Then Find the other

1.	Then Find the other
Answer» $\:\:\:\:\:\:\:\:\:\:\:\:\:\: \large\mathfrak{\dag \: Solution :- }\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ ❍GIVEN equation is :- $\sf \: {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 = 0$ As we KNOW that complex ROOTS OCCUR in conjugate pairs.Therefore, 1 - i will be the other ROOT of the given equation. $\sf \red{\: \:\bigg( x - (1 + i)\bigg) \bigg(x - (1 - i) \bigg)} \\$ $\sf \: :\implies \: (x - 1 - i)(x - 1 + i)\\$ $\sf \: :\implies {(x - 1)}^{2} - {i}^{2} \\$ $\sf \: :\implies {x}^{2} + 1 - 2x - ( - 1) \: \: \: \: \: ( \sf \red{\because \: {i}^{2} = - 1)}\\$ $\sf \: :\implies {x}^{2} - 2x + 2\\$ $\sf \:\red{ So\: {x}^{2} - 2x + 2 }\: is \: a \: factor \: of \: given \: equation.\\\\$ ❍On Dividing :- $\sf \: {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 \: by \: {x}^{2} - 2x + 2 \: we \: will \: get \:$ $\sf \: \pink{ {x}^{2} + 4x + 1}$ $\sf \green {[For \: details \: see \: the \:attachment ]}$ $\sf \: {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 = 0\:\\$ $\sf :\implies\: ({x}^{2} - 2x + 2)( {x}^{2} + 4x + 1) = 0\\$ $\sf :\implies \: {x}^{2} + 4x + 1 = 0\\\\$ ❍On comparing it with ax² + bx + c = 0, we have a = 1 b = 4 c = 1 ❍Roots of the equation is given by:- $\sf :\implies\:\sf \: x \: = \: \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$ $\sf :\implies \:x \: = \: \dfrac{ - 4 \: \pm \: \sqrt{ {4}^{2} - 4 \times 1 \times 1} }{2 \times 1}$ $\sf :\implies \:x = \dfrac{ - 4 \: \pm \: \sqrt{12} }{2}$ $\sf :\implies \:x = \dfrac{ - 4 \: \pm \: 2\sqrt{3} }{2}$ $\sf :\implies \purple{ \:x \: = \: - 2 \: \pm \: \sqrt{3} }\\\\$ $\therefore \sf \: The \: roots \: of \: \sf \: {x}^{4} + {2x}^{3} - {5x}^{2} + 6x + 2 = 0\: \: are$ $\sf\purple{\: 1 + i},\red{ \: 1 - i},\green{ \: - 2 + \sqrt{3} } \: and\pink {\: - 2 - \sqrt{3}}$ ⠀⠀⠀ ━━━━━━━━━━━━━━━━━━━⠀,

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$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \large\mathfrak{\dag \: Solution :- }\:\:\:\:\:\:\:\:\:\:\:\:\:\:$