Question

4 is a zero of the cubic polynomial p(x)=x³–3x²-6x+8. Find the remaining zeros of p(x).

Answer 1

Given p(x) = x^3 - 3x^2 - 6x + 8.

Given x = 4 is a zero of p(x).

= > x - 4 will be a factor of p(x).

Divide p(x) by x - 4, we get

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x - 4) x^3 - 3x^2 - 6x + 8 ( x^2 + x - 2

x^3 - 4x^2

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x^2 - 6x + 8

x^2 - 4x

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-2x + 8

-2x + 8

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0

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Now,

= > p(x) = (x - 4)(x^2 + x - 2)

= > (x - 4)(x^2 - x + 2x - 2)

= > (x - 4)(x(x - 1) + 2(x - 1))

= > (x - 4)(x + 2)(x - 1).

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Therefore, the remaining zeroes are : -2,1.

Hope this helps!

Answer 2

Hi ,

Let p( x ) = x³ - 3x² - 6x + 8

It is given that 4 is the zero of p( x ).

( x - 4 ) is a factor of p( x ).

x-4) x³ - 3x² - 6x + 8 ( x²+x-2
*****x³ -4x²
_______________
********* x² - 6x
********* x² - 4x
_______________
************* - 2x + 8
************* - 2x + 8
________________
Remainder = 0

p( x ) = ( x - 4 )( x² + x - 2 )

= ( x - 4 )[ x² + 2x - x - 2 ]

= ( x - 4 )[ x( x + 2 ) - 1( x + 2 ) ]

= ( x - 4 )( x + 2 ) ( x - 1 )

To find other zeroes ,

Take p( x ) = 0 ,


x + 2 = 0 or x - 1 = 0

x = -2 or x = 1

I hope this helps you.

: )