.
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can be simplified as the product of an integer and the square root of 2:
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is the product of two binomials:
.Binomial Two: 
![({\bf 8} - {\bf \sqrt{2}}) \cdot {(2 + 2\; \sqrt{2})}\\= [{\bf 8} \cdot {(2 + 2\; \sqrt{2})}] - [{\bf \sqrt{2}} \cdot {(2 + 2\; \sqrt{2})}]](/tpl/images/0386/2883/b787a.png)
![[8 \cdot ({\bf 2} + {\bf 2\; \sqrt{2}})] - [\sqrt{2} \cdot ({\bf 2} + {\bf 2\; \sqrt{2}})]\\= [8 \times {\bf 2} + 8 \times {\bf 2\;\sqrt{2}}] - [\sqrt{2} \times {\bf 2} + \sqrt{2} \times {\bf 2\; \sqrt{2}}]](/tpl/images/0386/2883/38dc0.png)
.
.![[8 \times 2 + 8 \times 2\;\sqrt{2}] - [\sqrt{2} \times 2 + 2 \sqrt{2} \times \sqrt{2}]\\ =[16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 4]\\= 16 + 16\;\sqrt{2} - 2\; \sqrt{2} - 4](/tpl/images/0386/2883/a10e3.png)
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. 
.![(8 - \sqrt{2}) \cdot (2 + 2\; \sqrt{2})\\= [8 \cdot (2 + 2\; \sqrt{2})] - [\sqrt{2} \cdot (2 + 2\; \sqrt{2})]\\= [8 \times 2 + 8 \times 2\;\sqrt{2}] - [\sqrt{2} \times 2 + \sqrt{2} \times 2 \;\sqrt{2}]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 2 \times (\sqrt{2})^{2}]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 2 \times 2]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 4]\\= 16 + 16\;\sqrt{2} - 2\; \sqrt{2} - 4\\= (16 - 4) + (16 \; \sqrt{2} - 2\; \sqrt{2})\\](/tpl/images/0386/2883/e9899.png)
.What is (8-√2)(2+√8)?
show work. i've been stuck on this question for a long time and can't...
Mathematics, 22.11.2019 10:31 leannaadrian
What is (8-√2)(2+√8)?
show work. i've been stuck on this question for a long time and can't figure it out.
Answers: 1
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