3
−5x+k, and the remainder when f, of, xf(x) is divided by x, minus, 1x−1 is 1414, 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 the function
\[ f(x) = 2x^3 - 5x + k \]
and we know that the remainder when \( f(x) \) is divided by \( x - 1 \) is 14. According to the Remainder Theorem:
\[
f(1) = 14
\]
Now, let's evaluate \( f(1) \):
\[
f(1) = 2(1)^3 - 5(1) + k
\]
Calculating the individual terms:
\[
f(1) = 2(1) - 5 + k
\]
\[
f(1) = 2 - 5 + k
\]
\[
f(1) = -3 + k
\]
Setting this equal to the remainder:
\[
-3 + k = 14
\]
Now, solve for \( k \):
\[
k = 14 + 3
\]
\[
k = 17
\]
Check Answer:
Now we can verify by substituting \( k = 17 \) back into our function \( f(x) \):
