Answered

Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

If [tex]$x-3$[/tex] is a factor of [tex]$x^3 - kx^2 + x - 6$[/tex], find the value of [tex]$k$[/tex].

Sagot :

GinnyAnswer
Certainly! To find the value of [tex]\( k \)[/tex] such that [tex]\( x-3 \)[/tex] is a factor of the polynomial [tex]\( x^3 - kx^2 + x - 6 \)[/tex], we can use the Factor Theorem. The Factor Theorem states that [tex]\( x-3 \)[/tex] is a factor of a polynomial if and only if substituting [tex]\( x = 3 \)[/tex] into the polynomial gives zero.

Let's substitute [tex]\( x = 3 \)[/tex] into the polynomial and set it equal to zero:

[tex]\[ P(x) = x^3 - kx^2 + x - 6 \][/tex]

Substituting [tex]\( x = 3 \)[/tex]:

[tex]\[ P(3) = 3^3 - k(3^2) + 3 - 6 \][/tex]

Now, simplify it step-by-step:

1. Calculate [tex]\( 3^3 \)[/tex]:
[tex]\[ 3^3 = 27 \][/tex]

2. Calculate [tex]\( k \cdot 3^2 \)[/tex]:
[tex]\[ k \cdot 3^2 = 9k \][/tex]

3. Calculate the remaining constants ( [tex]\( 3 \)[/tex] and [tex]\( -6 \)[/tex] ):
[tex]\[ 3 - 6 = -3 \][/tex]

Now, combine these steps:

[tex]\[ 27 - 9k + 3 - 6 = 0 \][/tex]

Simplify the equation:

[tex]\[ 27 - 9k - 3 = 0 \][/tex]

[tex]\[ 24 - 9k = 0 \][/tex]

Solving for [tex]\( k \)[/tex]:

[tex]\[ 24 = 9k \][/tex]

[tex]\[ k = \frac{24}{9} \][/tex]

[tex]\[ k = \frac{8}{3} \][/tex]

Therefore, the value of [tex]\( k \)[/tex] that makes [tex]\( x-3 \)[/tex] a factor of [tex]\( x^3 - kx^2 + x - 6 \)[/tex] is:

[tex]\[ k = \frac{8}{3} \][/tex]
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.