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Question
Find k so that x2 + 2x + k is a factor of 2x4 + x3 – 14 x2 + 5x + 6. Also find all the zeroes of the two polynomials.
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Solution
Using division algorithm, we get
`2x^2 - 3x + (-8 - 2k)`
`x^2 + 2x + k")"overline(2x^4 + x^3 - 14x^2 + 5x + 6)`
2x4 + 4x3 + 2kx2
(–) (–) (–)
–3x3 – (2k + 14)x2 + 5x + 6
–3x3 – 6x2 – 3kx
(+) (+) (+)
(–8 – 2k)x2 + (3k + 5)x + 6
(–8 – 2k)x2 + 2(–8 – 2k)x + k(–8 – 2k)
(–) (–) (–)
(7k + 21)x + (6 + 8k + 2k2)
Since, (x2 + 2x + k) is a factor of 2x4 + x3 – 14x2 + 5x + 6
So, remainder should be zero.
⇒ (7k + 21 )x + (2k2 + 8k + 6) = 0 – x + 0
⇒ 7k + 21 = 0 and 2k2 + 8k + 6 = 0
⇒ k = –3 or k2 + 4k + 3 = 0
⇒ k2 + 3k + k + 3 = 0
⇒ (k + 1)(k + 3) = 0
⇒ k = –1 or –3
For k = –3, remainder is zero.
∴ The required value of k = –3.
Also, dividend = Divisor x Quotient + Remainder
⇒ 2x4 + x3 – 14x2 + 5x + 6 = (x2 + 2x – 3)(2x2 – 3x – 2)
= (x2 + 3x – x – 3)(2x2 – 4x + x – 2)
= (x – 1)(x + 3)(x – 2)(2x + 1)
Hence, the zeroes of x2 + 2x – 3 are 1, –3 and the zeroes of 2x2 – 3x – 2 are `2, - 1/2`.
Thus, the zeroes of 2x4 + x3 – 14x2 + 5x + 6 are 1, –3, `- 1/2`.
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