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Algebraic Identities Of Polynomials

December 7, 2020 by Veerendra

Algebraic Identities Of Polynomials

algebraic-identities

You can also read https://www.aplustopper.com/ncert-solutions-for-class-10-maths-chapter-2/ for more solved examples.

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Algebraic Identities Of Polynomials Example Problems With Solutions

Example 1:    Expand each of the following
\(\left( \text{i} \right)\text{ }{{\left( \text{3x — 4y} \right)}^{\text{2}}}\text{ }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{ }\left( \text{ii} \right)\text{ }\!\!~\!\!\text{  }{{\left( \frac{x}{2}+\frac{y}{3} \right)}^{2}}\)
Solution:   (i) We have,
Algebraic Identities Of Polynomials 1

Example 2:    Find the products
(i) (2x + 3y) (2x – 3y)
\(\left( \text{ii} \right)\text{ }\left( x-\frac{1}{x} \right)\left( x+\frac{1}{x} \right)\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)\)
Solution:    (i) We have,
Algebraic Identities Of Polynomials 2

Example 3:    Evaluate each of the following by using identities
(i) 103 × 97         (ii) 103 × 103
(iii) (97)2             (iv) 185 × 185 – 115 × 115
Solution:    (i) We have,
Algebraic Identities Of Polynomials 3

Example 4:    \(\text{If   }x+\frac{1}{x}=6,\text{   find }{{x}^{4}}+\frac{1}{{{x}^{4}}}\)
Solution:    We have,
Algebraic Identities Of Polynomials 4

Example 5:     \(\text{If   }{{x}^{2}}+\frac{1}{{{x}^{2}}}=27,\text{   find the value of the  }x-\frac{1}{x}\)
Solution:    We have,
Algebraic Identities Of Polynomials 5

Example 6:    If x + y = 12 and xy = 32, find the value of x2 + y2
Solution:    We have,
Algebraic Identities Of Polynomials 6

Example 7:    Prove that:
2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = [(a – b)2 + (b – c)2 + (c – a)2]
Solution:    We have,
Algebraic Identities Of Polynomials 7

Example 8:    If a2 + b2 + c2 – ab – bc – ca = 0, prove that a = b = c.
Solution:    We have,
Algebraic Identities Of Polynomials 8

Example 9:    Write the following in expanded form :
(i) (9x + 2y + z)2            (ii) (3x + 2y – z)2
(iii) (x – 2y – 3z)2         (iv) (–x + 2y + z)2
Solution:    Using the identity
Algebraic Identities Of Polynomials 9
Algebraic Identities Of Polynomials 10
Algebraic Identities Of Polynomials 11

Example 10:    If a2 + b2 + c2 = 20 and a + b + c = 0, find ab + bc + ca.
Solution:    
Algebraic Identities Of Polynomials 12

Example 11:    If a + b + c = 9 and ab + bc + ca = 40, find a2 + b2 + c2.
Solution:    We know that
Algebraic Identities Of Polynomials 13

Example 12:    If a2 + b2 + c2 = 250 and ab + bc + ca = 3, find a + b + c.
Solution:    We know that
Algebraic Identities Of Polynomials 14

Example 13:    Write each of the following in expanded form:
(i) (2x + 3y)3     (ii) (3x ­– 2y)3
Solution:    
Algebraic Identities Of Polynomials 15

Example 14:    If x + y = 12 and xy = 27, find the value of x3 + y3.
Solution:    We know that
Algebraic Identities Of Polynomials 16

Example 15:    If x – y = 4 and xy = 21, find the value of x3 – y3.
Solution:    We know that
Algebraic Identities Of Polynomials 17

Example 16:    \(\text{If   }x+\frac{1}{x}=7,\text{   find the value of }{{x}^{3}}+\frac{1}{{{x}^{3}}}\)
Solution:    We have,
Algebraic Identities Of Polynomials 18

