y)
(ii)
(x – y)2 = x2 – 2xy + y2
2
Ex.
4y - 4y + 1
2
2
2
2
Ans: 4y - 4y + 1 = (2y) - (2×2y×1) + 1 = (2y - 1) = (2y - 1) (2y - 1)= (2y - 1)
(iii)
(iv)
2
2
x2 – y2 = (x + y) ( x – y)
Ex.
104 × 96
Ans:
104 × 96 = (100 + 4) (100 - 4) = (100) - (4) = 10000 - 16 = 9984
2
2
(x + a) (x + b) = x2 + (a + b) x + ab
Ex. (x + 4) (x + 10)
Ans: In (x + 4) (x + 10), a = 4 and b = 10
2
2
Now, (x + 4) (x + 10) = x + (4 + 10)x + (4 × 10) => x + 14x + 40
(v)
(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
(vi)
(x + y)3 = x3 + y3 + 3xy (x + y)
(vii) (x – y)3 = x3 – y3 – 3xy (x – y)
(viii) x3 + y3 + z3 – 3xyz = (x + y + z)
(x2 + y2 + z2 –xy - yz – zx)
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IX ACADEMIC QUESTIONS
Subjective
Assignment - 5
Question 1 - Factorize:
1.
25x2 – 64y2
12. 25p2 + 4q2 + 9r2 – 20pq – 12qr + 30 pr
2.
2a5 – 32a
13. x6 – y6
3.
x4 – 625
14. a3 - 2 2 b3
4.
108 a2 – 3 (b – c)2
5.
(a + b)3 – a – b
2
15. 2a3 – 128 a
16. 8 x3 
2
6.
27a – 48 b
7.
2 – 50x2
8.
(3x + 5y)2 – 4z2
9.
4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
1
27 y 3
17. 2 2 a3  8b3  27c3  18 2 abc
18. a3 – b3 + 1 + 3ab
19. a3 – 8b3 + 64c3 + 24abc.
10. 9x2 + 16y2 + 4z2 – 24xy + 16yz – 12xz
2
2
2
11. 2x  y  8z  2
20. (2x + 3y)3 – (2x – 3y)3
2 xy  4 2 yz  8xz
Question 2 – Find the Product
(i)
( x + y – z) (x2 + y2 + z2 – zy + yz + zx)
(ii) (x–2y + 3) (x2+4y2 + 2xy – 3x + 6y + 9)
(iii) (a– b – c) (a2 + b2 + c2 + ab + ac – bc)
3.
If (x + y + z) = 0, prove that
(x3 + y3 + z3) = 3xyz
4.
x + y + 4 = 0, find the value of
5.
Evaluate by using identities:-
(x 3 + y3 – 12xy + 64)
(i)
(106)3
(v)
95  96
(ii)
105  102
(vi)
(x + 8) (x – 10 )
(iii)
(997)2
(vii) (999)3
(iv)
(103)  (97)
(viii) (x – 3) (x + 5)
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(ix)
(105)3
(x)
(2x – 4 )(2x + 4 )
6. If P = 2 – a, prove that
a3 + 6ap + p3 – 8 = 0.
(iv) (3a – 2ab)3
7.
(v)
Expand:
(i)
(3x + 2)3
(ii)
4

