Question

Match each binomial with its factors. Tiles 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 Pairs

Answer 1

You can do this in two ways. The easiest way using the special case formulas since these are perfect squares. Another would be multiplying the pairs and seeing what matches the other.   
Here are the formulas for special cases:

$a^{2} + 2ab + b^{2} = (a+b)^{2}$
$a^{2} - 2ab + b^{2} = (a-b)^{2}$
$a^{2} - b^{2} = (a+b)(a-b)$

1. (2x+1)(2x-1) This follows the third formula where a=2x and b = 1
$(2x+1)(2x-1) = 4x^{2}-1^{2} = 4x^{2}-1$

2. (2x + 3)(2x − 3) This also follows the third formula where a=2x and b = 3
$(2x+3)(2x-3) = 4x^{2}-3^{2} = 4x^{2}-9$

3.  4(2x + 1)(2x − 1) This follows the third formula but the result should be multiplied by 4. We can do this by combining the pairs first before multiplying it by 4. The associative property of multiplication allows this.
$4(2x+1)(2x-1) = 4(4x^{2}-1^{2}) = 4(4x^{2}-1)$

From here we can use the distributive property:
$4(4x^{2}-1) = 16x^{2}-4$

4. (4x − 1)(4x + 1) This also uses the third formula where a= 4x and b= 1
$(4x-1)(4x+1) = 16x^{2}-1^{2} = 16x^{2}-1$

So let's match up the pairs with their results:
$16x^{2}-1=(4x-1)(4x+1)$
$16x^{2}-4 = 4(2x + 1)(2x − 1)$
$4x^{2}-1 = (2x+1)(2x-1)$
$4x^{2}-9 = (2x+3)(2x-3)$