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A student uses the division shown to divide –3x4 15x3 – x 5 by x – 5, and concludes that x – 5 is not a factor of the polynomial. describe two errors the student made. is x – 5 a factor? explain.

Respuesta :

watermelonfairy
The student used k = –5 instead of k = 5. And the student did not use a placeholder zero for the x2 term. Also, x – 5 is a factor because there is no remainder when the division is performed correctly.

The synthetic division of the polynomial can be used to determine the

factors of the polynomial.

The two errors are;

1) Error in the test zero sign

2) Error in missing coefficient of 0 for x²

Yes x - 5 is a factor of -3x⁴ + 15·x³ - x + 5

Required:

To describe two errors made.

Solution:

The given synthetic division is presented as follows;

-3x⁴ + 15·x³ - x + 5 divided by x - 5

[tex]-5\left|\begin{array}{cccc}-3&15&-1&5\\&15&-150&755\\\end{array} }}\\ \\\overline{\hspace {23}-3 \ \ \ 30 \ \ -151 \ \ \ 760}}[/tex]

Two errors made are;

  • 1) The value to be used as the test zero is +5, because (x - 5) = 0 gives x = +5
  • 2) The coefficient of x² should be included as 0

Required:

The determine if x - 5 is a factor

Solution:

x - 5 is a factor of f(x) = -3·x⁴ + 15·x³ - x + 5, if f(5) = 0

f(5) = -3·(5)⁴ + 15·(5)³ - 5 + 5 = -1875 + 1875 - 5 + 5 = 0

Therefore, (x - 5) is a factor of -3·x⁴ + 15·x³ - x + 5

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