Which of the following is a factor of x4+6x3−29x2−114x+280

, given that x−4
is one of the linear factors?(1 point)
Responses

x−5
x minus 5

x2+7x+10
x squared plus 7 x plus 10

x−7
x minus 7

x−2

1 answer

To determine which of the given options is a factor of the polynomial \( x^4 + 6x^3 - 29x^2 - 114x + 280 \) given that \( x - 4 \) is a factor, we can perform polynomial long division or use synthetic division with \( x - 4 \) to find the quotient.

Let's start with synthetic division using \( x = 4 \):

  1. Set up the coefficients of the polynomial: \( [1, 6, -29, -114, 280] \).
  2. Use \( 4 \) in synthetic division:
   4 |  1   6   -29  -114   280
     |      4    40    44    -280
     --------------------------
       1   10   11   -70    0

The quotient we obtain is \( x^3 + 10x^2 + 11x - 70 \).

Now we need to factor \( x^3 + 10x^2 + 11x - 70 \) further. To find a root, we can test the potential rational roots. We start with integer values:

Let's test \( x = 2 \):

\[ f(2) = 2^3 + 10(2^2) + 11(2) - 70 = 8 + 40 + 22 - 70 = 0. \]

Since \( x = 2 \) is a root, we divide \( x^3 + 10x^2 + 11x - 70 \) by \( x - 2 \):

Perform synthetic division again with \( x = 2 \):

   2 |  1   10   11   -70
     |      2    24     70
     ---------------------
       1   12   35     0

This means \( x^3 + 10x^2 + 11x - 70 \) factors as \( (x - 2)(x^2 + 12x + 35) \).

Next, we can factor \( x^2 + 12x + 35 \):

\[ x^2 + 12x + 35 = (x + 5)(x + 7). \]

Thus, the full factorization of the polynomial \( x^4 + 6x^3 - 29x^2 - 114x + 280 \) is:

\[ (x - 4)(x - 2)(x + 5)(x + 7). \]

Given the options, the factors we found, specifically \( x - 2 \) and \( x + 5 \) match with the potential choices. However, since \( x - 2 \) is explicitly listed, it is the correct answer.

Thus, the factor of \( x^4 + 6x^3 - 29x^2 - 114x + 280 \) is:

x - 2.