Question

question
if $f(x) = x^3 - 12x^2 + 41x - 42$ and $x - 7$ is a factor of $f(x)$, then find all of the zeros of $f(x)$ algebraically.
answer attempt 1 out of 2
add a zero no zeros
$x = $ submit answer

Explanation:

Step1: Perform Polynomial Division or Use Synthetic Division

Since \( x - 7 \) is a factor of \( f(x)=x^{3}-12x^{2}+41x - 42 \), we can use synthetic division to divide \( f(x) \) by \( x - 7 \).
The coefficients of \( f(x) \) are \( 1, - 12,41,-42 \) and we use \( x = 7 \) for synthetic division.
Bring down the first coefficient: \( 1 \)
Multiply \( 1\times7 = 7 \), add to the second coefficient: \( - 12+7=-5 \)
Multiply \( - 5\times7=-35 \), add to the third coefficient: \( 41+( - 35)=6 \)
Multiply \( 6\times7 = 42 \), add to the fourth coefficient: \( - 42 + 42=0 \)
So the quotient polynomial is \( x^{2}-5x + 6 \)

Step2: Factor the Quotient Polynomial

Factor \( x^{2}-5x + 6 \). We need two numbers that multiply to \( 6 \) and add to \( - 5 \). The numbers are \( - 2 \) and \( - 3 \).
So \( x^{2}-5x + 6=(x - 2)(x - 3) \)

Step3: Find the Zeros

The factors of \( f(x) \) are \( (x - 7)(x - 2)(x - 3) \). To find the zeros, set \( f(x)=0 \)
\( (x - 7)(x - 2)(x - 3)=0 \)
Using the zero - product property, \( x-7 = 0 \) or \( x - 2=0 \) or \( x - 3=0 \)
So \( x = 7 \), \( x = 2 \), \( x = 3 \)

Answer:

The zeros of \( f(x) \) are \( x = 2 \), \( x = 3 \), \( x = 7 \)