what is the formula to find a expression for a cubic polynomial with alpha,beta,gama as zeroes
Answers
ex :- 3x^3 + 4x^2 +5x+ 6 = 0
The general form of a cubic equation is ax^3+bx^2+cx+d=0
The graph of cubic equation is also a curve having 2 turns and cutting the x axis at 3 points. These 3 points of intersection are known as the roots of the cubic equation.There are three roots of a cubic equation given by α (Alpha), β (Beta) and γ (Gamma).
α + β + γ = -b/a i.e sum of roots = -b/a
α β + β γ + αγ = c/a i.e products of roots taken 2 at a time = c/a
αβγ = -d/a i.e. Product of roots = -d/a
A cubic equation may also be written as :-
x^3-(sum of roots) x^2+(Product of roots taken 2 at a time)x+product of roots=0
Let us consider a general case of a polynomial:
ax^3+bx^2+cx+dax
3
+bx
2
+cx+d
Suppose \alphaα , \betaβ and \gammaγ are zeros.
Then we have:
\begin{lgathered}\alpha + \beta + \gamma = -\frac{b}{a} \\ \\ \\ \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} \\ \\ \\ \alpha \beta \gamma = -\frac{d}{a}\end{lgathered}
α+β+γ=−
a
b
αβ+βγ+γα=
a
c
αβγ=−
a
d
_____________________________
In this question, we are also given that \alpha , \, \, \beta \, \, and \, \, \gammaα,βandγ are zeros.
Let us assume the cubic polynomial to be:
ax^3+bx^2+cx+dax
3
+bx
2
+cx+d
We are given:
\begin{lgathered}\alpha + \beta + \gamma = 3 \\ \\ \implies -\frac{b}{a} = 3 \\ \\ \text{Let us assume a = 1} \\ \\ \implies -\frac{b}{1} = 3 \\ \\ \implies b = -3\end{lgathered}
α+β+γ=3
⟹−
a
b
=3
Let us assume a = 1
⟹−
1
b
=3
⟹b=−3
Also,
\begin{lgathered}\alpha \beta + \beta \gamma + \gamma \alpha = 6 \\ \\ \implies \frac{c}{a} = 6 \\ \\ \text{We have assumed a = 1} \\ \\ \implies \frac{c}{1} = 6 \\ \\ \implies c = 6\end{lgathered}
αβ+βγ+γα=6
⟹
a
c
=6
We have assumed a = 1
⟹
1
c
=6
⟹c=6
And then:
\begin{lgathered}\alpha \beta \gamma = -20 \\ \\ \implies -\frac{d}{a} = -20 \\ \\ \implies \frac{d}{1} = 20 \\ \\ \implies d =20\end{lgathered}
αβγ=−20
⟹−
a
d
=−20
⟹
1
d
=20
⟹d=20
So, we have:
\begin{lgathered}\boxed{\begin{array}{ccc} a & = & 1 \\ b & = & -3 \\ c & = & 6 \\ d & = & 20\end{array}}\end{lgathered}
a
b
c
d
=
=
=
=
1
−3
6
20
Now, we can substitute these values in the general form and find the polynomial:
\begin{lgathered}ax^3 + bx^2 + cx + d \\ \\ = x^3-3x^2+6x+20\end{lgathered}
ax
3
+bx
2
+cx+d
=x
3
−3x
2
+6x+20
If we name this polynomial as P(x)P(x) , our answer becomes:
\boxed{P(x)=x^3-3x^2+6x+20}
P(x)=x
3
−3x
2
+6x+20
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