If two zeros of the polynomial (x4-6x³-26x²+138x-35) are (2+√3) &(2-√3). Find the other zeros....
Answers
SOLUTION:-
Given,
p(x)= 2±√3.
and,
x- (2±√3) are the factors of p(x).
Now,
{x- (2+√3)}{x-(2-√3)}
=) {(x-2) -√3}{(x-2)+√3}
=) (x-2)² - (√3)²
=) x² -4x +1
Therefore,
Dividing p(x) by x² - 4x +1.
For this polynomial;
(x⁴ -6x³ -26x²+138x -35)
Above attachment divide.
Therefore,
p(x)= {x-(2+√3)} {x-(2-√3)} (x² -2x -35)
So,
x² - 2x -35
=) x² +5x -7x -35 =0
=) x(x+5) -7(x+5)=0
=) (x+5) (x-7) =0
=) x+5 =0 or x-7=0
=) x= -5 or x= 7
Hence,
Other two zeroes of p(x) are -5 & 7.
Hope it helps ☺️
Answer:
∴7 and −5 are zeros of polynomial
Step-by-step explanation:
Answer
P(n)=x
4
−6x
3
−26x
2
+138−35
as (2+
3
) and (2−
3
) are joas of polynomial of P(a)
So, we divid (R−(2+
3
))(x−(2
3
)) then find quotient
x
2
−2x−
3
x−2x−
3
x+1=x
2
−4x+1
x
2
+4x+1)
x
4
−6x
3
−26x
2
+138x−35
(n
2
−2x−35
x
4
−4x
3
+x
2
___________________________
−2x
3
−27x
2
+138x
−2x
3
+8x
2
−2a
_____________________________
−35x
2
+140x−35
−35x
2
+140x−35
____________________________
0
x
2
−2x−35=x
2
−7x+5x−35=x(x−7)+5(x−7)=0
x=−5,7
∴7 and −5 are zeros of polynomial