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Using the identity (x + a) (x + b) = x 2 + (a + b)x + ab We have (2y + 5 ) (2y + 5) = (2y) 2 + ( 5 + 5)2y + ( 5 x 5) = (2y) 2 + (10)2y + 25 = 4y 2 + 20y + 25

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Algebraic Expressions and Identities

Short Answer Type

91.

Consider the special case of Identity (IV) with b = –a. What do you get? It is related to Identity (III).

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92.

Use a suitable identity to get each of the following products.

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93.
Use a suitable identity to get each of the following products.

(2y + 5) (2y + 5)

Using the identity
(x + a) (x + b) = x²+ (a + b)x + ab

We have

(2y + 5 ) (2y + 5) = (2y)²+ ( 5 + 5)2y + ( 5 x 5)
                        = (2y)² + (10)2y + 25
                        = 4y² + 20y + 25

106 Views

94.

Use a suitable identity to get each of the following products.

(2a - 7) (2a - 7)

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95.

Use a suitable identity to get each of the following products.

$space space space space space space space space space space space space space space space space space space space space open parentheses 3 a minus 1 half close parentheses open parentheses 3 a minus 1 half close parentheses$

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96.

Use a suitable identity to get each of the following products.

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97.

Use a suitable identity to get each of the following products.

(a²+ b²) (-a² + b²)

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98.

Use a suitable identity to get each of the following products.

(6x - 7) (6x + 7)

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99.

Use a suitable identity to get each of the following products.

(-a + c) (-a + c)

84 Views

100.

Use a suitable identity to get each of the following products.

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