3
+kx−5, and x, plus, 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:
\( f(x) = x^3 + kx - 5 \)
and \( x + 1 \) is a factor of \( f(x) \).
Factor Theorem:
If \( (x + 1) \) is a factor of \( f(x) \), then:
\[ f(-1) = 0 \]
Substituting \( k = -6 \) back into \( f(x) \):
\[ f(x) = x^3 - 6x - 5 \]
Now we check if \( (x + 1) \) is a factor of \( f(x) \) by performing polynomial long division or direct substitution:
\[ f(-1) = (-1)^3 - 6(-1) - 5 \]
\[ f(-1) = -1 + 6 - 5 \]
\[ f(-1) = 0 \]
Since \( f(-1) = 0 \), \( (x + 1) \) is indeed a factor of \( f(x) \).
Conclusion:
The value of \( k \) is \( -6 \).
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