find a cubic polynomial with zeroes 0,-1 and 3
Answers
Step-by-step explanation:
Given -
- Zeroes are 0, -1 and 3
To Find -
- A cubic polynomial
As we know that :-
- α + β + γ = -b/a
→ -1 + 0 + 3 = -b/a
→ 2/1 = -b/a ........ (i)
And
- αβγ = -d/a
→ -1 × 0 × 3 = -d/a
→ 0/1 = -d/a ...... (ii)
And
- αβ + βγ + γα = c/a
→ -1×0 + 0×3 + 3×-1 = c/a
→ -3/1 = c/a ........ (ii)
Now, From (i), (ii) and (iii), we get :-
a = 1
b = -2
c = -3
d = 0
As we know that :-
For a cubic polynomial :-
- ax³ + bx² + cx + d
→ (1)x³ + (-2)x² + (-3)x + (0)
→ x³ - 2x² - 3x + 0
Hence,
The cubic polynomial is x³ - 2x² - 3x + 0
Verification :-
- α + β + γ = -b/a
→ -1 + 0 + 3 = -(-2)/1
→ 2 = 2
LHS = RHS
And
- αβγ = -d/a
→ -1 × 0 × 3 = -(0)/1
→ 0 = 0
LHS = RHS
And
- αβ + βγ + γα = c/a
→ -1×0 + 0×3 + 3×-1 = -3/1
→ -3 = -3
LHS = RHS
Hence,
Verified...
It shows that our answer is absolutely correct.
GIVEN :-
Zeros of a cubic polynomial are 0, -1 and 3.
TO FIND :-
The cubic polynomial.
FORMULAS TO BE USED :-
- Sum of the zeros(S) = -b/a
- Product of the zeros(P) = -d/a
- Product of zeros taken two at a time(D) = c/a
SOLUTION :-
◙ METHOD-1 :-
Let α = 0, β = -1 and γ = 3.
S = α + β + γ = 0 + (-1) + 3
⇒ -b/a = 2 ...(i)
P = αβγ = 0 × -1 × 3
⇒ -d/a = 0 ...(ii)
D = αβ + βγ + αγ = (0 × -1) + (-1 × 3) + (0 × 3)
⇒c/a = 0 + (-3) + 0
⇒c/a = -3 ...(iii)
From eq.(i), (ii) and (ii), we get :-
- a = 1
- b = -2
- c = -3
- d = 0
Required cubic polynomial = ax³ + bx² + cx + d.
⇒Required cubic polynomial = x³ + -2x² + -3x
⇒Required cubic polynomial = x³ - 2x² - 3x
◙ METHOD-2 :-
S = α + β + γ = 0 + (-1) + 3 = 2
P = αβγ = 0 × -1 × 3 = 0
D = αβ + βγ + αγ = (0 × -1) + (-1 × 3) + (0 × 3) = 0 + (-3) + 0 = -3
A cubic polynomial is of the form :-
x³ - (α + β + γ)x² + (αβ + βγ + αγ)x - αβγ
= x³ - Sx² + Dx - P
= x³ - 2x² + (-3)x - 0