if one of the zeroes of the cubic polynomial x3+ax2+bx+c is - 1 then find the product of other two zeroes.
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Answered by
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Solution :
_____________________________________________________________
Given :
One of the zeroes of the cubic polynomial x3+ax2+bx+c is - 1
_____________________________________________________________
To Find :
Other zeroes of the polynomial,.
_____________________________________________________________
We know that,.
The given polynomial is of the form :
ax³ + bx² + cx + d,.
If α,β & γ are the zeores of the polynomial
ax³ + bx² + cx + d, Then,.
It can also written as,
x³ + (α + β + γ)x² + (αβ + βγ + γα)x + αβγ
So,
If a = 1,
Then,
b = (α + β + γ),
c = (αβ + βγ + γα)
d = αβγ
Here,. a = 1, b = a, c = b & d = c,..
___________________________
We also know that,
Sum of the zeroes =
Let the zeroes be, α, β & -1.(as -1 is given as the zero of the given polynomial,)
⇒ α + β + (-1) =
⇒ α + β - 1 = -a
⇒ α + β = 1 - a ...(1)
__________________
We know that,.
Product of the zeroes =
⇒ αβγ =
⇒ αβ(1) = -c
⇒ αβ = -c,.(2)
_______________________
So,
We know that,
⇒ (a + b)² = a² + 2ab + b²
Substituting zeroes of the polynomial,
We get,.
⇒ (α + β)² = α² + 2αβ + β²
⇒ (1 - a)² = α² + 2(-c) + β²
⇒ 1 - 2a + a² = α² + β² - 2c
⇒ a² - 2a + 1 -2c = α² + β² ...(3)
_____________________________
Hence,
⇒ (a - b)² = a² - 2ab + b²
⇒ (α - β)² = α² - 2αβ + β²
⇒ (α - β)² = α² + β² - 2αβ
⇒ (α - β)² = a² - 2a + 1 -2c - 2(-c)
⇒ (α - β)² = a² - 2a + 1 - 2c + 2c
⇒ (α - β)² = a² - 2a + 1
⇒ (α - β)² = a² - a - a + 1
⇒ (α - β)² = a(a - 1) - 1(a - 1)
⇒ (α - β)² = (a - 1)(a - 1)
⇒ (α - β)² = (a - 1)²
⇒ ∴ α - β = a - 1...(4)
Adding Equation 1 & 4,
We get,
⇒ (α + β) + (α - β) = (1 - a) + (a - 1)
⇒ α + α + β - β = 1 - a + a - 1
⇒ 2α = 1 - 1 + a - a
⇒ 2α = 0
⇒ ∴ α = 0,.
_____________
Substituting value of α in Equation 1,
We get,.
⇒ α + β = 1 - a
⇒ 0 + β = 1 - a
⇒ ∴ β = 1 - a
∴ The other zeroes of the given polynomial are 0 & 1 - a,.
_____________________________________________________________
Hope it Helps !!
⇒ Mark as Brainliest,.. (If,possible,.)
Sorry, the answer is lengthy & (little bit - complex),.
_____________________________________________________________
Given :
One of the zeroes of the cubic polynomial x3+ax2+bx+c is - 1
_____________________________________________________________
To Find :
Other zeroes of the polynomial,.
_____________________________________________________________
We know that,.
The given polynomial is of the form :
ax³ + bx² + cx + d,.
If α,β & γ are the zeores of the polynomial
ax³ + bx² + cx + d, Then,.
It can also written as,
x³ + (α + β + γ)x² + (αβ + βγ + γα)x + αβγ
So,
If a = 1,
Then,
b = (α + β + γ),
c = (αβ + βγ + γα)
d = αβγ
Here,. a = 1, b = a, c = b & d = c,..
___________________________
We also know that,
Sum of the zeroes =
Let the zeroes be, α, β & -1.(as -1 is given as the zero of the given polynomial,)
⇒ α + β + (-1) =
⇒ α + β - 1 = -a
⇒ α + β = 1 - a ...(1)
__________________
We know that,.
Product of the zeroes =
⇒ αβγ =
⇒ αβ(1) = -c
⇒ αβ = -c,.(2)
_______________________
So,
We know that,
⇒ (a + b)² = a² + 2ab + b²
Substituting zeroes of the polynomial,
We get,.
⇒ (α + β)² = α² + 2αβ + β²
⇒ (1 - a)² = α² + 2(-c) + β²
⇒ 1 - 2a + a² = α² + β² - 2c
⇒ a² - 2a + 1 -2c = α² + β² ...(3)
_____________________________
Hence,
⇒ (a - b)² = a² - 2ab + b²
⇒ (α - β)² = α² - 2αβ + β²
⇒ (α - β)² = α² + β² - 2αβ
⇒ (α - β)² = a² - 2a + 1 -2c - 2(-c)
⇒ (α - β)² = a² - 2a + 1 - 2c + 2c
⇒ (α - β)² = a² - 2a + 1
⇒ (α - β)² = a² - a - a + 1
⇒ (α - β)² = a(a - 1) - 1(a - 1)
⇒ (α - β)² = (a - 1)(a - 1)
⇒ (α - β)² = (a - 1)²
⇒ ∴ α - β = a - 1...(4)
Adding Equation 1 & 4,
We get,
⇒ (α + β) + (α - β) = (1 - a) + (a - 1)
⇒ α + α + β - β = 1 - a + a - 1
⇒ 2α = 1 - 1 + a - a
⇒ 2α = 0
⇒ ∴ α = 0,.
_____________
Substituting value of α in Equation 1,
We get,.
⇒ α + β = 1 - a
⇒ 0 + β = 1 - a
⇒ ∴ β = 1 - a
∴ The other zeroes of the given polynomial are 0 & 1 - a,.
_____________________________________________________________
Hope it Helps !!
⇒ Mark as Brainliest,.. (If,possible,.)
Sorry, the answer is lengthy & (little bit - complex),.
poorniragp89:
Is their any shortcuts
You can complete in 4 steps, I just explained this Problem,.much
(Probably too much,.)
Answered by
0
Answer:
Step-by-step explanation:
Answer:
Product of other two zeroes =c
Step-by-step explanation:
Compare p(x) with Ax³+Bx²+Cx+D , we get
A=1, B=a , C = b , D = c
Therefore,
Product of other two zeroes =c
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