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POLYNOMIALS
Important Points
1. A polynomial p(x) in one variable x is an algebraic expression in x of the form
p(x) = anxn+an-1xn-1+…………………a2x2+a1x1+a0x0
where a0, a1, a2 ……… an are constants and an ≠ 0.
a0, a1, a2 ……… an are respectively the coefficients of x0, x1, x2, ……..xn and n is
called the degree of the polynomial. Each of anxn, an-1xn-1, ……a0x0with an ≠ 0, is
called a term of the polynomial p(x).
2.
3.
4.
5.
6.
7.
A
A
A
A
A
A
polynomial
polynomial
polynomial
polynomial
polynomial
polynomial
of
of
of
of
of
of
one term is called a monomial.
two terms is called a binomial.
three terms is called a trinomial.
degree one is called a linear polynomial.
degree two is called a quadratic polynomial.
degree three is called a cubic polynomial.
8. A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0. In this case, a is also
called a root of the equation p(x) = 0.
9. Every linear polynomial in one variable has a unique zero, a non-zero constant
polynomial has no zero, and every real number is a zero of the zero polynomial.
10. Remainder Theorem: If p(x) is any polynomial of degree greater than or equal
to 1 and p(x) is divided by the linear polynomial x – a, then the remainder is p(a).
Proof: Let p(x) be any polynomial with degree greater than or equal to 1. Suppose
that when p(x) is divided by x – a, the quotient is q(x) and the remainder is r(x),
i.e.,
p(x) = (x – a) q(x) + r(x)
Since the degree of x – a is 1 and the degree of r(x) is less than the degree of x – a,
the degree of r(x) = 0. This means that r(x) is a constant, say r.
So, for every value of x,
r(x) = r.
Therefore,
p(x) = (x – a) q(x) + r
In particular, if x = a, this equation gives us
p(a) = (a – a) q(a) + r = r,
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which proves the theorem.
Example 9: Find the remainder when x4 + x3 – 2x2 + x + 1 is divided by x – 1.
Solution: Here,
p(x) = x4 + x3 – 2x2 + x + 1,
and the zero of x – 1 is 1.
So,
p(1) = (1)4 + (1)3 – 2(1)2 + 1 + 1= 2
So, by the Remainder Theorem, 2 is the remainder when x4 + x3 – 2x2 + x + 1 is
divided by x – 1.
11. Factor Theorem: x – a is a factor of the polynomial p(x), if p(a) = 0. Also, if x
–a is a factor of p(x), then p(a) = 0.
Example: Examine whether x + 2 is a factor of x3 + 3x2 + 5x + 6 and of 2x + 4.
Solution: The zero of x + 2 is –2.
Let p(x) = x3 + 3x2 + 5x + 6
and s(x) = 2x + 4
Then,
p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6
= –8 + 12 – 10 + 6
=0
So, by the Factor Theorem,
x + 2 is a factor of x3 + 3x2 + 5x + 6
Again,
s(–2) = 2(–2) + 4 = 0
So, x + 2 is a factor of 2x + 4.
