Question

If two zeroes are 2+3 and 2-3 of the p(x) = 2x4-9x3+5x2+3x-1 find other two zeroes

Answer 1

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Answer 2

Answer:

All zeroes of p ( x ) are 1 , - 1 / 2 , 2 + √ 3 and 2 - √ 3

Step-by-step explanation:

Given :

p ( x ) = 2 x⁴ - 9 x³ + 5 x² + 3 x - 1

Two zeroes of p ( x ) :

2 + √ 3 and 2 - √ 3

p ( x ) = ( x - 2 - √ 3 ) ( x - 2 + √ 3 ) × g ( x )

g ( x ) = p ( x ) ÷ ( x² - 4 x + 1 )

On dividing ( 2 x⁴ - 9 x³ + 5 x² + 3 x - 1 ) by ( x² - 4 x + 1 )

$\displaystyle{\begin{tabular}{c | c | c}(x^2-4x+1) & \overline{2x^4-9x^3+5x^2+3x-1}} & 2x^2-x-1\\ & 2x^4-8x^3+2x^2 \ \ \ \ \ \ \ \ \ \ \ \\ \\ & ( - ) ( + ) ( - ) \ \ \ \ \ \ \ \\ \\ & \overline{-x^3+3x^2+3x} \\ \\ & - x^3+4x^2-x\\ \\ & ( + ) ( - ) ( + ) \\ \\ & \overline{-x^2 + 4x-1} \\ \\ & -x^2 + 4x-1 \\ \\ & ( + ) ( - ) ( + ) \\ \\ & \overline{0}\endtabular$

We got ,

g ( x ) = 2 x² - x - 1

By splitting mid term

2 x² - 2 x + x - 1

2 x ( x - 1 ) + ( x - 1 )

( x - 1 ) ( 2 x + 1 )

Now zeroes of g ( x )

x = 1 or x = - 1 / 2 .