This is the CORRECT WAY to do both problems….

PROBLEM 4A:

sqrt 2 * ( 3sqrt10 + 2sqrt2)….

= (sqrt2 * 3sqrt 10) + (sqrt2 * 2sqrt2)

** see what I just did? each of the terms in the ( )’s get multiplied by the sqrt 2 in front of the ( )’s **

…so when you do that…. now it’s just multiplication…

** hint…. sqrt 10 is the same as (sqrt2)(sqrt5) **

= (sqrt 2 * 3 sqrt 2 * sqrt 5) + (sqrt2 * 2sqrt2)

** hint… rearranging the terms….**

= (3 * sqrt 2 * sqrt 2 * sqrt 5) + (2 * sqrt 2 * sqrt 2)

** hint…. sqrt 2 * sqrt 2 = 2, right? **

= (3 * 2 * sqrt 5) + (2 * 2)

= (6 sqrt 5) + 4

= 4 + 6sqrt5 FINAL ANSWER

PROBLEM 4B:

(sqrt 2 + sqrt 7)(sqrt 3 – sqrt 7)….

**Hint: Use FOIL method to find the resulting polynomial**

You are given: (sqrt 2 + sqrt 7)(sqrt 3 – sqrt 7)

If you multiply… using the FOIL method… the…

first terms… you get (sqrt 2)(sqrt 3) = ? (a)

outer terms… you get (sqrt 2)(-sqrt 7) = ?? (b)

inner terms… you get (+sqrt 7)(+sqrt 3) = ??? (c)

last terms…. you get (+sqrt 7)(-sqrt 7) = ???? (d)

….then when you add all those products up… like this…

? + ?? + ??? + ????

….you get… (a) + (b) + (c) + (d)…..

Now, you “may” have some “like” terms…. in which case you would combine those….. and then you will have your final answer!

Hope these hints help!!!

So if you did everything correctly… this is what you should have come up with…

first terms…. (sqrt 2)(sqrt 3) = sqrt (2 * 3) = sqrt 6 (a)

outer terms… (sqrt 2)(-sqrt 7) = – sqrt (2 * 7) = – sqrt 14 (b)

inner terms… (+sqrt 7)(+sqrt 3) = + sqrt (7 * 3) = + sqrt 21 (c)

last terms…. (+sqrt 7)(-sqrt 7) = -sqrt (7*7) = – sqrt 49 = -7 (d)

….then when you add all those products up… like this…

(a) + (b) + (c) + (d)….

….you get…

(a) + (b) + (c) + (d) = (sqrt 6 – sqrt 14 + sqrt 21 – 7) ANSWER

PLEASE REMEMBER….. to pay attention to your ” + ” and your ” – ” signs!!!! Because forgetting about them will definitely affect your final answer….

to multiply radicals, use distributive property as with numbers

when you get an even number of the same numbers inside a radical take half of them out and multiply each of them by the coefficient. also, only add radicals if they have same number inside radical

4a) sqrt(2)*[3*sqrt(10) + 2*sqrt(2)]

3*sqrt(10*2) + 2*sqrt(2*2)

12*sqrt(5) + 8*sqrt

4b) [sqrt(2) + sqrt(7)]*[sqrt(3) – sqrt(7)]

first outer inner last

so sqrt(2) tiimes sqrt(3)

sqrt(2) times sqrt(7)

sqrt(7) times sqrt(3)

sqrt(7) times sqrt(7)

sqrt(2) tiimes sqrt(3) equals to sqrt(6)

sqrt(2) times sqrt(7) equals to sqrt(14)

sqrt(7) times sqrt(3) equals to sqrt(21)

sqrt(7) times sqrt(7) equals to 7

add them together:

sqrt(6) + sqrt(14) + sqrt(21) + 7

to multiply radicals, use distributive property as with numbers as quickly as you get an staggering form of the comparable numbers indoors an huge take 0.5 of them out and multiply each and each and each and each of them by using the coefficient. additionally, purely upload radicals in the event that they have comparable huge sort indoors radical 4a) sqrt(2)*[3*sqrt(10) + 2*sqrt(2)] 3*sqrt(10*2) + 2*sqrt(2*2) 12*sqrt(5) + 8*sqrt 4b) [sqrt(2) + sqrt(7)]*[sqrt(3) – sqrt(7)] first outer inner very final so sqrt(2) tiimes sqrt(3) sqrt(2) situations sqrt(7) sqrt(7) situations sqrt(3) sqrt(7) situations sqrt(7) sqrt(2) tiimes sqrt(3) equals to sqrt(6) sqrt(2) situations sqrt(7) equals to sqrt(14) sqrt(7) situations sqrt(3) equals to sqrt(21) sqrt(7) situations sqrt(7) equals to 7 upload them mutually: sqrt(6) + sqrt(14) + sqrt(21) + 7