"Question 2 Use the identity (x + a) (x + b) = x^2 + (a + b)x + ab to find the following products. (i) (x + 3) (x + 7) (ii) (4x +5) (4x + 1) (iii) (4x − 5) (4x − 1) (iv) (4x + 5) (4x − 1) (v) (2x +5y) (2x + 3y) (vi) (2a^2 +9) (2a^2 + 5) (vii) (xyz − 4) (xyz − 2)
Class 8 Algebraic Expressions and Identities Page 151"
Answers
An identity is true only for certain values of its variables. An equation is not an identity.
The following are the identities
(a + b)² = a² + 2ab + b²
(a – b)² = a² – 2ab + b²
(a – b)(a + b) = a² – b²
Another useful identity is
(x + a) (x + b) = x² + (a + b) x + ab
If the given expression is the difference of two squares we use the formula
a² –b² = (a+b)(a-b)
• The above four identities are useful in carrying out squares and products of algebraic expressions. They also allow easy alternative methods to calculate products of numbers and so on.
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Solution:
1) (x + 3) (x + 7)
x² + (3+7)x + 21
= x² + 10x + 21
2) (4x + 5) (4x + 1)
(4x + 5) (4x + 1)
= 16x² + (5 + 1)4x + 5
= 16x² + 24x + 5
3) (4x – 5) (4x – 1)
= 16x²+ (-5-1)4x + 5
= 16x² - 24x + 5
4) (4x + 5) (4x – 1)
= 16x²+ (5-1)4x - 5
= 16x² +16x – 5
5) (2x + 5y) (2x + 3y)
= 4x² + (5y + 3y)2x + (5y×3y)
=4x² + (8y)2x + 15y²
= 4x² + 16xy + 15y²
6) (2a²+ 9) (2a²+ 5)
= 4a⁴ + (9+5)2a² + 9×5
=4a⁴ + (14)2a² + 9×5
= 4a⁴ + 28a² + 45
7) (xyz – 4) (xyz – 2)
(xyz)²+(-4-2)xyz+ (-4×-2)
= x²y²z² + (-6)xyz + 8
= x²y²z² - 6xyz + 8
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Hope this will help you.....
Answer:
Hey mate your answer is here.
Step-by-step explanation:
(i) (x + 3) (x + 7)
= x² + (3+7)x + 21
= x²+ 10x + 21
ii) (4x + 5) (4x + 1)
= 16x2 + (5 + 1)4x + 5
= + 24x + 5
iii) (4x – 5) (4x – 1)
= 16x2 – 4x – 20x + 5
= 16x2 – 24x + 5
iv) (4x + 5) (4x – 1)
= 16x2 + (5-1)4x – 5
= 16x2 +16x – 5
v) (2x + 5y) (2x + 3y)
= 4x2 + (5y + 3y)2x + 15y2
= 4x2 + 16xy + 15y2
vi) (2a2+ 9) (2a2+ 5)
= 4a4 + (9+5)2a2 + 45
= 4a4 + 28a2 + 45
vii) (xyz – 4) (xyz – 2)
= x2y2z2 + (-4 -2)xyz + 8
= x2y2z2 – 6xyz + 8