5/(x^2-4)+2/x=2/(x-2) if we factor the leading denominator, it is a "difference of squares" of the form (a^2-b^2) which always factors to (a-b)(a+b) so we have:
5/((x-2)(x+2))+2/x=2/(x-2), to add/subtract fractions we need a common denominator, so in this case we need a common denominator of x(x-2)(x+2) so
5(x)+2(x+2)(x-2)=2(x)(x+2), perform indicated operations...
5x+2(x^2-4)=2x(x+2)
5x+2x^2-8=2x^2+4x subtract 2x from both sides
5x-8=4x subtract 4x from both sides
x-8=0 add 8 to both sides
x=8
check...
5/(x^2-4)+2/x=2/(x-2) when x=8 so
5/(64-4)+2/8=2/(8-2)
5/60+1/4=2/6
5/60+15/60=20/60
20/60=20/60
Answer:
x^7/6
7 square root x^6 would be x ^7/6
Answer:
Step-by-step explanation:
Here we are given that a polynomial has zeros as 2 , i and -i . We need to find out the cubic polynomial . In general we know that if 
are the zeros of the cubic polynomial , then ,
Here in place of the Greek letters , substitute 2,i and -i , we get ,
Now multiply (x-i) and (x+i ) using the identity (a+b)(a-b)=a² - b² , we have ,
Simplify using i = √-1 ,
Multiply by distribution ,
Simplify by opening the brackets ,
Rearrange ,
8x - 16y = a common factor would be 8...so factor out 8
8(x - 2y)