Example 17:    If a + b = 10 and a2 + b2 = 58, find the value of a3 + b3.
Solution:    We know that
Algebraic Identities Of Polynomials 19

Example 18:    \(\text{If   }{{x}^{2}}+\frac{1}{{{x}^{2}}}=7,\text{   find the value of }{{x}^{3}}+\frac{1}{{{x}^{3}}}\)
Solution:    We have,
Algebraic Identities Of Polynomials 20
Algebraic Identities Of Polynomials 21

Example 19:    \(\text{If   }{{x}^{4}}+\frac{1}{{{x}^{4}}}=47,\text{   find the value of }{{x}^{3}}+\frac{1}{{{x}^{3}}}\)
Solution:   We know that
Algebraic Identities Of Polynomials 22
Algebraic Identities Of Polynomials 23

Example 20:    If a + b = 10 and ab = 21, find the value of a3 + b3.
Solution:    We know that
Algebraic Identities Of Polynomials 24

Example 21:     If a – b = 4 and ab = 45, find the value of a3 – b3.
Solution:   We have,
Algebraic Identities Of Polynomials 25

Example 22:    If a + b + c = 0, then prove that a3 + b3 + c3 = 3abc  
Solution:   We know that
Algebraic Identities Of Polynomials 26

Example 23:    Find the following product:
(x + y + 2z) (x2 + y2 + 4z2 – xy – 2yz – 2zx)
Solution:    We have,
Algebraic Identities Of Polynomials 27

Example 24:     If a + b + c = 6 and ab + bc + ca = 11, find the value of a3 + b3 + c3 – 3abc.
Solution:  
We know that
a3 + b3 + c3 – 3abc
= (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
⇒ a3 + b3 + c3 – 3abc =
(a + b + c) {(a2 + b2 + c2) – (ab + bc + ca)}…(i)
Clearly, we require the values of a + b + c,
a2 + b2 + c2 and ab + bc + ca to obtain the value of a3 + b3 + c3 – 3abc. We are given the values of a + b + c and ab + bc + ca. So, let us first obtain the value of a2 + b2 + c2.
We know that
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
⇒ (a + b + c)2 = (a2 + b2 + c2) + 2(ab + bc + ca)
⇒ 62 = a2 + b2 + c2 + 2 × 11
[Putting the values of a + b + c and ab + bc + ca]
⇒ 36 = a2 + b2 + c2 + 22
⇒ a2 + b2 + c2 = 36 – 22
⇒ a2 + b2 + c2 = 14
Now, putting a + b + c = 6, ab + bc + ca = 1 and a2 + b2 + c2 = 14 in (i), we get
a3 + b33 + c3 – 3abc = 6 × (14 – 11)
= 6 × 3 = 18.

Example 25:    If x + y + z = 1, xy + yz + zx = –1 and xyz = –1, find the value of x3 + y3 + z3.
Solution:    
We know that :
x3 + y3 + z3 – 3xyz
= (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
⇒ x3 + y3 + z3 – 3xyz
= (x + y + z) (x2 + y2 + z2 + 2xy + 2yz + 2zx – 3xy – 3yz – 3zx)
[Adding and subtracting 2xy + 2yz + 2zx]
⇒ x3 + y3 + z3 – 3xyz
= (x + y + z) {(x + y + z)2 – 3(xy + yz + zx)}
⇒ x3 + y3 + z3 – 3 × –1 = 1 × {(1)2 – 3 × –1}
[Putting the values of x + y + z, xy + yz + zx and xyz]
⇒ x3 + y3 + z3 + 3 = 4
⇒ x3 + y3 + z3 = 4 – 3
⇒ x3 + y3 + z3 = 1

Filed Under: Mathematics Tagged With: Algebraic Identities, Algebraic Identities Example Problems, Algebraic Identities Examples, Algebraic Identities Of Polynomials, Algebraic Identities Of Polynomials Examples, Polynomials

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