If two zeroes of the polynomial
[tex] {x}^{4} - 6 {x}^{3} - 26 {x}^{2} + 138x - 35[/tex]
are
[tex] 2 + - \sqrt{3} [/tex]
find the other zeroes.
Plz give in full details
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If two zeroes of the polynomial
[tex] {x}^{4} - 6 {x}^{3} - 26 {x}^{2} + 138x - 35[/tex]
are
[tex] 2 + - \sqrt{3} [/tex]
find the other zeroes.
Plz give in full details
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Given :-
To Find :-
Concept used :-
Solution :-
→ sum of Roots = (2 + √3) + (2 - √3) = 4
→ Product of Roots = (2 + √3)(2 - √3) = 2² - (√3)² = 4 - 3 = 1
Hence,
→ Quadratic Equation is = x² - 4x + 1 = 0
_______________
Now,
Divide :-
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, Splitting The Quotient we get,
→ x² - 2x - 35 = 0
→ x² - 7x + 5x - 35 = 0
→ x(x - 7) + 5(x - 7) = 0
→ (x - 7)(x + 5) = 0
→ x = 7 & (-5). (Ans.)
Hence, Other zeros of given Polynomial are 7 & (-5).
Verified answer
Correct Question :
If two zeros of the polynomial 'x⁴ - 6x³ - 26x² + 138x - 35' are 2 ± √3 .Find the other zeros.
Solution :
According to remainder theorem, if two zeros of a polynomial are given then, (x - a)(x - b) will completely divide the polynomial and the remainder will be = 0 & quotient = 0
Now, finding divisor :
➝ [x + (- 2 - √3)][(x + (- 2 + √3)] = Divisor
➝ x² + (- 2 - √3 - 2 + √3)x + (-2 - √3) × (-2 + √3) = Divisor
➝ x² + (-4)x + (-2)² - (√3)² = Divisor
➝ x² - 4x + 4 - 3 = Divisor
➝ x² - 4x + 1 = Divisor
Now dividing the polynomial by (x² - 4x + 1)
x² - 4x + 1)x⁴ - 6x³ - 26x² + 138x - 35(x²-2x - 35
x⁴ - 4x³ + x²
(-) (+) (-)
-2x³ - 27x² + 138x - 35
-2x³ + 8x² - 2x
(+) (-) (+)
-35x² + 140x - 35
-35x² + 140x - 35
(+) (-) (-)
0
So, divisor = x² - 2x - 35 .
Therefore, x² - 2x - 35 = 0
➻ x² - 7x + 5x - 35 = 0
➻ x(x - 7) + 5(x - 7) = 0
➻ (x + 5)(x - 7) = 0
➻ x = -5 or, x = 7
➥ Other two zeros = -5 & 7
Therefore,
All the zeros of polynomial are (2 + √3),(2 - √3), -5 & 7 .