Answer: x1=1 x2=-2 and x3=2
Step-by-step explanation:
1st x1=1 is 1 of the roots , so
F(1)=1-1-4+4=0 - true
So lets divide x^3-x^2-4x+4 by (x-x1), i.e (x^3-x^2-4x+4) /(x-1)=(x^2-4)
x^2-4 can be factorized as (x-2)*(x+2)
So x^3-x^2-4x+4=(x-1)*(x^2-4)=(x-1)(x-2)*(x+2)
So there are 3 dofferent roots:
x1=1 x2=-2 and x3=2
Answer:
They answer is D.
Step-by-step explanation:
Hope this helps. Have a nice day!
Answer:
Find a common denominator on the right side of the equation
Step-by-step explanation:
The equation before the problem is
X² + b/a(x) + (b/2a)²= -c/a + b²/4a²
The next step in solving the above equation is to fibd tge common denominator on the right side of the equation.
X² + b/a(x) + (b/2a)²= -c/a + b²/4a²
X² + b/a(x) + (b/2a)²= -4ac/4a² + b²/4a²
X² + b/a(x) + (b/2a)²=( b²-4ac)/4a²
The right side of the equation now has a common denominator
The next step is to factorize the left side of the equation.
(X+b/2a)²= ( b²-4ac)/4a²
Squaring both sides
X+b/2a= √(b²-4ac)/√4a²
Final equation
X=( -b+√(b²-4ac))/2a
Or
X=( -b-√(b²-4ac))/2a
8r and -6r are like terms. What is 8r-6r?
Answer:
the length of missing side is 28
