please solve ques 4 in the attachment
Answers
Comparing the polynomial
Given :
Polynomial = x⁴ - 6 x³ - 26 x² + 138 x - 35
Comparing with a x⁴ + b x³ + c x² + d x + e x , we get :
a = 1
b = -6
c = - 26
d = 138
e = - 35
Calculating the roots
Given :
2 roots are 2 + √3 and 2 - √3
Let other zeroes be p and q
Sum of roots
We know that sum of roots = - b / a
= - ( - 6 ) / 1
= 6
So : p + q + 2 + √3 + 2 - √3 = 6
==> p + q + 4 = 6
==> p + q = 6 - 4
==> p + q = 2 ...........................(1)
Product of roots
We know that product of roots = e / a
[ for even degree [ here 4 ] take last term / first term
for odd take ( - last term / first term ) ]
So : p q ( 2 + √3 )( 2 - √3 ) = e/a
==> p q ( 2² - 3 ) = - 35 [ using ( a + b )( a - b ) = a² - b² ]
==> p q ( 4 - 3 ) = 35
==> p q = 35
==> p = 35/q .......................(2)
Putting this in (1) we get :
35/q + q = 2
==> 35 + q² = 2 q [ Multiplying both sides by q ]
==> q² - 2 q + 35 = 0
==> q² - 7 q + 5 q + 35 = 0
==> q ( q - 7 ) + 5 ( q - 7 ) = 0
==> ( q - 7 )( q + 5 ) = 0
Either
q - 7 = 0
==> q = 7 or ,
q + 5 = 0
==> q = - 5
When q = 7 , p + q = 2
==> p + 7 = 2
==> p = 2 - 7
==> p = -5
Same numbers will come when q = - 5 :
The zeroes are -5 and 7 .
Hope it helps :-)
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Answer:
-5,7
Step-by-step explanation:
Given, polynomial is p(x) = x⁴ - 6x³ - 26x² + 138x - 35.
Let x = 2 ±√3 are the zeroes of p(x).
∴ x - (2 ± √3) are the factors of p(x)
⇒ {x - (2 + √3)}{x - (2 - √3)}
⇒ {(x - 2) - √3)}{(x - 2) + √3)}
⇒ (x - 2)² - (√3)²
⇒ x² + 4 - 4x - 3
⇒ x² - 4x + 1.
∴ Now, Dividing p(x) by x² - 4x + 1, we get
x² - 4x + 1) x⁴ - 6x³ - 26x² + 138x - 35(x² - 2x - 35
x⁴ - 4x³ + x²
-------------------------------------
-2x³ - 27x² + 138x
-2x³ + 8x² - 2x
--------------------------------------
-35x² + 140x - 35
-35x² + 140x - 35
--------------------------------------
0
Now,
we find zeroes of x² - 2x - 35
⇒ x² - 7x + 5x - 35 = 0
⇒ x(x - 7) + 5(x - 7) = 0
⇒ (x + 5)(x - 7) = 0
⇒ x = -5,7.
Therefore, the other zeroes are -5,7.
Hope it helps!