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Sagot :
To find the value of [tex]\( k \)[/tex] for which the polynomial [tex]\( h(x) \)[/tex] is exactly divisible by the polynomial [tex]\( g(x) \)[/tex], we need to perform polynomial division of [tex]\( h(x) \)[/tex] by [tex]\( g(x) \)[/tex].
Given polynomials:
[tex]\[ h(x) = x^3 + 8x^2 + 12x + 1 \][/tex]
[tex]\[ g(x) = x^2 + 6x + 9 \][/tex]
### Step 1: Perform Polynomial Division
1. Divide the leading term of [tex]\( h(x) \)[/tex] by the leading term of [tex]\( g(x) \)[/tex]:
[tex]\[ \frac{x^3}{x^2} = x \][/tex]
So, the first term of the quotient is [tex]\( x \)[/tex].
2. Multiply [tex]\( g(x) \)[/tex] by [tex]\( x \)[/tex] and subtract from [tex]\( h(x) \)[/tex]:
[tex]\[ x(x^2 + 6x + 9) = x^3 + 6x^2 + 9x \][/tex]
[tex]\[ h(x) - (x^3 + 6x^2 + 9x) = (x^3 + 8x^2 + 12x + 1) - (x^3 + 6x^2 + 9x) \][/tex]
[tex]\[ = (x^3 - x^3) + (8x^2 - 6x^2) + (12x - 9x) + 1 \][/tex]
[tex]\[ = 2x^2 + 3x + 1 \][/tex]
3. Now take the result [tex]\( 2x^2 + 3x + 1 \)[/tex] and divide by the leading term of [tex]\( g(x) \)[/tex]:
[tex]\[ \frac{2x^2}{x^2} = 2 \][/tex]
So, the next term of the quotient is 2.
4. Multiply [tex]\( g(x) \)[/tex] by 2 and subtract from the result:
[tex]\[ 2(x^2 + 6x + 9) = 2x^2 + 12x + 18 \][/tex]
[tex]\[ (2x^2 + 3x + 1) - (2x^2 + 12x + 18) = (2x^2 - 2x^2) + (3x - 12x) + (1 - 18) \][/tex]
[tex]\[ = -9x - 17 \][/tex]
### Step 2: Check the remainder
For [tex]\( h(x) \)[/tex] to be exactly divisible by [tex]\( g(x) \)[/tex], the remainder must be 0. However, we have found that the remainder is:
[tex]\[ -9x - 17 \][/tex]
### Step 3: Solve for [tex]\( k \)[/tex]
In this case, the division process does not involve any [tex]\( k \)[/tex]. The absence of [tex]\( k \)[/tex] in the result means there's no value of [tex]\( k \)[/tex] within our constraints that makes [tex]\( h(x) \)[/tex] exactly divisible by [tex]\( g(x) \)[/tex].
Thus, the quotient is [tex]\( x + 2 \)[/tex], the remainder is [tex]\( -9x - 17 \)[/tex], and there is no value of [tex]\( k \)[/tex] that sets the remainder to 0, resulting in [tex]\( k \)[/tex]:
[tex]\[ k = \text{No solution found} \][/tex]
Hence, there is no value of [tex]\( k \)[/tex] for which the polynomial [tex]\( h(x) \)[/tex] is exactly divisible by [tex]\( g(x) \)[/tex].
Given polynomials:
[tex]\[ h(x) = x^3 + 8x^2 + 12x + 1 \][/tex]
[tex]\[ g(x) = x^2 + 6x + 9 \][/tex]
### Step 1: Perform Polynomial Division
1. Divide the leading term of [tex]\( h(x) \)[/tex] by the leading term of [tex]\( g(x) \)[/tex]:
[tex]\[ \frac{x^3}{x^2} = x \][/tex]
So, the first term of the quotient is [tex]\( x \)[/tex].
2. Multiply [tex]\( g(x) \)[/tex] by [tex]\( x \)[/tex] and subtract from [tex]\( h(x) \)[/tex]:
[tex]\[ x(x^2 + 6x + 9) = x^3 + 6x^2 + 9x \][/tex]
[tex]\[ h(x) - (x^3 + 6x^2 + 9x) = (x^3 + 8x^2 + 12x + 1) - (x^3 + 6x^2 + 9x) \][/tex]
[tex]\[ = (x^3 - x^3) + (8x^2 - 6x^2) + (12x - 9x) + 1 \][/tex]
[tex]\[ = 2x^2 + 3x + 1 \][/tex]
3. Now take the result [tex]\( 2x^2 + 3x + 1 \)[/tex] and divide by the leading term of [tex]\( g(x) \)[/tex]:
[tex]\[ \frac{2x^2}{x^2} = 2 \][/tex]
So, the next term of the quotient is 2.
4. Multiply [tex]\( g(x) \)[/tex] by 2 and subtract from the result:
[tex]\[ 2(x^2 + 6x + 9) = 2x^2 + 12x + 18 \][/tex]
[tex]\[ (2x^2 + 3x + 1) - (2x^2 + 12x + 18) = (2x^2 - 2x^2) + (3x - 12x) + (1 - 18) \][/tex]
[tex]\[ = -9x - 17 \][/tex]
### Step 2: Check the remainder
For [tex]\( h(x) \)[/tex] to be exactly divisible by [tex]\( g(x) \)[/tex], the remainder must be 0. However, we have found that the remainder is:
[tex]\[ -9x - 17 \][/tex]
### Step 3: Solve for [tex]\( k \)[/tex]
In this case, the division process does not involve any [tex]\( k \)[/tex]. The absence of [tex]\( k \)[/tex] in the result means there's no value of [tex]\( k \)[/tex] within our constraints that makes [tex]\( h(x) \)[/tex] exactly divisible by [tex]\( g(x) \)[/tex].
Thus, the quotient is [tex]\( x + 2 \)[/tex], the remainder is [tex]\( -9x - 17 \)[/tex], and there is no value of [tex]\( k \)[/tex] that sets the remainder to 0, resulting in [tex]\( k \)[/tex]:
[tex]\[ k = \text{No solution found} \][/tex]
Hence, there is no value of [tex]\( k \)[/tex] for which the polynomial [tex]\( h(x) \)[/tex] is exactly divisible by [tex]\( g(x) \)[/tex].
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