Question

10. If 4 is the zero of the cubic polynomial p(x)=x²-3x² - 6x + 8 then find the other zeroes of p(x)​

Answer 1

$\boxed{\colorbox{Answer}}$

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.