Given f, of, x, equals, 2, x, squared, plus, k, x, minus, 1f(x)=2x

Question

2
+kx−1, and the remainder when f, of, xf(x) is divided by x, minus, 2x−2 is 1111, 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)

Answer 1

To solve the problem, we will follow a similar approach to the given example.

Given: \[ f(x) = 2x^2 + kx - 1 \]
and the remainder when \( f(x) \) is divided by \( x - 2 \) is \( 1111 \). We want to find the value of \( k \).

Remainder Theorem: According to the Remainder Theorem, the remainder when a polynomial \( f(x) \) is divided by \( x - a \) is equal to \( f(a) \). Here, we need to find \( f(2) \), and set it equal to \( 1111 \).

Calculating \( f(2) \): \[ f(2) = 2(2^2) + k(2) - 1 \] Calculating \( 2^2 \): \[ f(2) = 2(4) + 2k - 1 \] This simplifies to: \[ f(2) = 8 + 2k - 1 \] \[ f(2) = 2k + 7 \]

Setting the remainder equal to 1111: According to the problem, we know the remainder is \( 1111 \). Thus: \[ 2k + 7 = 1111 \]

Solving for \( k \): Subtract 7 from both sides: \[ 2k = 1111 - 7 \] \[ 2k = 1104 \] Now, divide both sides by 2: \[ k = \frac{1104}{2} \] \[ k = 552 \]

Final Answer: The value of \( k \) is: \[ \boxed{552} \]

Check the answer: To ensure our solution is correct, we can check our calculation of \( f(2) \): \[ f(2) = 2(2^2) + 552(2) - 1 \] Calculating: \[ f(2) = 2(4) + 1104 - 1 = 8 + 1104 - 1 = 1111 \] This confirms our value of \( k = 552 \) is correct since the remainder indeed equals \( 1111 \). Thus, the solution is verified: \[ \boxed{552} \]