If two zeroes of the polynomial x4-6x3-26x2+138x-35 are 2 √3, find the other roots
Answers
★ refer to the attachment for your answer
________________________________________________________________________________
Therefore the other roots of the polynomial are -5 and 7.
If two zeroes of the polynomial x⁴ - 6x³ - 26x² + 138x - 35 are 2 ± √3,
We have to find the other roots of the given polynomial.
we have to roots of the polynomial are ; 2 + √3 and 2 - √3.
∴ the factors are the polynomial are ; (x - 2 + √3) and (x - 2 - √3)
we get, (x - 2 + √3)(x - 2 - √3) = (x - 2)² - (√3)² = x² - 4x + 1 , is a factor of x⁴ - 6x³ - 26x² + 138x - 35.
Using division method, We can find the other factor of polynomial.
x² - 4x + 1 )x⁴ - 6x³ - 26x² + 138x - 35 (x² - 2x - 35
x⁴ - 4x³ + x²
----------------------------
-2x³ - 27x² + 138x
-2x³ + 8x² - 4x
---------------------------------------
-35x² + 142x - 35
- 35x² + 142x - 35
------------------------------------
0
Hence, it is clear that the other factor of polynomial is x² - 2x - 35
∴ x² - 2x - 35 = 0
⇒ x² - 7x + 5x - 35 = 0
⇒ x(x - 7) + 5(x - 7) = 0
⇒ (x - 7) (x + 5) = 0
⇒ x = 7 and - 5