Posted by **lee** on Sunday, November 25, 2012 at 3:15pm.

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

- algebra -
**Henry**, Tuesday, November 27, 2012 at 10:27pm
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).

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