SOLUTION: I sent this in earlier, I've now included my work. Need help. Thanks (3+sqrt-5)(7-sqrt-10) (3+sqrt-5)^2(7-sqrt-10)^2=0 (9+5i)(49-10i)=0 441-90i+245i-50i^2=0 441+155i-50(-1)

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I sent this in earlier, I've now included my work. Need help. Thanks (3+sqrt-5)(7-sqrt-10) (3+sqrt-5)^2(7-sqrt-10)^2=0 (9+5i)(49-10i)=0 441-90i+245i-50i^2=0 441+155i-50(-1)      Log On


   

Question 192031This question is from textbook Colleg Algebra A Graphing Approach
: I sent this in earlier, I've now included my work.
Need help. Thanks
(3+sqrt-5)(7-sqrt-10)
(3+sqrt-5)^2(7-sqrt-10)^2=0
(9+5i)(49-10i)=0
441-90i+245i-50i^2=0
441+155i-50(-1)=0
441+155i+50=0
answer 491+155i
a+bi form This question is from textbook Colleg Algebra A Graphing Approach

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%283%2Bsqrt%28-5%29%29%287-sqrt%28-10%29%29 Start with the given expression.


Now let's FOIL the expression.


Remember, when you FOIL an expression, you follow this procedure:


%28highlight%283%29%2Bsqrt%28-5%29%29%28highlight%287%29-sqrt%28-10%29%29 Multiply the First terms:%283%29%2A%287%29=21.


%28highlight%283%29%2Bsqrt%28-5%29%29%287%2Bhighlight%28-sqrt%28-10%29%29%29 Multiply the Outer terms:%283%29%2A%28-sqrt%28-10%29%29=-3%2Asqrt%28-10%29=-3i%2Asqrt%2810%29.


%283%2Bhighlight%28sqrt%28-5%29%29%29%28highlight%287%29-sqrt%28-10%29%29 Multiply the Inner terms:%28sqrt%28-5%29%29%2A%287%29=7%2Asqrt%28-5%29=7i%2Asqrt%285%29.


%283%2Bhighlight%28sqrt%28-5%29%29%29%287%2Bhighlight%28-sqrt%28-10%29%29%29 Multiply the Last terms:%28sqrt%28-5%29%29%2A%28-sqrt%28-10%29%29=5%2Asqrt%282%29.


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So we have the terms: 21, -3i%2Asqrt%2810%29, 7i%2Asqrt%285%29, 5%2Asqrt%282%29


21-3i%2Asqrt%2810%29%2B7i%2Asqrt%285%29%2B5%2Asqrt%282%29 Now add every term listed above to make a single expression.


21%2B5%2Asqrt%282%29%2B7i%2Asqrt%285%29-3i%2Asqrt%2810%29 Rearrange the terms


%2821%2B5%2Asqrt%282%29%29%2B%287%2Asqrt%285%29-3%2Asqrt%2810%29%29%2Ai Group the terms and factor out the term "i"


So %283%2Bsqrt%28-5%29%29%287-sqrt%28-10%29%29 FOILs to %2821%2B5%2Asqrt%282%29%29%2B%287%2Asqrt%285%29-3%2Asqrt%2810%29%29%2Ai.


In other words, .


The number is now in standard form a%2Bbi where a=21%2B5%2Asqrt%282%29 and b=7%2Asqrt%285%29-3%2Asqrt%2810%29