Difference of 2 perfec squares is
(a^2)-(b^2)
if the exponents are both even and the coeficient (the number in front) are perfect squares, then it is difference t 2 perfect squares
first one
8 is not perfect square
2nd one
(4e^4)^2-(9g^2)^2
third
25 is odd, so it cannot be split up into 2 nice numbers
4th
(11m^9)^2-(3n^5)^2
<span>97.68 - 32.3=65.38 :)</span>
First, find any zero of the polynomial. Since you didn't ask for work, I'll assume it's okay if I use my calculator. Your given polynomial has only one real root which is x=-4.
Now we use the rule that x-a divides the polynomial where a is a zero of said polynomial.
So x+4 divides 2x^3+2x^2-19x+20.
<span>(2x^3+2x^2-19x+20) / (x+4 equals 2x^2-6x+5).
If we factor out a two, we can use the quadratic formula.
2(x^2-3x+2.5) so we have x = (-(-3)+/-(9-4*1*2.5)^(1/2))/2*1)=(3+i)... or (3-i)/2 Where i is the square root of negative one. final answer:
2x^3+2x^2-19x+20=0
then x=-4, (3+i)/2, or (3-i)/2
</span>we have two imaginary number.
I hope it helped you
Answer:
x = 2
Step-by-step explanation:
6(x-4) + 7 = -5
6(x-4) + 7 - 7 = -5 - 7
6(x-4) = -12
6(x-4)/6 = -12/6
x - 4 = -2
x - 4 + 4 = -2 + 4
x = 2
