Can someone please explain these problems to me I have tried and cannot figure them out.
1)Use synthetic division to find p(-3) for p(x)=x^4-2x^3-4x+4
2)Find all the zeroes of the equation
x^4-6x^2-7x-6=0
1. X = -3.
x+3 = 0.
Using synthetic division, divide the given expression by x+3. You will get
x^3 - 5x^2 + 15x - 49. + 151 Remainder.
The fact that we got a remainder, means that x+3 is not a factor of the given
expression and -3 is not a zero. Which
means it will not give zero output when plugged into the Eq.
2. All values of X that satisfy the given Eq is a "zero" of the Eq.
x^4 - 6x^2 - 7x - 6 = 0.
It was determined by trial and error that -2 satisfies the given Eq.
X = -2.
x+2 = 0
Divide the given Eq by x+2 and get
x^3 - 2x^2 - 2x -3. + 0 Remainder.
(x+2)(x^3-2x^2-2x-3) = 0.
3 satisfies the cubic polynomial:
x = 3.
x-3 = 0.
Divide the cubic polynomial by x-3 and
get x^2+x+1 + 0 Remainder.
Our Eq 1s: (x+2)(x-3)(x^2+x+1) = 0.
The 2nd degree polynomial cannot be
factored and has no zeroes, because
B^2 < 4AC. Therefore, the original Eq
has only 2 zeroes: -2, and 3.
OR
(x,y)
(-2,0)
(3,0).