 5 x  2


(iii)
2

 3 x  1


8.
(2a + 3b + 4c)
(vi) (3a – 5b – c)2
3
1
1
(vii)  a  b  2 
2
4
2

3
Verify: -
(i) x3 + y3 = (x + y) (x2 – xy + y2 )
(ii) x3 – y3 = (x – y) (x2 + xy + y2 )
(iii) x3 – y3 = (x – y) (x2 + xy + y2 )
Verify that : x3 + y3 – 3xyz =
9.
1
2
10. (x + y + z) ( x  y)2  ( y  z )2  ( z  x)2
11. Without actually calculating the cubes, find the value of each of the following:
(i) (– 12)3 + (7)3 + (5)3
(ii) (28)3 + (– 15)3 + (– 13)3
12. If x = 2y + 6, find the value of (x3 – 8y3 – 36 xy – 216) ]
13. Without actual division, prove that (2x4 – 6x3 + 3x2 + 3x – 2) is exactly divisible by
3x + 2)
(x 2 –
14. If polynomials (2x3 + ax2 + 3x – 5) and (x3 + x2 – 2x + a) leave the same remainder when divided by
(x – 2), find the value of a. Also, find the remainder in each case.
15. The polynomial ax3 + 3x2 – 13 and 2x3 – 5x + a are divided by x + 2. If the remainder in each case in
the same, find the value of a.
16. If f (x) = x4 – 2x3 + 3x2 – ax + b is a polynomial such that when it is divided by
1) (x + 1), the remainders are respectively 5 and 19. Determine the remainder when
divided by ( x – 2)
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(x –
f (x) is
Polynomial
17. Find the remainder when f (x) is divided by g (x) and verify the result by actual division.
a. f (x)=9x3 – 3x2 + x – 5, g (x) = x 
b. f (x) = 3x4 + 2x3 
2
3
x 2
2
x2
, g (x) = x 
 
9 27
3
3
c. f (x) = x3 + 4x2 – 3x + 10, g (x) = x + 4
18. The polynomials ax3 + 3x2 – 3 and 2x3 – 5x + a when divided by (x – 4) leave the remainders R1 and
R2 respectively. Find the values of a is each of the following cases, if:
(i) R1 = R2
(ii) R1 + R2 = 0
(iii) 2R1 – R2 = 0
19. Find the values of a and b so that the polynomial x3 + 10x2 + ax + b is exactly divisible x – 1 as well as
x – 2.
20. If both x – 2 and x 
1
are factors of px2 + 5x + r show that p = r
2
21. If x2 – 1 is a factor of ax4 + bx3 + cx2 + dx + e, show that a + c + e = b + d = 0
22. Use factor theorem to verify that x + a is a factor of x2 + –a2 for any positive integer.
23. Verily whether 0 and 3 are the zeroes of the polynomial p (x) = x 2 – 3x.
24. Verify that – 1/2 is a zeroes of p (y) = 2 y + 1
25. If p (x) = (x – a) q (x) then prove that.
The degree of q (x) = the degree of p (x) – 1
26. Using factor theorem, show that x – y, y – z, z – x, are the factor of x (xy2 – z2) + y (z2 – x2) + z (x2 – y2)
27. Factories the p (x) = x4 + x2y2 + y4
28. Factories:
a2
b
2
2
b2
a2
29. Factories:
a2 – b2 – 4c2 + 4d2 – 4 (ad – bc)
30. Evaluate:
[(999)2  (1)2 ]
31. Factories:
4x2 – z2 + 9y2 – 4 p2 + 4pz – 12xy
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XI SCIENCE & DIP. ENTRANCE
Multiple Choice Questions
Assignment – 6
1.
A symbol which may be assigned different numerical value is known as
(a) Constant
2.
(b) variable
(b) Factors
(b) Degree
(b) Negative sign,
2
The coefficient of x is equation
(a) 0
8.
(d) Term
(c) Zero
(d) None of these
(c) Whole number
(d) Rational
(c) Both a and b (d) None of these
2
2x + 5x – 4 is:
(b) 2
(c) 5
(d) 4
The coefficient of x in 1 is (a) 1
9.
(c) Both a & b
Variable in a polynomial should not be under : (a) Radical sign
7.
(d) All of these
In a polynomial, exponent should be a
(a) Negative integer (b) Fraction
6.
(c) Polynomial
A combination of constants and variable connected by some or all of the operation is known as
(a) Coefficient
5.
(b) Term
The parts of an algebraic expression are called
(a) Operation
4.
(d) None of these
An Algebraic expression in which the variables involved have only non-negative integral powers is
called
(a) Algebraic identities
3.
(c) Expression
(b) 2
(c)
6
(d) None of these
The highest power of the variable is called the ______ of polynomial.
(a) Term
(b) Coefficient
(c) Operation
(d) Degree
(c) 0
(d) All of these
(c) 0
(d) 5
(c) 2
(d) 4
10. The degree of x in 2 is: (a) 10
(b) 2
11. Linear polynomial is a polynomial of degree: (a) 100
(b) 1
12. The degree of constant polynomial is always: (a) 0
(b) 1
13. A polynomial of degree 2 is called a: (a) Cubic polynomial
(b) Constant
(c) Biquadratic polynomial
(d) Quadratic polynomial
14. Biquadratic polynomial is a polynomial of degree
(a) 2
(b) 4
(c)
16
(d) Both b & c
15. Monomial is a polynomial containing _____ non-zero term:
(a) Two
(b) Zero
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16. Cubic polynomial is a polynomial of degree.
(a) 3
(b) 5
(c) 0
(d) 25
17. A polynomial containing one term only, consisting the variable to degree O is called.
(a) Constant polynomial
(b) Monomial
(c) Binomial
(d) Trinomial
(c) Trinomial
(d) Coefficient.
18. A polynomial containing three non – zero terms is: (a) Binomial
(b) Constant polynomial
19. Binomial is a polynomial containing _______ non-zero term is called: (a) Four
(b) Three
(c) Six
(d) Two
20. –5 is an example of ________ polynomial:
(a) Monomial
(b) Constant
(c) Linear
(d) both a & b
21. The degree of a non – zero constant polynomial is : (a) 1
(b) 2
(c) 0
(d) 4
22. A polynomial consisting of one term namely zero only is called a _________ polynomial : (a) Linear
(b) Quadratic
(c) Cubic
(d) Zero
(c) Not defined
(d) None of these
23. The degree of a zero polynomial is : (a) 0
(b) 1
24. Which of the following is the standard form of polynomial: (a) 1 – x + x2 + x3
(b) x2 + x3 + x4 – 0
(c) x3 + x2–x + 1
(d) None of these
25. Tick out the odd one: (a) 1
(b)
1
x2

