Given f, of, x, equals, 3, x, cubed, minus, 8, x, plus, kf(x)=3x

3
−8x+k, and x, minus, 1x−1 is a factor of f, of, xf(x), then what is the value of kk?

Solve the problem just like the example below
⬇️
Given
f, of, x, equals, 2, x, cubed, plus, k, x, minus, 7
f(x)=2x
3
+kx−7, and
x, minus, 1
x−1 is a factor of
f, of, x
f(x), then what is the value of
k
k?
Factor Theorem:
Factor Theorem:
left parenthesis, x, minus, 1, right parenthesis, is a factor of , f, of, x, colon
(x−1) is a factor of f(x):
f, of, 1, equals, 0
f(1)=0
2, left parenthesis, 1, right parenthesis, cubed, plus, k, left parenthesis, 1, right parenthesis, minus, 7, equals, 0
2(1)
3
+k(1)−7=0
2, left parenthesis, 1, right parenthesis, plus, k, minus, 7, equals, 0
2(1)+k−7=0
2, plus, k, minus, 7, equals, 0
2+k−7=0
k, minus, 5, equals, 0
k−5=
0
plus, 5, plus, 5
+5=
+5
k, equals, 5
k=
5
Check Answer:
Check Answer:
left parenthesis, 2, x, cubed, plus, 5, x, minus, 7, right parenthesis, divided by, left parenthesis, x, minus, 1, right parenthesis
(2x
3
+5x−7)÷(x−1)
x, minus, 1
x−1
2, x, cubed
2x
3

plus, 5, x
+5x
minus, 7
−7

1
1
2
2
0
0
5
5
minus, 7
−7

down arrow

2
2
2
2
7
7

2
2
2
2
7
7
0
0

When , k, equals, 5, , the remainder is , 0
When k=5, the remainder is 0
So , left parenthesis, x, minus, 1, right parenthesis, is a factor of , f, of, x
So (x−1) is a factor of f(x)

1 answer

Given \( f(x) = 3x^3 - 8x + k \)

and \( x - 1 \) is a factor of \( f(x) \),

then what is the value of \( k \)?

Factor Theorem: If \( (x - 1) \) is a factor of \( f(x) \), then \( f(1) = 0 \).

Calculating \( f(1) \): \[ f(1) = 3(1)^3 - 8(1) + k \] \[ = 3(1) - 8 + k \] \[ = 3 - 8 + k \] \[ = -5 + k \]

Setting \( f(1) \) equal to 0: \[ -5 + k = 0 \]

Solving for \( k \): \[ k = 5 \]

Check Answer: We can verify by plugging the value of \( k \) back into the function and confirming that \( (x - 1) \) is indeed a factor.

Calculating \( f(x) \) with \( k = 5 \): \[ f(x) = 3x^3 - 8x + 5 \]

Now, we will divide \( f(x) \) by \( x - 1 \): \[ (3x^3 - 8x + 5) \div (x - 1) \]

Using synthetic division: \[ \begin{array}{r|rrr} 1 & 3 & 0 & -8 & 5 \ & & 3 & 3 & -5 \ \hline & 3 & 3 & -5 & 0 \ \end{array} \]

The remainder is \( 0 \), confirming \( (x - 1) \) is a factor of \( f(x) \).

Thus, when \( k = 5 \), the remainder is \( 0 \).

So, the value of \( k \) is: \[ \boxed{5} \]