3
+7x+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
To solve the problem using the same format, let's start by using the given information and applying the Factor Theorem.
Given:
\[ f(x) = 2x^3 + 7x + k \]
and
\[ x - 1 \text{ is a factor of } f(x) \]
According to the Factor Theorem:
If \( x - 1 \) is a factor of \( f(x) \), then:
\[ f(1) = 0 \]
Now, let's calculate \( f(1) \):
\[
f(1) = 2(1)^3 + 7(1) + k
\]
\[
f(1) = 2(1) + 7 + k
\]
\[
f(1) = 2 + 7 + k
\]
\[
f(1) = 9 + k
\]
Setting \( f(1) = 0 \):
\[
9 + k = 0
\]
Now, solve for \( k \):
\[
k = -9
\]
Check Answer:
Now, let's check if substituting \( k = -9 \) makes \( x - 1 \) a factor of \( f(x) \):
Substituting \( k \) back into \( f(x) \):
\[
f(x) = 2x^3 + 7x - 9
\]
Now we divide \( f(x) \) by \( x - 1 \) to check if the remainder is indeed 0.
Using synthetic division or polynomial long division, we find:
Coefficients of \( f(x) \) are \( [2, 0, 7, -9] \).
Perform synthetic division using \( x = 1 \):
1 | 2 0 7 -9
| 2 2 9
-------------------
2 2 9 0
The remainder is 0.
So, when \( k = -9 \), the remainder is 0, confirming that \( (x - 1) \) is indeed a factor of \( f(x) \).