1
x 1

1
(c) 4x2 – 3x+7
2
(d) 3 x  2 x
26. Which of the polynomial has degree 0:
(a) 0
(c) x2 – 1
(b) 5
(d) x
27. Among the following, which is a monomials of degree 100:
(a) x100 – x90 + 1
(b) 2x100
(c) 0.x100
(d) All of these
28. The value of variable in the equation which may satisfy the equation, is called the ___________ of
polynomial: (a) Coefficient
(b) Zero
(c) Both a & b
(d) None of these
29. The zero of a polynomial is called _________ of polynomial.
(a) Real
(b) Root
(c) Coefficient
(d) All of these
30. A non-zero constant polynomial has _____ zero (s): (a) Two
(b) One
(c) No
(d) Both a & b
(c) Zero
(d) All of these
31. Every real number is a zero of the ______ p (x):
(a) Constant
(b) Monomial
32. Every _________ polynomial has only one zero: (a) Cubic
(b) Quadratic
(c) Both a & b
33. Number of zeroes is same as the _______ of p (x):
(a) Zero
(b) Coefficient
(c) Both a & b
34. (x – a) is a factor of p (x) is: (a) p (a)  0
(b) p (a)  0
(c) p (a)  0
2
2
(d) linear
(d) None of these
(d) p (a)  0
2
35. Product of (x + y + z ) (x + y + z – xy –yz – zx) =
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(a) (x2 + y2 + z2 – 3xyz
(c) x3 – y3 + z3 – 3xyz
36. x2 + 2xy + y2 =
(a) x2 + y2
(b) x2 + y2 + 2
37. If x + y + z = 0, then x3 + y3 + z3 =
(a) 3zyz
(b) (x + y + z)3
38. (105)3 can be expanded as: (a) (100 + 5)3
(b) (110 – 5)3
39. 95  96 can be expanded as
(a) (90 + 5) (100 – 4)
(c) (90 + 5) 96
40. (2)3 + (– 1)3 + (–1)3 is equal to
(a) 8
(b) (2)3
2
2
2
41. x + y + z + 2xy + 2yz + 2zx is =
(a) (2xy + 2zx + 2yz)2
(c) (z + y + x)
42. (x + a) (x + b) is same as:
(a) x2 – (– a – b) x + ab
(c) Both a & b
43. The zero of p (x) = x – 3 is
(a) – 3
(b) 0
44. The zero of p (x) = 2x – 1 is
(a) 2
(b) – 2
45. The zero of p (x) = x is
(a) 1
(b) 0
46. x6 – y6 can be solved by using identity: (a) x2 – y2
(b) x3 – y3
47. 0.x100 is a:(a) Monomial
(b) Binomial
(b) x3 – y3 – z3 + 3xyz
(d) x3 + y3 + z3 – 3xyz
(c) (x + y)2
(d) (x – y)2
(c) x2 + y2 + z2
(d) 2xyz ‘
(c) Both are correct
(d) None
(b) (90 + 5) (90 + 6)
(d) All of these
(d) (– 1)3
(c) 6
(b) (x2 – y2 –z2)
(d) None of these
(b) x2 + (a + b) x + ab
(d) None of these
(c) 3
(d) None of these
(c) – 1/2
(d) 1/2
(c) 10
(d) All are correct
(c) Both a & b
(d) None of these
(c) Zero polynomial
(d) constant
(c) t
(d) 3  2
(c) 1
(d) – 1
(c) 1
(d) 2
48. Coefficient of t is 3  t 2 is : (a) 3
(b)
3
2
2
49. Coefficient of x is x (x – 1) is
(a) 0
(b) 2
50. Degree of t in p (t) = 5t is
(a) 0
(b) 5
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ANSWERS
Assignment – 1
1.
2.
3.
4.
5.
6.
(i)
(ii)
(iv)
(v)
(vi)
(vii)
