please answer correct
Answers
Answer:
-5,7
Step-by-step explanation:
Given p(x) = x⁴ - 6x³ - 26x² - 138x - 35.
Given zeroes are (2 + √3) and (2 - √3).
∴ [x - (2 + √3)][(x - (2 - √3)]
= (x - 2 - √3)(x - 2 + √3)
= (x - 2)² - (√3)²
= x² + 4 - 4x - 3
= x² - 4x + 1
So, x² - 4x + 1 is a factor of p(x).
Now,
We have to divide p(x) by x² - 4x + 1.
Long Division Method:
x² - 4x + 1) x⁴ - 6x³ - 26x² + 138x - 35 (x² - 2x - 35
x⁴ - 4x³ + x²
--------------------------------------
-2x³ - 27x² + 138x
-2x³ + 8x² - 2x
--------------------------------------
-35x² + 140x - 35
-35x² + 140x - 35
------------------------------------------
0
Now,
We have to find other two zeroes of x² - 2x - 35.
= x² - 7x + 5x - 35
= x(x - 7) + 5(x - 7)
= (x + 5)(x - 7)
So, its zeroes are -5, 7.
Hence, all the zeroes of the polynomial are:
2 + √3, 2 - √3, -5,7
Hope it helps!
(x ± 2√3) are zeroes.
This means that (x-2-√3)(x-2+√3) are factors
(x - 2-√3)(x - 2+√3)
= (x - 2)^2 - (√3)^2
= x^2 - 4x +4 - 3
= x^2 - 4x + 1
x^2 - 4x + 1 is a factor of x^4 - 6x^3 - 26x^2 + 138x - 35.
Divide x^4 - 6x^3- 26x^2 + 138x - 35 with x^2- 4x+1
we get the answer as :-
x^2 - 2x - 35
x^2 - 7x + 5x - 35
x(x - 7) + 5(x - 7)
(x + 5)(x - 7)
The other zeroes are :-
-5 and 7
Hope it help you