Question

please solve this ques

Answer 1

Hey mate here is your answer,
Sol. Since 2±3–√, are two zeroes of the polynomial p(x)=x4−6x3−26x2+138x−35
Let x=2±3–√ ⇒ x−2=±3–√
Squaring we get x2−4x+4=3
⇒ x2−4x+1=0
Let us divide p(x) by x2−4x+1 to obtain other zeroes.

15

Therefore, p(x)=x4−6x3−26x2+138x−35
=(x2−4x+1)(x2−2x−35)
=(x2−4x+1)(x2−7x+5x−35)
=(x2−4x+1)[x(x−7)+5(x−7)]
=(x2−4x+1)(x+5)(x−7)
⇒ (x + 5) and (x – 7) are other factors of p(x).
Therefore, – 5 and 7 are other zeroes of the given polynomial.
HOPE IT IS HELP YOU

Answer 2

#RAM RAM ji ❤^_^


ĀNSWĒR ⏬⏬

other zeroes are ▶-5,7


THANKS ✌☺


#NAVI ❤ HARYANVI ♠