Please answer and explain your reasoning thanks!

Mathematics

2 answers:

Alla [95]3 years ago

7 0

The first one is the correct answer.
When you multiply, you end up with 49x^16+28xy^40-28x^40 -16y^64
The two middle terms cancel out so you end up with 49x^16-16^64.

Ksju [112]3 years ago

5 0

$a^2-b^2=(a-b)(a+b)$

This formula will be useful in helping us factor this.

$49x^{16}-16y^{64}$

This is the equation that we are going to factor.
Let's first change this equation to have the form $a^2-b^2$

In order to do that, here are two important exponent rules that was used.
$(a^b)^c=a^{bc}$
$(ab)^c=a^c b^c$

With those rules, we can change the equation into the following:
$(7x^8)^2-(4y^{32})^2$

Refer back to the formula I showed in the beginning of this answer.
$a^2-b^2=(a-b)(a+b)$

$(7x^8)^2-(4y^{32})^2=(7x^8-4y^{32})(7x^8-4y^{32})$

Thus, your answer is the first choice. Hope this helps! :)

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Which term is correct when simplifying √128? A.) 16 B.) 12 C.) 10 D.) 8√2

alina1380 [7]

When dealing with roots, we want to factor to see if it's possible

$128 = 2 \times 64$
$= 2 \times 2 \times 32$
$= 2 \times 2 \times 2 \times 16$
$= 2 \times 2 \times 2 \times 2 \times 8$
$= 2 \times 2 \times 2 \times 2 \times 2 \times 4$
$= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

now since we're dealing with square roots, we circle every pair of numbers we see in the final factoring
so..
$= (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times 2$

now for every circled pair, we take only 1 of the 2 circled and bring it outside
anything left uncircled stays inside the root
so..
$= 2 \times 2 \times 2 \times \sqrt{2}$

now simply multiply
$8 \sqrt{2}$