yes, since the variable power is non negative integer and degree = 2.
No
(iii) No
yes, since the variable power is non negative integer and degree = 2.
yes, since the variable power is non negative integer and degree = 2.
No
yes, since the variable power is non negative integer and degree = 50
(viii) yes, since the variable power is non negative integer and degree = 0.
(ix) yes, since the variable power is non negative integer and degree = 2.
(x) No
(i) 10
(ii) 0
(iii) 1
(iv) 3
(v) 1
(vi) 1
(vii) 3
(viii) 2
(ix) 1
(x) 3
(i) 0
(ii) 1
(iii) 0
(iv) 3
(v) 1/8 (vi) 1
(vii) 1
(viii) 2 (ix) 0
(x) 1
(i) x35 + x (ii) y100
(iii) x27 + x26 + x
(iv) 2x74 – 1
(v) 7z16
(i) Quadratic
(ii) Cubic
(iii) Quadratic
(iv) Cubic
(v) Cubic
(vi) Linear
(vii) Linear
(viii) Quadratic
(ix) Linear
(x) Cubic
(i) x2 + x + 1
(ii) 9x3 + 8x2 + x – 7
(iii) –3x3 – 5x2 + 4x + 6
(iv) 7x4 + 6x3 – 2 x2 – x – 4
(v) 9y4 + 5y3 + y2 – 7y – 11
Assignment – 2
1.
(i) 10
2.
(i) p(0) = –5, p(1) = 0, p(2) = 13
(iii) p(0) = 7, p(1) = 19, p(1) = 89
3.
a = 5, b = 0
4.
a = –2/7
5.
(i) –b/a
(ii) 1/6
(vii) –1/2
(viii) x = 0
6.
(i) No
(ii) Yes
7.
2 is not but –3 is the root of p(x).
(ii) – 108
(iii) 12
(ii) p(0) = –1, p(1) = 0, p(2) = 3
(iv) p(0) = 1, p(1) = 3, p(3) = 19
(iii) 7/5
(ix) x = 5
(iii) No
(iv) 1, –1
(x) 13/24
(iv) No
Assignment – 3 : (Remainder Theorem)
1) 1
2) 0
3) 3
4) a = 0 5) a = –16) a = 2
7) yes, f(x) is the multiple of 2x + 1.
8) k = 33
9) 0
(v) –3
(vi) 3/2
(v) Yes
(10) a = 5, b = 8
(Factor Theorem)
1. Yes x  2 is the factor
2. K 
2
1
or K 
3
3
3. m = –1
Assignment – 4 : 1. Factorize
1. (a – 1)(a – b) 2. (x + b)(a + c) 3. (2x + 3)(3x – 2)
4. (y – 2)(y – 3)
5. (2x + 5) (3x + 1)
6. (x + 1) (x + 3) 7. 2(2a + 3b) (6a + 9b – 4) 8. (b + c) (5a – 7b)
9. (x + 3 (x – 7) 10. (2x + 3) (3x – 1)
11. x(x – y) (x2 + xy + y2)
12. (x – 1) (x – 10) (x – 12)
13. (x + 1)(x + 2)(x + 3) 14. (y – 2)(y + 3) (2y – 7)
15. (x + 1)(x + 2)(x + 10)
16. (2x – 1)(2x + 1)(x + 1) 17. (x + 1)(x – 2) (x – 5)
18. (3x + 1) (2x – 1) (x + 2)
19. (x + 3)(x – 3)(x – 1) 20. (y – 1) (y + 1) (2y + 1)
3. a = –2
4. (i) (x  3 3)( 3x  2)
(ii) ( 3x  2)(4x  3)
(iii) (x  2)(9x  4)
(iv) (x + 3)(2x – 13)
(iv) (x + 3) (2x – 13
(v) (6x – 1)(4x – 1)
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Assignment – 5
1. Factorize :
2. 2a(a2 + 4) (a + 2) (a – 2)
3. (x2 + 25) (x + 5)(x – 5)
5. (a + b)(a + b + 1)(a + b – 1)
6. 3(3a + 4b)(3a – 4b)
8. (3x + 5y + 2z)(3x + 5y – 2z)
10. (–3x + 4y + 2z)(–3x + 4y + 2z)
1. (5x + 8y) (5x – 8y)
4. 3((6a + b – c) (6a – b + c)
7. 2(a – 5x)(a + 5x)
9. (2x + 3y – 4z)(2x + 3y – 4z)
11.
( 2x  y  2 2z)
12. (5p – 2q + 3r) (5p – 2q + 3r)
13. (x – y)(x + y)(x2 + 2b2 +
2 ab 14.