Important Formulae
(x + y)2 = x2 + 2xy + y2
(x – y) 2 = x2 – 2xy + y2
x2 – y2 = (x + y) (x – y)
(x + y + z)2= x2 + y2 + z2 + 2xy + 2yz + 2zx
(x + y)3 = x3 + y3 + 3xy(x + y)
(x – y) 3 = x3 – y3 – 3xy(x – y)
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
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1. Use suitable identities to find the following products:
(i) (x + 4) (x + 10)
Solution: (x+4)(x+10)
= x2+10x+4x+4 × 10
= x2+14x+40
(ii) (x + 8) (x – 10)
Solution: x2-10x+8x-80
= x2-2x-80
(iii) (3x + 4) (3x – 5)
Solution: 9x2-15x+12x-20
= 9x2-3x-20
(iv) (y2 +
3
3
) (y2- )
2
2
Solution: (a+b)(a-b) = a2-b2
Suppose, y2=a and
3
=b
2
Hence, the given equation can be written as follows:
y4 -
9
4
(v) (3 – 2x) (3 + 2x)
Solution: This can be solved as the earlier question
(3-2x)(3+2x) = 9-4x2
2. Evaluate the following products without multiplying directly:
(i) 103 × 107
Solution: 103× 107
= (100+3)(100+7)
= 1002+7 × 100+3 × 100+7 × 3
= 10000+700+300+21
= 11021
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(ii) 95 × 96
Solution: 95 × 96
= (100-5)(100-4)
= 1002-400-500+20
= 10000-900+20
= 9120
(iii) 104 × 96
Solution: 104 × 96
= (100+4)(100-4)
= 1002-42
= 10000-16
= 9984
3. Factorise the following using appropriate identities:
(i) 9x2 + 6xy + y2
Solution: 9x2+6xy+y2
= 32x2+3xy+3xy+y2 …………………………… (1)
= 3x(3x+y)+y(3x+y)
= (3x+y)(3x+y)
= (3x+y)²
Alternate way of solving this problem:
Equation 1 gives a hint that this can be solved through following formula:
(a+b)² = a²+2ab+b²
(ii) 4y² – 4y + 1
Solution: 2²y²-2 × 2y+1²
= (2y-1)²
(iii) x²-
y²
100
Solution: As you know a²-b² = (a+b)(a-b)
y²
10²
y
y
= (x+
) (x)
10
10
So, x²-
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4. Expand each of the following, using suitable identities:
(i) (x + 2y + 4z)²
Solution: As you know (x + y + z)2= x2 + y2 + z2 + 2xy + 2yz + 2zx
Using this formula in the given equation,
(x+2y+4z)²
= x²+4y²+16z²+4xy+16yz+8zx
(ii) (2x – y + z)²
Solution: (x-y+z)² = x²+y²+z²-2xy-2yz+2zx
So, (2x-y+z) ²
= 4x²+y²+z²-4xy-2yz+4zx
(iii) (–2x + 3y + 2z)²
Solution: (-2x+3y+2z)²
= 4x²+9y²+4z²-12xy+12yz-8zx
(iv) (3a – 7b – c)²
Solution: (x-y-z)²= x²+y²+z²-2xy-2yz-2zx
Hence, (3a-7b-c)²
= 9a²+49b²+c²-42ab-14bc-6ac
5. Factorise:
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
Solution: It is clear that this can be solved using
(x+y-z)² = x²+y²+z²+2xy-2yz-2zx
Hence, 4x²+9y²+16z²+12xy-24yz-2zx
= (2x+3y-4z)²
(ii) 2x² + y² + 8z² – 2
2 xy + 4 2 yz – 8xz
Solution: using (x+y+z)²
The value of x= 2 x, y=y and z= 2 2 z
Hence, given equation can be written as follows:
( 2 x+y+2 2 z)²
6. Write the following cubes in expanded form:
(i) (2x + 1)³
Solution: As you know, (x + y)3 = x3 + y3 + 3xy(x + y)
Hence, (2x+1)³
= 8x³+1+6xy(2x+1)
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(ii) (2a – 3b)³
Solution: As you know, (x – y) 3 = x3 – y3 – 3xy(x – y)
Hence, (2a-3b)³ = 8a³-27y³-18ab(2a-3b)
(iii) (
3
x + 1 )³
2
Solution: Using the formula which was used in question 6-(i)
We get,
27
9
3
x ³ + 1 + xy ( x + y )
8
2
2
7. Evaluate the following:
(i) 99³
Solution: 99³ can be written as (100-1)³
(100-1)³ can be solved through using (x-y)³
Now, (100-1)³= 100³-1³-300(100-1)
= 1000000-1-300(99)
= 1000000-1-29700
= 970299
(ii) 102³
Solution: 102³ can be written as (100+2)³ and can be solved using (x+y)³
Hence, (100+2) ³
= 100³+2³+600(100+2)
= 1000000+8+61200
= 1061208
8. Factorise each of the following:
(i) 8a³ + b³ + 12a²b + 6ab²
Solution:
= 8a³+b³+6ab(a+b)
=(2a+b)³
(ii) 8a³ – b³ – 12a²b + 6ab²
Solution:
8a³-b³-12a²b+6ab²
=8a³-b³-6ab(a+b)
=(2a-b)³
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(iii) 27 – 125a³ – 135a + 225a²
Solution: 27 – 125a³ – 135a + 225a²
= 3³-5³a³-3³5a+3²5²a²
= 3³-5³a³-3²5a(3-5a)
if x=3 and y=5a
hence, (3-5a)³= 3³-5³a³-3²5a(3-5a)
(iv) 64a³ – 27b³ – 144a²b + 108ab²
Solution: 64a³ – 27b³ – 144a²b + 108ab²
= 4³a³-3³b³-3 × 4a3b(4a-3b)
= (4a-3b)³
Note: Try to identify the values of x and y by carefully analysing the first
two terms of the equations. This will give you exact clue to the final answer.
