Find the real zeros of the following:
(a) f(x)= x^3+2x^2-4x-8
(b) g(x)= -x^2+2

  (a) -2 (multiplicity 2) and +2

  (b) ±√2

Step-by-step explanation:

(a) Factor by grouping:

  f(x) = (x^3 +2x^2) -(4x +8)

     = x^2(x +2) -4(x +2) . . . . . remove the common factor from each group

     = (x^2 -4)(x +2) . . . . . . . . .collect the factors

     = (x -2)(x +2)(x +2) . . . . . . factor the difference of squares

Zeros are the values of x that make these factors be zero: {2, -2, -2}.

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(b) g(x) = 0 = -x^2 +2

You can factor this as the difference of squares:

  (√2 -x)(√2 +x) = 0

The zeros are the values of x that make these factors be zero: {√2, -√2}.

the answer is a

step-by-step explanation:

because your just changing the position of where the shape is

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x