Since 16x³+kx²-2x+1 has 16x³+kx²-2x+1 as one of its factors, this means that if we divide the polynomial by 16x³+kx²-2x+1, we should get a factorable result.
Let's set up the division:
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16x³+kx²-2x+1 | 16x³ + 0x² + 0x + 0
To obtain a factorable result, the first term of the quotient should be 16x. Multiplying 16x by 16x³+kx²-2x+1 gives us 16x^4 + kx^3 - 32x² + 16x.
We then subtract this from the original polynomial:
16x³ + 0x² + 0x + 0
-(16x⁴ + kx³ - 32x² + 16x)
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16x³ - kx³ + 32x² - 16x
Now, for this to be a factorable result, the second term of the quotient should be -kx. Multiplying -kx by 16x³+kx²-2x+1 gives us -kx² - k^2x.
We subtract this from our previous result:
16x³ - kx³ + 32x² - 16x
-(-kx² - k^2x)
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16x³ - kx³ + 32x² - 16x + kx² + k^2x
For this to be factorable, we require the third term of the quotient to be 32x. Multiplying 32x by 16x³+kx²-2x+1 gives us 512x^2 + 32kx.
Subtracting this from the previous result:
16x³ - kx³ + 32x² - 16x + kx² + k^2x
-(512x² + 32kx)
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16x³ - kx³ + 32x² - 16x + kx² + k^2x - 512x² - 32kx
In order for the polynomial to be completely divisible, the remainder should be zero. Therefore, the coefficient of x in the remainder should be zero:
(-16 + k - 512 - 32) = 0
k - 528 = 0
k = 528
Therefore, k = 528.
The polynomial 16x³+kx²-2x+1 as one of its factors. Find k.
1 answer
1 answer