Respuesta :

sqdancefan

9514 1404 393

Answer:

  1. real: -1, 2; complex: +i, -i
  2. 1, 3, 4

Step-by-step explanation:

1. The graph (red) shows the only real zeros to be -1 and 2. When the corresponding factors are divided from the function, the remaining factor is the quadratic (x^2 +1), which has only complex roots. The quadratic is graphed in green.

The linear factorization is ...

  f(x) = (x +1)(x -2)(x -i)(x +i)

The roots are -1, 2, -i, +i.

__

2. The graph (blue) shows the zeros are 1, 3, 4.

You observe that the sum of coefficients is zero, so x=1 is a root. Factoring that out gives the quadratic (x^2 -7x +12), which you recognize factors as

  (x -3)(x -4) . . . zeros of 3 and 4

__

I have attached a spreadsheet that does synthetic division. There are web sites that will do this, too. The tables shown correspond to f1(x)/(x-2) and f2(x)/(x-1). When you fill in the zero and coefficients, the built-in formulas do the rest.

Ver imagen sqdancefan
Ver imagen sqdancefan
izzy4160

9514 1404 393

Answer:

real: -1, 2; complex: +i, -i

1, 3, 4

Step-by-step explanation:

1. The graph (red) shows the only real zeros to be -1 and 2. When the corresponding factors are divided from the function, the remaining factor is the quadratic (x^2 +1), which has only complex roots. The quadratic is graphed in green.

The linear factorization is ...

 f(x) = (x +1)(x -2)(x -i)(x +i)

The roots are -1, 2, -i, +i.

__

2. The graph (blue) shows the zeros are 1, 3, 4.

You observe that the sum of coefficients is zero, so x=1 is a root. Factoring that out gives the quadratic (x^2 -7x +12), which you recognize factors as

 (x -3)(x -4) . . . zeros of 3 and 4

__ The tables shown correspond to f1(x)/(x-2) and f2(x)/(x-1).

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