Which is the completely factored form of the following equation?

[tex]\[ 4a^4b^2 - 21a^2b + 5 = 0 \][/tex]

A. [tex]\(\left(4a^2b - 1\right)\left(a^2b - 5\right) = 0\)[/tex]

B. [tex]\(\left(2a^2b + 5\right)\left(2a^2b + 1\right) = 0\)[/tex]

C. [tex]\(\left(4a^2b - 1\right)(ab - 5) = 0\)[/tex]

D. [tex]\(\left(4a^2b + 1\right)\left(a^2b + 5\right) = 0\)[/tex]



Answer :

To factor the polynomial [tex]\( 4a^4 b^2 - 21a^2 b + 5 \)[/tex], let's follow a detailed process to arrive at the correct factored form.

Given the polynomial expression:
[tex]\[ 4a^4 b^2 - 21a^2 b + 5 \][/tex]

We need to factor this into two binomials. According to the result obtained, the expression can be factored as follows:
[tex]\[ (4a^2b - 1)(a^2b - 5) = 0 \][/tex]

To ensure this matches the polynomial, let's verify by expanding these choices:

1. Multiply the binomials:
[tex]\[ (4a^2b - 1)(a^2b - 5) \][/tex]

2. Use distributive property (FOIL method):
[tex]\[ (4a^2b \times a^2b) + (4a^2b \times (-5)) + ((-1) \times a^2b) + ((-1) \times (-5)) \][/tex]

3. Calculate each term:
[tex]\[ 4a^4b^2 - 20a^2b - a^2b + 5 \][/tex]

4. Combine like terms:
[tex]\[ 4a^4b^2 - 21a^2b + 5 \][/tex]

This matches our original polynomial [tex]\( 4a^4 b^2 - 21a^2 b + 5 \)[/tex].

Thus, the completely factored form of the polynomial [tex]\( 4a^4 b^2 - 21a^2 b + 5 \)[/tex] is:
[tex]\[ (4a^2b - 1)(a^2b - 5) = 0 \][/tex]

So the correct answer is:
A. [tex]\(\left(4a^2b-1\right)\left(a^2b-5\right)=0\)[/tex]

Other Questions