Question

The polynomial equation of degree 4 having real coefficients with three of its roots as $\left( 2\pm \sqrt{3} \right)\text{ and }\left( 1\pm 2i \right)$ is

& A.{{x}^{4}}-6{{x}^{3}}+14{{x}^{2}}+22x+5=0 \\\ & B.{{x}^{4}}-6{{x}^{3}}+19{{x}^{2}}+22x-5=0 \\\ & C.{{x}^{4}}-6{{x}^{3}}+19{{x}^{2}}-22x+5=0 \\\ & D.{{x}^{4}}-6{{x}^{3}}+14{{x}^{2}}-22x+5=0 \\\ \end{aligned}$$

Answer 1

To solve this question, we will first obtain a fourth root and for that we will use the fact that imaginary roots always occur in pairs. After this, we will make two different quadratic equations by using $\left( 2\pm \sqrt{3} \right)\text{ and }\left( 1\pm 2i \right)$ separately. Finally, after obtaining 2 quadratic equations we will multiply them to obtain a final 4 degree polynomial.

Complete step-by-step answer:
We have to find a 4 degree polynomial with real coefficient. Let the polynomial be denoted by p(x).
Since, the polynomial is 4 degrees, so it has 4 roots. The roots are given as $2+\sqrt{3}\text{,}2-\sqrt{3}\text{ and }1+2i$
So, we are given 3 roots but one of them is 1+2i and the imaginary root/complex root always occurs in pairs and the pair of (1+2i) is (1-2i).
Therefore, (1-2i) is also a root of our polynomial.
Therefore, we now have all 4 roots as $2+\sqrt{3}\text{,}2-\sqrt{3},1+2i\text{ and 1-2i}$
Consider, first two roots: $\Rightarrow \left( 2+\sqrt{3} \right)\text{ and }\left( 2-\sqrt{3} \right)$
Any quadratic equation whose roots are $\alpha \text{ and }\beta$ can be formed by writing as
$\Rightarrow {{x}^{2}}-\left( \text{sum of roots }\alpha \text{ and }\beta \right)x+\left( \text{product of roots }\alpha \text{ and }\beta \right)=0$
Quadratic polynomial whose roots are $\alpha \text{ and }\beta$ are given as
${{x}^{2}}-\left( \alpha +\beta \right)x\left( \alpha \beta \right)=0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}$
So, on the similar basis we will construct a quadratic polynomial having roots as $\Rightarrow \left( 2+\sqrt{3} \right)\text{ and }\left( 2-\sqrt{3} \right)$
Sum of roots $\begin{aligned} & \Rightarrow 2+\sqrt{3}+2-\sqrt{3}=4+\sqrt{3}-\sqrt{3} \\\ & \Rightarrow 4\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)} \\\ \end{aligned}$
And product of roots $\Rightarrow \left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right)=1$
Now, we know that $\Rightarrow \left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
Applying this in above we get:
Product of roots $\Rightarrow {{\left( 2 \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}=4-3=1\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)}$
Forming a quadratic equation whose roots are $\left( 2+\sqrt{3} \right)\text{ and }\left( 2-\sqrt{3} \right)$ by form of equation (i) using equation (ii) and (iii) we get:

& {{x}^{2}}-\left( 4 \right)x+1=0 \\\ & \Rightarrow {{x}^{2}}-4x+1=0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iv)} \\\ \end{aligned}$$ Similarly, we will try to obtain a quadratic polynomial whose roots are (1+2i) and (1-2i). Sum of roots $$\Rightarrow 1+2i+1-2\text{i}=2+2i-2i=2$$ And product of roots $$\Rightarrow \left( 1+2i \right)\left( 1-2\text{i} \right)$$ Again using $$\Rightarrow \left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$$ we get: Product of roots: $$\begin{aligned} & \Rightarrow \left( {{1}^{2}}-{{\left( 2i \right)}^{2}} \right)={{1}^{2}}-4{{i}^{2}} \\\ & \Rightarrow {{i}^{2}}=-1 \\\ & \text{Product}=1+4=5 \\\ \end{aligned}$$ So, the quadratic equation formed using roots (1+2i) and (1-2i) as given in equation (i) is as below: $${{x}^{2}}-2x+5=0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (v)}$$ So finally we have obtained two quadratic equations from (iv) and (v) equation. Multiplying (iv) and (v) to get p(x). $$\begin{aligned} & p\left( x \right)=\left( {{x}^{2}}-2x+5 \right)\left( {{x}^{2}}-4x+1 \right) \\\ & \Rightarrow {{x}^{2}}\left( {{x}^{2}}-4x+1 \right)-2x\left( {{x}^{2}}-4x+1 \right)+5\left( {{x}^{2}}-4x+1 \right) \\\ & \Rightarrow {{x}^{4}}-4{{x}^{3}}+{{x}^{2}}-2{{x}^{3}}+8{{x}^{2}}-2x+5{{x}^{2}}-20x+5 \\\ & \Rightarrow {{x}^{4}}-6{{x}^{3}}+14{{x}^{2}}-22x+5 \\\ \end{aligned}$$ So, our required polynomial is $$p\left( x \right)={{x}^{4}}-6{{x}^{3}}+14{{x}^{2}}-22x+5$$ which is option D. **So, the correct answer is “Option D”.** **Note:** The possibility of mistake in this question can be not trying to obtain the fourth root of p(x) and directly calculating the value of p(x). Always remember that, a ‘n’ degree polynomial has n roots whether real or imaginary. Also, imaginary or complex roots always occur in pairs if $\left( \alpha +i\beta \right)$ is a root then $\left( \alpha -i\beta \right)$ is also a root of the same polynomial.