Find the zeros of each function. Part 3. Please show work. 1. f(x)= x^4 - x^3 - x^2 - x - 2
2. f(x)= x^3 - 8x^2 + 19x - 12

9514 1404 393

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.

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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

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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.

14.3

step-by-step explanation:

add all the numbers together.

9 + 12 + 15 + 16 + 17 + 17 = 86

now, divide 86 by how many numbers we have: 6.

86 / 6 = 14.3