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Question: How do you write a polynomial in standard form given zeros \( - 1\) (multiplicity \(2\)), \( - 2 - i...

How do you write a polynomial in standard form given zeros −1 - 1 (multiplicity 22), −2−i - 2 - i (multiplicity 11)?

Explanation

Solution

First assume α=−1\alpha = - 1, β=−1\beta = - 1 and γ=−2−i\gamma = - 2 - i to be the zeroes of the cubic polynomial f(x)=ax3+bx2+cx+df\left( x \right) = a{x^3} + b{x^2} + cx + d, a≠0a \ne 0. Then, substitute the value of α,β,γ\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 α,β,γ\alpha ,\beta ,\gamma be the zeroes of a cubic polynomial f(x)=ax3+bx2+cx+df\left( x \right) = a{x^3} + b{x^2} + cx + d, a≠0a \ne 0. Then by factor theorem, x−αx - \alpha , x−βx - \beta and x−γx - \gamma are factors of f(x)f\left( x \right).
∴f(x)=k(x−α)(x−β)(x−γ)\therefore f\left( x \right) = k\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)…(i)
⇒ax3+bx2+cx+d=k(x−α)(x−β)(x−γ)\Rightarrow a{x^3} + b{x^2} + cx + d = k\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)
⇒ax3+bx2+cx+d=k[x3−(α+β+γ)x2+(αβ+βγ+γα)x−αβγ]\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]
⇒ax3+bx2+cx+d=kx3−k(α+β+γ)x2+k(αβ+βγ+γα)x−kαβγ\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 x3,x2,x{x^3},{x^2},x and constant terms on both sides, we get
a=ka = k, b=−k(α+β+γ)b = - k\left( {\alpha + \beta + \gamma } \right), c=k(αβ+βγ+γα)c = k\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) and d=−kαβγd = - k\alpha \beta \gamma …(ii)
⇒α+β+γ=−ba\Rightarrow \alpha + \beta + \gamma = - \dfrac{b}{a}
αβ+βγ+γα=ca\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}
And, αβγ=−da\alpha \beta \gamma = - \dfrac{d}{a}
Thus, Sum of zeroes of cubic polynomial, α+β+γ=−ba\alpha + \beta + \gamma = - \dfrac{b}{a}
And, Product of zeroes of cubic polynomial, αβγ=−da\alpha \beta \gamma = - \dfrac{d}{a}
Step by step solution:
Here, it is given that a polynomial has zeros −1 - 1 (multiplicity 22), −2−i - 2 - i (multiplicity 11).
We have to find the polynomial in standard form.
First, we have to assume α=−1\alpha = - 1, β=−1\beta = - 1 and γ=−2−i\gamma = - 2 - i to be the zeroes of the cubic polynomial f(x)=ax3+bx2+cx+df\left( x \right) = a{x^3} + b{x^2} + cx + d, a≠0a \ne 0.
Since, α=−1\alpha = - 1, β=−1\beta = - 1 and γ=−2−i\gamma = - 2 - i are zeros of the cubic polynomial f(x)=ax3+bx2+cx+df\left( x \right) = a{x^3} + b{x^2} + cx + d, a≠0a \ne 0.
Then by factor theorem, x−αx - \alpha , x−βx - \beta and x−γx - \gamma are factors of f(x)f\left( x \right).
⇒ x+1x + 1, x+1x + 1 and x+2+ix + 2 + i are factors of f(x)f\left( x \right).
Substitute the value of α,β,γ\alpha ,\beta ,\gamma in equation (i).
⇒f(x)=k(x+1)(x+1)(x+2+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.
⇒f(x)=k(x2+2x+1)(x+2+i)\Rightarrow f\left( x \right) = k\left( {{x^2} + 2x + 1} \right)\left( {x + 2 + i} \right)
⇒f(x)=k[x2(x+2+i)+2x(x+2+i)+(x+2+i)]\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]
⇒f(x)=k(x3+2x2+x2i+2x2+4x+2xi+x+2+i)\Rightarrow f\left( x \right) = k\left( {{x^3} + 2{x^2} + {x^2}i + 2{x^2} + 4x + 2xi + x + 2 + i} \right)
⇒f(x)=k[x3+(4+i)x2+(5+2i)x+2+i]\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[x3+(4+i)x2+(5+2i)x+2+i]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 - 1 (multiplicity 22), −2−i - 2 - i (multiplicity 11).
We have to find the polynomial in standard form.
First, we have to assume α=−1\alpha = - 1, β=−1\beta = - 1 and γ=−2−i\gamma = - 2 - i to be the zeroes of the cubic polynomial f(x)=ax3+bx2+cx+df\left( x \right) = a{x^3} + b{x^2} + cx + d, a≠0a \ne 0.
We have to find a,b,c,da,b,c,d.
Substitute the value of α,β,γ\alpha ,\beta ,\gamma in (ii) and find the value of a,b,c,da,b,c,d.
a=ka = k, b=−k(−1−1−2−i)b = - k\left( { - 1 - 1 - 2 - i} \right), c=k[(−1)(−1)+(−1)(−2−i)+(−2−i)(−1)]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(−1)(−1)(−2−i)d = - k\left( { - 1} \right)\left( { - 1} \right)\left( { - 2 - i} \right)
⇒ a=ka = k, b=k(4+i)b = k\left( {4 + i} \right), c=k(5+2i)c = k\left( {5 + 2i} \right) and d=k(2+i)d = k\left( {2 + i} \right)
Now, put the value of a,b,c,da,b,c,d in f(x)=ax3+bx2+cx+df\left( x \right) = a{x^3} + b{x^2} + cx + d.
⇒f(x)=k[x3+(4+i)x2+(5+2i)x+2+i]\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[x3+(4+i)x2+(5+2i)x+2+i]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.