It two zeroes of the polynomial x4 - 6x3 - 26x2 - 138x - 35 are 2-3- find other zeroes-

Thus, x^4 6x^3 26x^2+138x 35=(x (2+√(3)))(x (2 √(3)))(x α)(x β) RHS: ⇒ ((x 2)^2 3)(x α)(x β) =(x^2 4x+1)(x α)(x β) Divide the initial polynomial by ...

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Standard XII

Mathematics

Definition of Functions

It two zeroes...

Question

It two zeroes of the polynomial $x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35$ are $2 \pm \sqrt{3}$ , find other zeroes.

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Solution

Thus,
$x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35 = (x - (2 + \sqrt{3})) (x - (2 - \sqrt{3})) (x - α) (x - β)$

RHS:

$\Rightarrow ((x - 2)^{2} - 3) (x - α) (x - β)$

$= (x^{2} - 4 x + 1) (x - α) (x - β)$

Divide the initial polynomial by the first term of RHS:
$(x - α) (x - β) = x^{2} - 2 x - 35 = (x - 7) (x + 5)$

Thus, $α = 7, β = - 5$

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Similar questions

If two zeros of the polynomial f(x) = $x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35$ are $2 \pm \sqrt{3},$ find other zeros.

Q. If two zeros of the polynomial

f (x) = x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35

are

2 \pm \sqrt{3}

, find ohter zeros. [4 MARKS]

Q. If two zeroes of the polynomial

x^{2} - 6 x^{2} - 26 x^{2} + 138 x - 35 a r e 2 x \sqrt{3}

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. Find the remaining zeros of

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It two zeroes of the polynomial x4−6x3−26x2+138x−35 are 2±√3, find other zeroes.

Thus,x4−6x3−26x2+138x−35=(x−(2+√3))(x−(2−√3))(x−α)(x−β)RHS:⇒((x−2)2−3)(x−α)(x−β)=(x2−4x+1)(x−α)(x−β)Divide the initial polynomial by the first term of RHS:(x−α)(x−β)=x2−2x−35=(x−7)(x+5)Thus, α=7,β=−5

It two zeroes of the polynomial $x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35$ are $2 \pm \sqrt{3}$ , find other zeroes.