(v) 27p³-
1 9
1
- p²+ p
216 2
4
Solution: 27p³ can be written as 3³p³
Hence, x= 3p
1
1³
can be written as
216
6³
1
Hence, y=
6
And
So, the required answer will be (3p-
1
)³
6
Note: This step is to help you develop the problem solving skills. In exam
situation you have to write all steps to get full marks.
9. Verify :
(i) x³ + y³ = (x + y) (x² – xy + y²)
Solution: RHS (x+y)(x²-xy+y²)
= x³-x²y+xy²+x²y-xy²+y³
=x³+y³ LHS proved
(ii) x³ – y³ = (x – y) (x² + xy + y²)
Solution: RHS (x – y) (x² + xy + y²)
= x³+x²y+xy²-x²y-xy²-y³
= x³-y³ LHS proved
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10. Factorise each of the following:
(i) 27y³ + 125z³
Solution: From the previous question you can recall
x³+y³=(x+y)(x²+xy+y²)
Hence, 3³y³+5³z³ can be written as follows:
(27=3³ and 125=5³)
(3y+5z)(9y²+15yz+25z²
(ii) 64m³ – 343n³
Solution: As you know, x³-y³=(x-y)(x²+xy+y²)
Hence, 4³m³-7³n³ can be written as follows:
(64=4³ and 343=7³)
(4m-7n)(16n²+28mn+49n²)
11. Factorise : 27x³ + y³ + z³ – 9xyz
Solution: As you know,
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
Hence, 27x³ + y³ + z³ – 9xyz
= (3x+y+z) (9x²+y²+z²-3xy-yz-3zx)
12. Verify that x³ + y³ + z³ – 3xyz =
Solution: RHS
1
(x+y+z)
2
[
1
(x+y+z)
2
[
(x-y)²+(y-z)²+(z-x)²
(x-y)²+(y-z)²+(z-x)²
]
1
(x+y+z)(x²+y²-2xy+y²+z²-2yz+z²+x²-2zx)
2
1
=
(x+y+z) (2x²+2y²+2z²-2xy-2yz-2zx)
2
=
=(x+y+z) (x²+y²+z²-xy-yz-zx)
Now, the RHS satisfies the condition as per following identity:
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
13. If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.
Solution: As you know,
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
Now, as per question x+y+z=0,
Putting value of x+y+z=0 in the equation we get
x3 + y3 + z3 – 3xyz = 0 × (x2 + y2 + z2 – xy – yz – zx)
⇒ x3 + y3 + z3 – 3xyz = 0
⇒ x3 + y3 + z3 = 3xyz proved
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]
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14. Without actually calculating the cubes, find the value of each of the following:
(i) (–12)³ + (7)³ + (5)³
Solution: As you know,
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
⇒ x3 + y3 + z3 =(x + y + z) (x2 + y2 + z2 – xy – yz – zx)+3xyz
Hence, (–12)³ + (7)³ + (5)³
=(-12+7+5) -12²+7²+5²-(-12 × 7)-(7 × 5)-(-12 × 5) +3(-12 × 7 × 5)
[
[
=0 ×
-12²+7²+5²-(-12 × 7)-(7 × 5)-(-12 × 5)
= 0-1260 = -1260 answer
]
]
-1260
(ii) (28)³ + (–15)³ + (–13)³
Solution: This question can be solved in the same way as above.
Here, value of (x+y+z) = (28-15-13) = 0
Hence, you need to calculate the value of 3xyz
3 × 28 × -15 × -13 = 16380
Hence, the required answer = -16380
But, while practicing at home try following every step for better learning.
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