How do you evaluate $(6 + \sqrt{7}) (4 - \sqrt{7})$ $(6+ \sqrt { 7} ) ( 4- \sqrt { 7} )$ ?

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How do you evaluate $(6 + \sqrt{7}) (4 - \sqrt{7})$ $(6+ \sqrt { 7} ) ( 4- \sqrt { 7} )$ ?

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Answer 1 · 2017-09-18T17:58:38

You can use the F.O.I.L. method but that only applies to binomials (such as these). A more general method is to write each term of the first factor as a multiplier of the second factor.

Explanation:

Write each term of the first factor as a multiplier of the second factor:

$(6 + \sqrt{7}) (4 - \sqrt{7}) = 6 (4 - \sqrt{7}) + \sqrt{7} (4 - \sqrt{7})$ $(6+ sqrt7) (4-sqrt 7) = 6(4-sqrt7) + sqrt7(4-sqrt7)$

Use the distributive property on both terms of the right side:

$(6 + \sqrt{7}) (4 - \sqrt{7}) = 24 - 6 \sqrt{7} + 4 \sqrt{7} - \sqrt{7} \sqrt{7}$ $(6+ sqrt7) (4-sqrt 7) = 24-6sqrt7 + 4sqrt7-sqrt7sqrt7$

The last term becomes -7:

$(6 + \sqrt{7}) (4 - \sqrt{7}) = 24 - 6 \sqrt{7} + 4 \sqrt{7} - 7$ $(6+ sqrt7) (4-sqrt 7) = 24-6sqrt7 + 4sqrt7-7$

Combine like terms:

$(6 + \sqrt{7}) (4 - \sqrt{7}) = 17 - 2 \sqrt{7}$ $(6+ sqrt7) (4-sqrt 7) = 17-2sqrt7$

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How do you evaluate $(6 + \sqrt{7}) (4 - \sqrt{7})$ $(6+ \sqrt { 7} ) ( 4- \sqrt { 7} )$ ?

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How do you evaluate (6+√7)(4−√7)(6+ \sqrt { 7} ) ( 4- \sqrt { 7} )?

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How do you evaluate $(6 + \sqrt{7}) (4 - \sqrt{7})$ $(6+ \sqrt { 7} ) ( 4- \sqrt { 7} )$ ?