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JAMB Mathematics 2021 Paper

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Question : 35

Total: 40

Factorize completely 81a

^4

^4

(3a + 2b)(2a - 3b)(9a² + 4b²)
(3a - 2b)(2a - 3b)(4a² - 9b²)
(3a - 2b)(3a + 2b)(9a² + 4b²)
(6a - 2b)(8a - 3b)(4a³ - 9b²)

Solution:

Concept:

Use the difference of two squares repeatedly:

x^2 - y^2 = (x - y)(x + y)

.

Explanation:

Write

81a^4

as

(9a^2)^2

and

16b^4

as

(4b^2)^2

.

So

81a^4 - 16b^4 = (9a^2)^2 - (4b^2)^2

.

Factor the first difference of squares:

= (9a^2 + 4b^2)(9a^2 - 4b^2)

.

Now factor

9a^2 - 4b^2

again as a difference of squares:

9a^2 - 4b^2 = (3a - 2b)(3a + 2b)

.

Thus the complete factorization is:

(3a - 2b)(3a + 2b)(9a^2 + 4b^2)

.

Answer:

Option C:

(3a - 2b)(3a + 2b)(9a^2 + 4b^2)

.

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