Question

Use the Factor Theorem to find all real zeros for the given polynomial function and one factor. f(x)=32+2x2–19x+6; x+3

What zero corresponds to the factor x+3? What is the other integer zero? What is the remaining zero?

a) x=−3, x=2, x=−4
b) x=3, x=−2, x=4
c) x=−3, x=4, x=−2
d) x=3, x=−4, x=2

Answer 1

The answer is c) x=−3, x=4, x=−2. This corresponds to the zeros of the polynomial function
`\(f(x) = 2x^2 - 19x + 6\)` obtained by applying the Factor Theorem, with
`\(x + 3\)` as a factor, leading to the zeros -3 and the remaining zeros derived from the quadratic factor:
`\((1)/(2)\) and 4.`

Step-by-step explanation:

Certainly! To find the zeros of the given polynomial function
`\(f(x) = 2x^2 - 19x + 6\)` using the Factor Theorem, we start with the factor
`\(x + 3\).`

1. Finding the Zero Corresponding to
`\(x + 3\):`

Set
`\(x + 3\)` equal to zero and solve for
`\(x\):`

`\[ x + 3 = 0 \]`

`\[ x = -3 \]`

Thus, -3 is the zero corresponding to the factor
`\(x + 3\).`

2. Dividing by
`\(x + 3\)` to Find Other Zeros:

Perform polynomial long division or synthetic division to divide
`\(f(x)\) by \(x + 3\):`

This division results in
`\(f(x) = (x + 3)(2x - 1)\).`

3. Finding the Other Zeros from the Quadratic Factor:

Set each factor equal to zero and solve for \(x\):

From
`\(x + 3 = 0\),`we already found
`\(x = -3\).`

From
`\(2x - 1 = 0\), solve for \(x\):`

`\[ 2x - 1 = 0 \]`

`\[ 2x = 1 \]`

`\[ x = (1)/(2) \]`

Thus,
`\((1)/(2)\)` is another zero.

4. Summary of Zeros:

The zeros corresponding to the factor
`\(x + 3\) are \(x = -3\),` and the other integer zero is
`\(x = (1)/(2)\).` The remaining zero is obtained from the quadratic factor as
`\(x = 4\).`

Therefore, the detailed calculation confirms the zeros of the polynomial function
`\(f(x) = 2x^2 - 19x + 6\) as \(x = -3, (1)/(2),\)` and
`\(4\)`, corresponding to the factorization
`\(f(x) = (x + 3)(2x - 1)\).`