Question

If x = 3 + i, then find the value of the expression x⁴ - 4x³ + 4x² - 16x+60​

Answer 1

$\large\underline{ \bold{\sf{Given- }}}$

$\rm :\longmapsto\:x \: = \: 3 + i$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\: {x}^{4} - {4x}^{3} + {4x}^{2} - 16x + 60$

$\large\underline{\sf{Solution-}}$

Given that

$\bf :\longmapsto\:x = 3 + i$

$\rm :\longmapsto\:x - 3 = i$

On squaring both sides, we get

$\rm :\longmapsto\: {(x - 3)}^{2} = {(i)}^{2}$

$\rm :\longmapsto\: {x}^{2} + 9 - 6x = {i}^{2}$

$\rm :\longmapsto\: {x}^{2} + 9 - 6x = - 1 \: \: \: \: \: \: \: \: \: \{ \because \: {i}^{2} = - 1 \}$

$\rm :\longmapsto\: {x}^{2} - 6x + 10 = 0$

Now, Using long division, we have

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\: {x}^{2} + 2x + 6\:\:}}}\\ {\underline{\sf{ {x}^{2} - 6x + 10}}}& {\sf{\: {x}^{4} - {4x}^{3} + {4x}^{2} - 16x + 60 \:}} \\{\sf{}}& \underline{\sf{\: - {x}^{4} + 6 {x}^{3} - {10x}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\:}}\\{\sf{}}&{\sf{\: {2x}^{3} - {6x}^{2} - 16x \:\:}}\\{\sf{}}&\underline{\sf{\: { - 2x}^{3} + 12 {x}^{2} - 20x \: \: \:\:}}\\{\sf{}}&{\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {6x}^{2} - 36x + 60\:\:}}\\{\sf{}}&\underline{\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - {6x}^{2} + 36x - 60 \:\:}}\\{\sf{}}&\underline{\sf{\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \:\:}}\end{array}\end{gathered}\end{gathered}\end{gathered}$

By division Algorithm, we have

Dividend = Divisor × Quotient + Remainder

So,

$\bf :\longmapsto\: {x}^{4} - {4x}^{3} + {4x}^{2} - 16x + 60$

$\rm \: = \: \: ( {x}^{2} - 6x + 10)( {x}^{2} + 2x + 6) + 0$

$\rm \: = \: \: ( 0) \times ( {x}^{2} + 2x + 6)$

$\rm \: = \: \: 0$

Hence,

$\: \: \: \: \underbrace{ \boxed{\bf :\longmapsto\: {x}^{4} - {4x}^{3} + {4x}^{2} - 16x + 60 = 0}}$

Additional Information :-

$\green{\boxed{ \bf \:i = \sqrt{ - 1} \: }}$

$\green{\boxed{ \bf \: {i}^{2} = - \: 1 \: }}$

$\green{\boxed{ \bf \: {i}^{3} = - \: i \: }}$

$\green{\boxed{ \bf \: {i}^{4} = \: 1\: }}$

$\green{\boxed{ \bf \:\dfrac{1}{i} = - \: i \: }}$