Question

How do you write a polynomial in standard form given zeros $ - 1$ (multiplicity $2$), $ - 2 - i$ (multiplicity $1$)?

Answer 1

Hint: First assume $\alpha = - 1$, $\beta = - 1$ and $\gamma = - 2 - i$ to be the zeroes of the cubic polynomial $f\left( x \right) = a{x^3} + b{x^2} + cx + d$, $a \ne 0$. Then, substitute the value of $\alpha ,\beta ,\gamma $ in equation (i) and simplify the right side of the equation. We will get the standard form of polynomial for given zeros.

Complete step by step solution:
Let $\alpha ,\beta ,\gamma $ be the zeroes of a cubic polynomial $f\left( x \right) = a{x^3} + b{x^2} + cx + d$, $a \ne 0$. Then by factor theorem, $x - \alpha $, $x - \beta $ and $x - \gamma $ are factors of $f\left( x \right)$.
$\therefore f\left( x \right) = k\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)$…(i)
$ \Rightarrow a{x^3} + b{x^2} + cx + d = k\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)$
$ \Rightarrow a{x^3} + b{x^2} + cx + d = k\left[ {{x^3} - \left( {\alpha + \beta + \gamma } \right){x^2} + \left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right)x - \alpha \beta \gamma } \right]$
$ \Rightarrow a{x^3} + b{x^2} + cx + d = k{x^3} - k\left( {\alpha + \beta + \gamma } \right){x^2} + k\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right)x - k\alpha \beta \gamma $
Comparing the coefficients of ${x^3},{x^2},x$ and constant terms on both sides, we get
$a = k$, \[b = - k\left( {\alpha + \beta + \gamma } \right)\], $c = k\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right)$ and $d = - k\alpha \beta \gamma $…(ii)
\[ \Rightarrow \alpha + \beta + \gamma = - \dfrac{b}{a}\]
$\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}$
And, $\alpha \beta \gamma = - \dfrac{d}{a}$
Thus, Sum of zeroes of cubic polynomial, \[\alpha + \beta + \gamma = - \dfrac{b}{a}\]
And, Product of zeroes of cubic polynomial, $\alpha \beta \gamma = - \dfrac{d}{a}$
Step by step solution:
Here, it is given that a polynomial has zeros $ - 1$ (multiplicity $2$), $ - 2 - i$ (multiplicity $1$).
We have to find the polynomial in standard form.
First, we have to assume $\alpha = - 1$, $\beta = - 1$ and $\gamma = - 2 - i$ to be the zeroes of the cubic polynomial $f\left( x \right) = a{x^3} + b{x^2} + cx + d$, $a \ne 0$.
Since, $\alpha = - 1$, $\beta = - 1$ and $\gamma = - 2 - i$ are zeros of the cubic polynomial $f\left( x \right) = a{x^3} + b{x^2} + cx + d$, $a \ne 0$.
Then by factor theorem, $x - \alpha $, $x - \beta $ and $x - \gamma $ are factors of $f\left( x \right)$.
⇒ $x + 1$, $x + 1$ and $x + 2 + i$ are factors of $f\left( x \right)$.
Substitute the value of $\alpha ,\beta ,\gamma $ in equation (i).
$ \Rightarrow f\left( x \right) = k\left( {x + 1} \right)\left( {x + 1} \right)\left( {x + 2 + i} \right)$
Simplify the right side of the equation.
$ \Rightarrow f\left( x \right) = k\left( {{x^2} + 2x + 1} \right)\left( {x + 2 + i} \right)$
$ \Rightarrow f\left( x \right) = k\left[ {{x^2}\left( {x + 2 + i} \right) + 2x\left( {x + 2 + i} \right) + \left( {x + 2 + i} \right)} \right]$
$ \Rightarrow f\left( x \right) = k\left( {{x^3} + 2{x^2} + {x^2}i + 2{x^2} + 4x + 2xi + x + 2 + i} \right)$
$ \Rightarrow f\left( x \right) = k\left[ {{x^3} + \left( {4 + i} \right){x^2} + \left( {5 + 2i} \right)x + 2 + i} \right]$

Hence, $k\left[ {{x^3} + \left( {4 + i} \right){x^2} + \left( {5 + 2i} \right)x + 2 + i} \right]$ is the standard form of polynomial for given zeros.

Note: We can directly find the standard form of polynomial by finding the sum and product of given zeros.
Step by step solution:
Here, it is given that a polynomial has zeros $ - 1$ (multiplicity $2$), $ - 2 - i$ (multiplicity $1$).
We have to find the polynomial in standard form.
First, we have to assume $\alpha = - 1$, $\beta = - 1$ and $\gamma = - 2 - i$ to be the zeroes of the cubic polynomial $f\left( x \right) = a{x^3} + b{x^2} + cx + d$, $a \ne 0$.
We have to find $a,b,c,d$.
Substitute the value of $\alpha ,\beta ,\gamma $ in (ii) and find the value of $a,b,c,d$.
$a = k$, \[b = - k\left( { - 1 - 1 - 2 - i} \right)\], $c = k\left[ {\left( { - 1} \right)\left( { - 1} \right) + \left( { - 1} \right)\left( { - 2 - i} \right) + \left( { - 2 - i} \right)\left( { - 1} \right)} \right]$ and $d = - k\left( { - 1} \right)\left( { - 1} \right)\left( { - 2 - i} \right)$
⇒ $a = k$, \[b = k\left( {4 + i} \right)\], $c = k\left( {5 + 2i} \right)$ and $d = k\left( {2 + i} \right)$
Now, put the value of $a,b,c,d$ in $f\left( x \right) = a{x^3} + b{x^2} + cx + d$.
$ \Rightarrow f\left( x \right) = k\left[ {{x^3} + \left( {4 + i} \right){x^2} + \left( {5 + 2i} \right)x + 2 + i} \right]$
Final solution: Hence, $k\left[ {{x^3} + \left( {4 + i} \right){x^2} + \left( {5 + 2i} \right)x + 2 + i} \right]$ is the standard form of polynomial for given zeros.