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Ray Of Light [21]
3 years ago
8

Obtain the zeroes of polynomial

Physics
1 answer:
morpeh [17]3 years ago
3 0

I assume you meant to say

f(x)=2x^4+3x^3-5x^2-9x-3

Given that <em>x</em> = √3 and <em>x</em> = -√3 are roots of <em>f(x)</em>, this means that both <em>x</em> - √3 and <em>x</em> + √3, and hence their product <em>x</em> ² - 3, divides <em>f(x)</em> exactly and leaves no remainder.

Carry out the division:

\dfrac{2x^4+3x^3-5x^2-9x-3}{x^2-3} = 2x^2+3x+1

To compute the quotient:

* 2<em>x</em> ⁴ = 2<em>x</em> ² • <em>x</em> ², and 2<em>x</em> ² (<em>x</em> ² - 3) = 2<em>x</em> ⁴ - 6<em>x</em> ²

Subtract this from the numerator to get a first remainder of

(2<em>x</em> ⁴ + 3<em>x</em> ³ - 5<em>x</em> ² - 9<em>x</em> - 3) - (2<em>x</em> ⁴ - 6<em>x</em> ²) = 3<em>x</em> ³ + <em>x</em> ² - 9<em>x</em> - 3

* 3<em>x</em> ³ = 3<em>x</em> • <em>x</em> ², and 3<em>x</em> (<em>x</em> ² - 3) = 3<em>x</em> ³ - 9<em>x</em>

Subtract this from the remainder to get a new remainder of

(3<em>x</em> ³ + <em>x</em> ² - 9<em>x</em> - 3) - (3<em>x</em> ³ - 9<em>x</em>) = <em>x</em> ² - 3

This last remainder is exactly divisible by <em>x</em> ² - 3, so we're left with 1. Putting everything together gives us the quotient,

2<em>x </em>² + 3<em>x</em> + 1

Factoring this result is easy:

2<em>x</em> ² + 3<em>x</em> + 1 = (2<em>x</em> + 1) (<em>x</em> + 1)

which has roots at <em>x</em> = -1/2 and <em>x</em> = -1, and these re the remaining zeroes of <em>f(x)</em>.

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