1  2 1 2x 
16.  2x 
 4x  2 


3y  
9y 3y 

15. 2a(a + 8)(a – 8)
17.
(a  2b)(a 2  2b2  2ab)
( 2a  2b  3c)(2a 2  4b2  ac2  2 2ab  6bc  3 2ac)
18. (a  b 1)(a 2  b2 1 ab  b  a)
19. (a – 2b + 4c) ( a 2 + 4b2 + 16c2 + 2ab + 8bc – 4ac)
20.6y(4x2 + 9y2)
2. Find the Product:
(i) x3 + y3 – z3 + 3xyz
(ii) x3 – 8y2 + 27 + 18xy
(iii) a3 – b3 – c3 – 3abc
4. 0
5. Evaluate
(i) 119106
(vi) x2 – 2x – 80
(ii) 10710
(iii) 994009
(iv) 9991
(vii) 997002999 (viii) x2 + 2x – 15 (ix) 1157625
(v) 9120
(x) 4x2 – 16
7. Expand:
(i) 27x3 + 8 + 54x2 + 36x
3
(iv) 27a 
(ii)
1
27a 2
9a


3
4b 16b 2
64b
64 3
96
48
x  8  x2  x
125
25
5
8 3
4
x  1  x 2  2x
27
3
(v) 4a2 + ab2 + 16c2 + 12ab + 24bc + 16ac
(vi) 9a2 + 25b2 + c2 – 30ab + 10bc – 6ac
10. (i) –1180
14. a = 5/9
16. (i) –3
17. (i) a = 1
(iii)
(vii)
(ii) 16380
15. 7
(ii) 0
(ii) a = –153/65
1 2 1 2
1
a  b  4  ab  b  2a
4
16
4
11. 0
13. a = –3 and remainder = 5
(iii) 92
(iii) a = 18/127
a b
18. a = –37, b = 26
26. (x + y + xy)(x + y – xy)
27.   
b a
28. (a + b – 2c – 2d) (a – b + 2c – 2b)
29. 998000
30 (2x – 3y + 2p – z) (2x – 3y – 2p + z)
2
2
2
5.c
17.b
29.b
41.d
6.c
18.c
30.c
42.c
2
2
Assignment – 6
1. b
13.d
25.d
37.a
49.c
2.c
14.b
26.b
38.c
50.a
3.b
15.c
27.b
39.b
4.d
16.a
28.b
40.c
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7.b
19.d
31.a
43.c
8.d
20.d
32.d
44.d
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9.d
21.c
33.d
45.b
10.c
22.d
34.d
46.c
11.b
23.c
35.d
47.c
CLASS - IX
Mathematics
12.a
24.d
36.c
48.b
Polynomial