Question

Find all the zeroes of the polynomial x^4-3x^3+6x-4 , if two of its zeroes are root 2 and -root 2

Answer 1

ANSWER:

given

$p(x) = {x}^{4} - 3 {x}^{3} + 6x - 4$

two of the zeroes are

√2,-√2

REQUIRED TO FIND;

other two zeroes of the polynomial

METHOD:

in the attachment

→first multiply (x-√2)(x+√2) and let it be g(x)

→Now divide the given polynomial by g(x)

→Then we get a quotient,we need to factorise that to get to get other two zeroes

IDENTITIES USED:

$> (x + y)(x - y) = {x}^{2} - {y}^{2}$

Answer 2

$\bf Question:-\\ \tt Find\ all\ the\ zeroes\ of\ the\ polynomial\ x^4-3x^3+6x-4=0\\ \tt if\ two\ of\ its\ zeroes\ are\ \sqrt2\ and\ -\sqrt2.$

Step-by-step explanation:

$\rm Given,\ two\ zeroes\ are\ \sqrt2\ and\ -\sqrt2.\\ \rm So,\ these\ can\ also\ be\ written\ as\ (x-\sqrt2)\ and\ (x+\sqrt2)\ respectively.\\ \rm Now\ multiplying\ both.\ We\ get\ (x^2-2)\qquad \{(a+b)(a-b)=a^2-b^2\}\\ \rm Now,\ divide\ x^4-3x^3+6x-4\ by\ x^2-2\qquad (the\ division\ is\ in\ the\ attachment)\\ \rm After\ division\ you'll\ get\ this\ equation\\ x^2-3x+2\\ \rm We've\ to\ work\ on\ this\ equation\ by\ factorising\ it\\ x^2-3x+2=0\\ \implies \rm x^2-3x+2=0\\ \implies \rm x(x-2) -1(x-2)=0\\ \implies \rm (x-1) (x-2)=0\\ \implies \rm \therefore x=1\ and\ 2.\\ \therefore \bf The\ two\ zeroes\ of\ the\ polynomial\ are\ 1\ and\ 2.$