Question

if 2 zeroes of the polynomial p(x) =x^4-6x^3-26x^2 +138x-35are 2+√3,2-√3, find the other zetoes
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Answer 1

Given: roots 2 + $\sqrt{3}$ and 2 - $\sqrt{3}$

To find: other two zeroes

Step-by-step explanation:

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

Since x = 2 + $\sqrt{3}$ is a zero,

  So, x- 2 - $\sqrt{3}$

Since x = 2 - $\sqrt{3}$ is a zero,

  So, x- 2 + $\sqrt{3}$

⇒ (x- 2 - $\sqrt{3}$ ) × (x- 2 + $\sqrt{3}$ ) is a factor too

⇒ [(x-2) - $\sqrt{3}$ ] × [(x-2) + $\sqrt{3}$ ]

⇒ $(x-2)^{2}$ - $(\sqrt{3} )^{2}$

⇒ $x^{2} + 2^{2} - 4x - 3$

⇒ $x^{2} - 4x + 1$

$x^{2} - 4x + 1$ is a factor of p(x)

Now by dividing p(x) by $x^{2} - 4x + 1$,

= $\frac{x^{4} + 6x^{3} - 26x^{2} + 138x - 35 }{x^{2} - 4x + 1}$

= $x^{2} - 2x - 35$

= $x^{2} - 7x + 5x - 35 = 0$

= $x (x-7) + 5 (x-7) = 0$

= $(x+5) (x-7) = 0$

  $x = -5$  and $x = 7$ are the other two zeroes

Answer 2

Given :- if 2 zeroes of the polynomial p(x) =x^4-6x^3-26x^2 +138x-35are 2+√3,2-√3, find the other zeroes ?

Answer :-

since , 2+√3, 2-√3 are the 2 zeroes of the polynomial p(x) .

so,{(x - (2 + √3)} * {(x - (2 - √3)} will be a factor of given polynomial p(x)

→ {(x - (2 + √3)} * {(x - (2 - √3)}

→ (x - 2 - √3) * (x - 2 + √3)

→ {(x - 2) - √3} * {(x - 2) + √3}

using (a - b)(a + b) = a² - b²

→ (x - 2)² - (√3)²

→ x² - 4x + 4 - 3

→ x² - 4x + 1

dividing the given polynomial by (x² - 4x + 1) we get,

x²-4x+1 ) x⁴ - 6x³ - 26x² +138x - 35 ( x² - 2x - 35

x² - 4x³ + x²

-2x³ - 27x² + 138x

-2x³ + 8x² - 2x

-35x² + 140x - 35

-35x² + 140x - 35

0

then,

→ Quotient = x² - 2x - 35

factorising the quotient now,

→ x² - 2x - 35 = 0

→ x² - 7x + 5x - 35 = 0

→ x(x - 7) + 5(x - 7) = 0

→ (x - 7)(x + 5) = 0

→ x = 7 and (-5)

therefore, the other zeroes of the polynomial are 7 and (-5) .

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