2 zeroes of polynomial f(x) is such that the sum of product of zeroes is -6. if f(x)= x4+x3-12x2-6x+36. then find all the zeroes. Pls help me out with this question!!
Answers
Step-by-step explanation:
Let the 2 zeroes of the polynomial be α,β.
Given, f(x) = x⁴ + x³ - 12x² - 6x + 36.
Given that Sum of zeroes is 0 and Product is -6.
α + β = 0 and αβ = -6
The Equation thus formed is:
x² - (Sum of zeroes)x + Product of zeroes
⇒ x² - 6
∴ x² - 6 is a factor of f(x).
On dividing f(x) by x² - 6, we can find out the factors.
x² - 6) x⁴ + x³ - 12x² - 6x + 36 (x² + x - 6
x⁴ - 6x²
-----------------------------------
x³ - 6x² - 6x
x³ - 6x
-----------------------------------
-6x² + 36
-6x² + 36
------------------------------------
0
∴ f(x) = 0
⇒ x⁴ + x³ - 12x² - 6x + 36 = 0
⇒ (x² - 6)(x² + x - 6) = 0
⇒ (x² - (√6)²)(x² + 3x - 2x - 6) = 0
⇒ (x + √6)(x - √6)[x(x + 3) - 2(x + 3)] = 0
⇒ (x + √6)(x - √6)(x - 2)(x + 3) = 0
⇒ x = -√6, √6, 2, -3
Therefore, Zeroes are : √6, -√6, 2, -3
Hope it helps!
Answer:
Step-by-step explanation:
x⁴ + x³ - 12x² - 6x + 36 = 0
(x² - 6)(x² + x - 6) = 0
(x² - (√6)²)(x² + 3x - 2x - 6) = 0
(x + √6)(x - √6){x(x + 3) - 2(x + 3)}=0
[(x + √6)(x - √6)(x - 2)(x + 3)] =0
One of the zero :-
x = -√6
Another zero:-
X=√6
Another zero:-
X=2
Another zero:-
X=-2
Hence you got four zeros for the given polynomial.
Hope it will help you with
✌️Sai