Math, asked by reyhanaanu, 1 year ago

find all the zeros of the polynomial 2x4-8x3+5x2+4x-3,whose two zeroes are 1/root2 and - 1/root 2

Answers

Answered by gudia5000
1

Step-by-step explanation:

given two zeroes of p(x)=1/root2 and -1/root2

By factor theorem

(x-1/root2) (x+1/root2)=0

using identity (a-b) (a+b) = a^2-b^2

x^2-(1/root2)^2=0

x^2-1/2=0

2x^2/2-1/2=0

2x-1 =0

by long division method

2x^4-8x^3+5x^2+ 4x-3÷2x^2-1

q(x)=x^2-4x+3

for other 2 zeroes let p(x) =0

therefore, x^2-4x+3=0

by splitting middle terms

x^2-(3+1)x+3=0

x^2-3x-x+3=0

x(x-3)-1(x-3)=0

(x-1) (x-3) =0

either x-1=0 or x-3=0

x=1 or x= 3

therefore, all the zeros of p(x) are 1,3,1/root2 and -1 /root2

Similar questions