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What is the product?

[tex]\[ 6\left(x^2-1\right) \cdot \frac{6x-1}{6(x+1)} \][/tex]

A. [tex]\( 6(x-1)^2 \)[/tex]

B. [tex]\( 6\left(x^2-1\right) \)[/tex]

C. [tex]\( (x+1)(6x-1) \)[/tex]

D. [tex]\( (x-1)(6x-1) \)[/tex]



Answer :

GinnyAnswer
To solve the given expression:

[tex]\[ 6\left(x^2 - 1\right) \cdot \frac{6x - 1}{6(x + 1)} \][/tex]

let's break it down step-by-step.

### Step-by-Step Solution:

1. Factorize [tex]\( x^2 - 1 \)[/tex]:

Notice that [tex]\( x^2 - 1 \)[/tex] can be factorized using the difference of squares:
[tex]\[ x^2 - 1 = (x - 1)(x + 1) \][/tex]

2. Substitute the factorized form:

Now rewrite the expression by replacing [tex]\( x^2 - 1 \)[/tex]:
[tex]\[ 6\left( (x - 1)(x + 1) \right) \cdot \frac{6x - 1}{6(x + 1)} \][/tex]

3. Simplify the expression:

Next, observe that we have [tex]\( (x + 1) \)[/tex] in both the numerator and the denominator:
[tex]\[ 6(x - 1)(x + 1) \cdot \frac{6x - 1}{6(x + 1)} \][/tex]

The [tex]\( (x + 1) \)[/tex] terms cancel out:
[tex]\[ 6(x - 1) \cdot \frac{6x - 1}{6} \][/tex]

4. Further simplification:

Now notice that there's a factor of 6 in the numerator and denominator which cancels out:
[tex]\[ (x - 1) \cdot (6x - 1) \][/tex]

So, the product of the given expression is:

[tex]\[ (x - 1)(6x - 1) \][/tex]

Hence, the correct answer is:

[tex]\[(x-1)(6x-1)\][/tex]

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