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Mrac [35]
2 years ago
15

Factor completely 49q^4 - 4​

Mathematics
2 answers:
nadezda [96]2 years ago
7 0

Step-by-step explanation:

49q^4-4

(7q^2)^2-(2)^2

(7q^2+2)(7q^2-2)

Brrunno [24]2 years ago
5 0

Rewrite it in the form a^2 - b^2, where a = 7q^2 and b = 2

(7q^2)^2 - 2^2

Use the Difference of Squares; a^2 - b^2 = (a + b)(a - b)

<u>= (7q^2 + 2)(7q^2 - 2)</u>

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1) (2\times3^-2)^3 (5\times3^2)^2 / (3^-2)(5\times2)^2 = 2

2) (3^3) (4^0)^2 (3\times2)^-3 (2^2) = 1/2

3) (3^7\times4^7) (2\times5)^-3 (5)^2 / (12^7) (5^-1) (2^-4) = 2

4) (2.3)^-1 (2^0) / (2.3)^-1 = 1

<u>Step-by-step explanation</u>:

Step 1 :

(2\times3^-2)^3 (5\times3^2)^2 / (3^-2)(5\times2)^2

⇒ (2^3) (3^-6) (5^2) (3^4) / (3^-2) (10^2)

⇒ (2.2^2.5^2) (3^-6.3^4) / (3^-2) (10^2)

⇒ (2)(10^2) (3^-2) / (3^-2) (10^2)

⇒ 2

Step 2 :

(3^3) (4^0)^2 (3\times2)^-3 (2^2)

Any number with power 'zero' is 1.

⇒ (3^3) (1^2) (3^-3) (2^-3) (2^2)

⇒ (2^(-3+2))

⇒ 2^-1 = 1/2

Step 3 :

(3^7\times4^7) (2\times5)^-3 (5)^2 / (12^7) (5^-1) (2^-4)

⇒ (12^7) (2^-3) (5^-3) (5^2) / (12^7) (5^-1) (2^-4)

⇒ (12^7) (2^-3) (5^-3) (5^2) / (12^7) (5^-1) (2^-4)

⇒ (2^-3) (5^(-3+2)) / (5^-1) (2^-4)

⇒ (2^-3) (5^-1) / (5^-1) (2^-4)

⇒ 1 / (2^-1)

⇒ 2

Step 4 :

(2.3)^-1 (2^0) / (2.3)^-1

⇒ (2^0)

⇒ Any number with power 'zero' is 1

⇒ 1

8 0
3 years ago
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