F(x)=x^5 + 5*x^4 - 5*x^3 - 25*x^2 + 4*x + 20
By examining the coefficients of the polynomial, we find that
1+5-5-25+4+20=0 => (x-1) is a factor
Now, reverse the sign of coefficients of odd powers,
-1+5+5-25-4+20=0 => (x+1) is a factor
By the rational roots theorem, we can continue to try x=2, or factor x-2=0
2^5+5(2^4)-5(2^3)-25(2^2)+4(2)+20=0
and similarly f(-2)=0
So we have found four of the 5 real roots.
The remainder can be found by synthetic division as x=-5
Answer: The real roots of the given polynomial are: {-5,-2,-1.1.2}
Answer:
is rational number.......
We have that
y = 2x²<span> + 2
step 1
</span><span>exchange the value of x for y and the value of y for x
</span>y = 2x² + 2------> x=2y²+2
step 2
clear y variable
x=2y²+2----> 2y²=x-2----> y²=[(x-2)/2]-----> y=(+/-)√[(x-2)/2]
step 3
the inverse is
f(x)-1=(+/-)√[(x-2)/2]
In the general case, it is (x, y+3), where y = f(x).
(14 1/2)/(1 1/4) change both to improper fractions
(29/2)/(5/4) invert second function (or the denominator fraction) and multiply.
(29/2)*(4/5) multiply...
58/5 change improper fraction back to mixed fraction
11 3/5 or a decimal 11.6
