If =1 and =1+i√2, where i=-1 are two roots of the equation x^3+ax^2+bx+c=0,a,b, R, then _1…

Question

If $\alpha =1$ and $\beta =1+i\sqrt{2}$, where $i=\sqrt{-1}$ are two roots of the equation

${x}^{3}+a{x}^{2}+bx+c=0,a,b,\in R$, then ${\int }_{1}^{-1}({x}^{3}+a{x}^{2}+bx+c)dx$ is equal to:

Accepted Answer

Since the coefficients $a,b,c$ are real, complex roots occur in conjugate pairs.

Given two roots are:

$\alpha =1,\, \beta =1+i\sqrt{2}$

So the third root must be:

$1-i\sqrt{2}$

Hence the cubic polynomial is:

${x}^{3}+a{x}^{2}+bx+c=(x-1)(x-(1+i\sqrt{2}))(x-(1-i\sqrt{2}))$

Now first multiply the conjugate factors:

$(x-(1+i\sqrt{2}))(x-(1-i\sqrt{2}))$

This is of the form:

$(A-iB)(A+iB)={A}^{2}+{B}^{2}$

with

$A=x-1,\, B=\sqrt{2}$

So,

$(x-(1+i\sqrt{2}))(x-(1-i\sqrt{2}))=(x-1{)}^{2}+2$

Therefore,

${x}^{3}+a{x}^{2}+bx+c=(x-1)((x-1{)}^{2}+2)$

Let $y=x-1$. Then:

$(x-1)((x-1{)}^{2}+2)=(x-1{)}^{3}+2(x-1)$

Now expand:

$(x-1{)}^{3}={x}^{3}-3{x}^{2}+3x-1$

So,

${x}^{3}+a{x}^{2}+bx+c={x}^{3}-3{x}^{2}+3x-1+2x-2$

$={x}^{3}-3{x}^{2}+5x-3$

Now evaluate:

${\int }_{1}^{-1}({x}^{3}+a{x}^{2}+bx+c)\, dx={\int }_{1}^{-1}({x}^{3}-3{x}^{2}+5x-3)\, dx$

Using symmetry:

${x}^{3}$ is an odd function, so

${\int }_{1}^{-1}{x}^{3}\, dx=0$

$5x$ is also odd, so

${\int }_{1}^{-1}5x\, dx=0$

Thus,

${\int }_{1}^{-1}({x}^{3}-3{x}^{2}+5x-3)\, dx={\int }_{1}^{-1}(-3{x}^{2}-3)\, dx$

Now,

${\int }_{1}^{-1}{x}^{2}\, dx={[\frac{{x}^{3}}{3}]}_{1}^{-1}=\frac{1}{3}-(-\frac{1}{3})=\frac{2}{3}$

Therefore,

${\int }_{1}^{-1}(-3{x}^{2})\, dx=-3\cdot \frac{2}{3}=-2$

Also,

${\int }_{1}^{-1}(-3)\, dx=-3(2)=-6$

So total value:

$-2-6=-8$

Hence,

${\int }_{1}^{-1}({x}^{3}+a{x}^{2}+bx+c)\, dx=-8$

The correct option is:

$-8$

So, Option C is correct.

Solutions · 1