How do you use synthetic division to show that x=1+sqrt3x=1+√3 is a zero of x^3-3x^2+2=0x3−3x2+2=0?

1 Answer
maganbhai P.
Aug 1, 2018

Remainder =0=>x=1+sqrt3 =0⇒x=1+√3 is a zero of (x^3-3x^2+2)(x3−3x2+2)

Explanation:

Using synthetic division :

diamond(x^3-3x^2+2)div(x-1-sqrt3)⋄(x3−3x2+2)÷(x−1−√3)

We have , p(x)=x^3-3x^2+0x+2 and "divisor :"x=1+sqrt3p(x)=x3−3x2+0x+2anddivisor :x=1+√3

We take ,coefficients of p(x) to 1,-3,0,2p(x)→1,−3,0,2

1+sqrt3 |1+√3∣∣ 1color(white)(...........)-3color(white)(...........)0color(white)(...............)2
ulcolor(white)(..........)| ul(0color(white)( .......)1+sqrt3color(white)(.......)1-sqrt3color(white)(......)-2
color(white)(.............)1color(white)(...)-2+sqrt3color(white)(.....)1-sqrt3color(white)(........)color(violet)(ul|0|
We can see that , quotient polynomial :

q(x)=x^2+(-2+sqrt3)x+1-sqrt3 and"the Remainder"=0

Hence ,

(x^3-3x^2+2)=(x-1-sqrt3)[x^2+(-2+sqrt3)x+1-sqrt3]

Remainder =0=>x=1+sqrt3 is a zero of (x^3-3x^2+2)

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