Find k such that x2 +2x +k is a factor of 2x4 +x3- 14x2 + 5x +6 . Also, find the zeroes of the two polynomials.
Answers
Step-by-step explanation:
Given factor: x2 + 2x + k = 0
Given polynomial: 2x4 + x3 -14x2 + 5x + 6
Divide the polynomial by the factor
x2 + 2x + k ) 2x4 + x3 -14x2 + 5x + 6 ( 2x2 - 3x +(- 8 - 2k)
2x4 + 4x3 +2kx2 ( substract)
------------------------------
- 3x3 +(-14 - 2k)x2 + 5x
- 3x3 - 6x2 - 3kx ( substract)
------------------------------
(- 8 - 2k) x2 +( 5 + 3k)x + 6
(- 8 - 2k) x2 +(-16 - 4k)x + (- 8k - 2k2) ( substract)
-----------------------------------------------------------------
( 21 + 7k)x + (6 + 8k + 2k2)
The remainder is: ( 21 + 7k)x + (6 + 8k + 2k2) = 0
21 + 7k = 0 ⇒ k = -3.
The factors are x2 + 2x - 3 = 0 and 2x2 - 3x - 2 = 0
x2 + 3x - x - 3 = 0 and 2x2 - 4x + x - 2 = 0
x( x + 3 )-1( x + 3) = 0 and 2x (x - 2) + 1(x - 2) = 0
(x - 1)( x + 3) = 0 and (2x + 1)(x - 2) = 0
x = 1 ,3 ,-1 / 2 and 2.
The zeros are 1 ,3 ,-1 / 2 and 2.
Hope it helps
mrk me as BRAINLIEST
Given : x² + 2x +K is a factor of 2x⁴ + x³ – 14x² + 5x + 6.
To find : all the zeros of the two polynomials
Solution:
2x² - 3x - (8 + 2K)
x² + 2x +K _| 2x⁴ + x³ – 14x² + 5x + 6 |_
2x⁴ + 4x³+ 2Kx²
________________
-3x³ -(14 +2K)x² + 5x + 6
-3x³ - 6x² -3kx
__________________
- (8 + 2K)x² + (5 +3k)x + 6
- (8 + 2K)x² - (16 +4k)x - (8K + 2K²)
___________________________
(21 + 7k)x + (2K² + 8K + 6)
x coefficients and constant term must be zero
21 + 7K = 0
=> K = - 3
2K² + 8K + 6 = ( K + 3)(2K + 2)
Value of K = - 3 as this is common solution
x² + 2x +K = x² + 2x - 3
=> (x + 3)(x - 1) => Zeroes are -3 , 1
2x² - 3x - (8 + 2K) = 2x² - 3x - 2 = 2x² - 4x + x - 2
= 2x(x - 2) + 1(x - 2)
= (2x + 1)(x - 2)
=> x = -1/2 , 2
Zeroes are - 3 , -1/2 , 1 , 2
Value of k = - 3
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