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If (x+1) is a factor of the polynomial ...

If `(x+1)` is a factor of the polynomial ` (2x^2+kx)` then the value of k is

A

`-2`

B

`-3`

C

`2`

D

`3`

Text Solution

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The correct Answer is:
To find the value of \( k \) such that \( (x + 1) \) is a factor of the polynomial \( (2x^2 + kx) \), we can use the Factor Theorem. According to the Factor Theorem, if \( (x + 1) \) is a factor of the polynomial, then substituting \( x = -1 \) into the polynomial should yield zero. ### Step-by-step Solution: 1. **Set up the polynomial**: We have the polynomial \( P(x) = 2x^2 + kx \). 2. **Substitute \( x = -1 \)**: According to the Factor Theorem, we substitute \( x = -1 \) into the polynomial: \[ P(-1) = 2(-1)^2 + k(-1) \] 3. **Calculate \( P(-1) \)**: \[ P(-1) = 2(1) - k = 2 - k \] 4. **Set the polynomial equal to zero**: Since \( (x + 1) \) is a factor, we set \( P(-1) = 0 \): \[ 2 - k = 0 \] 5. **Solve for \( k \)**: \[ k = 2 \] ### Final Answer: The value of \( k \) is \( 2 \). ---
To find the value of \( k \) such that \( (x + 1) \) is a factor of the polynomial \( (2x^2 + kx) \), we can use the Factor Theorem. According to the Factor Theorem, if \( (x + 1) \) is a factor of the polynomial, then substituting \( x = -1 \) into the polynomial should yield zero. ### Step-by-step Solution: 1. **Set up the polynomial**: We have the polynomial \( P(x) = 2x^2 + kx \). 2. **Substitute \( x = -1 \)**: According to the Factor Theorem, we substitute \( x = -1 \) into the polynomial: \[ ...
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Knowledge Check

  • If ( x+1) is a factor of the polynomial (2x^(2)+kx) then k = ?

    A
    4
    B
    `-3`
    C
    2
    D
    `-2`

  • For a polynomial p(x) of degree ge1, p(a)=0 , where a is a real number, then (x-a) is a factor of the polynomial p(x) For what value of k, the polynomial 2x^(4)+3x^(3)+2kx^(2)+3x+6 is exactly divisible by (x+2) ?

    A
    `0`
    B
    `-1`
    C
    `1`
    D
    `2`

  • If (x - 2) is a factor of polynomial p(x) = x^(3) + 2x^(2) - kx + 10 , then the value of k is:

    A
    10
    B
    11
    C
    12
    D
    13

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