Obtain all other zeroes of 3x4+6x3−2x2−10x−5, if two of its zeroes are √53 and −√53.
Two zeroes are √53 and −√53
So we can write it as, x = √53 and x = −√53
we get x−√53=0 and x+√53=0
Multiply both the factors we get,
x2−53=0
Multiply by 3 we get
3x2−5=0 is the factor of 3x4+6x3−2x2−10x−5
Now divide, 3x4+6x3−2x2−10x−5 by 3x2−5=0 we get,
Quotient is x2+2x+1=0
Compare the equation with quadratic formula,
x2−(Sum of root)x+(Product of root)=0
⇒Sum of root =2
⇒Product of the root =1
So, we get⇒x2+x+x+1=0
⇒x(x+1)+1(x+1)=0
⇒x+1=0,x+1=0
⇒x=−1,x=−1
So, our zeroes are −1,−1, √53 and −√53