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drag the tiles to the boxes to form correct pairs. not all tiles will b…

Question

drag the tiles to the boxes to form correct pairs. not all tiles will be used. match each binomial with its factors. 16x² - 1 (2x + 1)(2x - 1) 16x² - 4 (2x + 3)(2x - 3) 16x² + 1 4(2x + 1)(2x - 1) 4x² - 1 (4x - 1)(4x + 1) 4x² - 9

Explanation:

Step1: Recall difference - of - squares formula

The difference - of - squares formula is \(a^{2}-b^{2}=(a + b)(a - b)\).

Step2: Factor \(16x^{2}-1\)

We can rewrite \(16x^{2}-1\) as \((4x)^{2}-1^{2}\). Using the difference - of - squares formula, we get \((4x + 1)(4x - 1)\).

Step3: Note \(16x^{2}+1\)

\(16x^{2}+1\) cannot be factored over the real numbers using real - valued binomial factors.

Step4: Factor \(4x^{2}-9\)

Rewrite \(4x^{2}-9\) as \((2x)^{2}-3^{2}\). By the difference - of - squares formula, it factors to \((2x + 3)(2x - 3)\).

Step5: Factor \(16x^{2}-4\)

First, factor out the common factor 4: \(16x^{2}-4 = 4(4x^{2}-1)\). Then factor \(4x^{2}-1=(2x + 1)(2x - 1)\), so \(16x^{2}-4=4(2x + 1)(2x - 1)\).

Step6: Factor \(4x^{2}-1\)

Rewrite \(4x^{2}-1\) as \((2x)^{2}-1^{2}\). Using the difference - of - squares formula, we get \((2x + 1)(2x - 1)\).

Answer:

\(16x^{2}-1\) - \((4x - 1)(4x + 1)\)
\(16x^{2}+1\) - No real - valued binomial factors
\(4x^{2}-9\) - \((2x + 3)(2x - 3)\)
\(16x^{2}-4\) - \(4(2x + 1)(2x - 1)\)
\(4x^{2}-1\) - \((2x + 1)(2x - 1)\)