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Question: Find the value of (i) \({{x}^{3}}+{{x}^{2}}-x-22\) when \(x=1+2i\). (ii) \({{x}^{3}}-3{{x}^{2}}...

Find the value of
(i) x3+x2−x−22{{x}^{3}}+{{x}^{2}}-x-22 when x=1+2ix=1+2i.
(ii) x3−3x2−8x+15{{x}^{3}}-3{{x}^{2}}-8x+15 when x=3+ix=3+i.

Explanation

Solution

We first assume the given polynomial as a function form of y=f(x)y=f\left( x \right). We have complex values of the variable x to put in the polynomial. We find the value of y(x)y\left( x \right) for the particular value using identities.

Complete step by step answer:
We have a cubic polynomial and a specific complex value of the variable is given.
We take y=f(x)y=f\left( x \right). Then we put the value of the x to find value of y.
As the given value of x is complex, we use identities like i2=−1,i3=−i,i4=1{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1.
We use the binomial and cubic identity formula of sum of two numbers.
We have (a+b)3=a3+b3+3a2b+3ab2,(a+b)2=a2+b2+2ab{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}},{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab.
(i) x3+x2−x−22{{x}^{3}}+{{x}^{2}}-x-22 when x=1+2ix=1+2i.
Let’s take y=f(x)=x3+x2−x−22y=f\left( x \right)={{x}^{3}}+{{x}^{2}}-x-22. Value of x is x=1+2ix=1+2i.
We need to find y=f(x=1+2i)y=f\left( x=1+2i \right).
y(x)=x3+x2−x−22 ⇒y(1+2i)=(1+2i)3+(1+2i)2−(1+2i)−22 \begin{aligned} & y\left( x \right)={{x}^{3}}+{{x}^{2}}-x-22 \\\ & \Rightarrow y\left( 1+2i \right)={{\left( 1+2i \right)}^{3}}+{{\left( 1+2i \right)}^{2}}-\left( 1+2i \right)-22 \\\ \end{aligned}
We now simplify the equation and get
y(1+2i)=(1+2i)3+(1+2i)2−(1+2i)−22 ⇒y(1+2i)=1+6i+12i2+8i3+1+4i+4i2−1−2i−22 ⇒y(1+2i)=8i+16i2+8i3−21 \begin{aligned} & y\left( 1+2i \right)={{\left( 1+2i \right)}^{3}}+{{\left( 1+2i \right)}^{2}}-\left( 1+2i \right)-22 \\\ & \Rightarrow y\left( 1+2i \right)=1+6i+12{{i}^{2}}+8{{i}^{3}}+1+4i+4{{i}^{2}}-1-2i-22 \\\ & \Rightarrow y\left( 1+2i \right)=8i+16{{i}^{2}}+8{{i}^{3}}-21 \\\ \end{aligned}
We use identities like i2=−1,i3=−i,i4=1{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1.
y(1+2i)=8i+16i2+8i3−21 ⇒y(1+2i)=8i−16−8i−21=−37 \begin{aligned} & y\left( 1+2i \right)=8i+16{{i}^{2}}+8{{i}^{3}}-21 \\\ & \Rightarrow y\left( 1+2i \right)=8i-16-8i-21=-37 \\\ \end{aligned}

So, the solution of the value is -37.

(ii) x3−3x2−8x+15{{x}^{3}}-3{{x}^{2}}-8x+15 when x=3+ix=3+i.
Let’s take y=f(x)=x3−3x2−8x+15y=f\left( x \right)={{x}^{3}}-3{{x}^{2}}-8x+15. Value of x is x=3+ix=3+i.
We need to find y=f(x=3+i)y=f\left( x=3+i \right).
y(x)=x3−3x2−8x+15 ⇒y(3+i)=(3+i)3−3(3+i)2−8(3+i)+15 \begin{aligned} & y\left( x \right)={{x}^{3}}-3{{x}^{2}}-8x+15 \\\ & \Rightarrow y\left( 3+i \right)={{\left( 3+i \right)}^{3}}-3{{\left( 3+i \right)}^{2}}-8\left( 3+i \right)+15 \\\ \end{aligned}
We now simplify the equation and get
y(3+i)=(3+i)3−3(3+i)2−8(3+i)+15 ⇒y(3+i)=9+27i+9i2+i3−27−18i−3i2−24−8i+15 ⇒y(3+i)=i+6i2+i3−27 \begin{aligned} & y\left( 3+i \right)={{\left( 3+i \right)}^{3}}-3{{\left( 3+i \right)}^{2}}-8\left( 3+i \right)+15 \\\ & \Rightarrow y\left( 3+i \right)=9+27i+9{{i}^{2}}+{{i}^{3}}-27-18i-3{{i}^{2}}-24-8i+15 \\\ & \Rightarrow y\left( 3+i \right)=i+6{{i}^{2}}+{{i}^{3}}-27 \\\ \end{aligned}
We use identities like i2=−1,i3=−i,i4=1{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1.
y(3+i)=i+6i2+i3−27 ⇒y(3+i)=i−6−i−27=−33 \begin{aligned} & y\left( 3+i \right)=i+6{{i}^{2}}+{{i}^{3}}-27 \\\ & \Rightarrow y\left( 3+i \right)=i-6-i-27=-33 \\\ \end{aligned}

So, the solution of the value is -33.

Note: We don’t need to find the zero of the polynomials even though in some cases the value of the polynomial will become 0. That particular value becomes the root of the polynomial. In other cases, we are just finding the image of the value for the given function. Here the preimages are x=1+2ix=1+2i, x=3+ix=3+i.