Question

Find the integral zeroes of the polynomial x cube +x square +x-3

Answer 1

Step-by-step explanation:

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Answer 2

The given question is we have to find the integral zeroes of the polynomial

The given polynomial is

${x}^{3} + {x}^{2} + x - 3$

we have to find the integral zeroes of the above expression.

let us substitute x= 1 in the above equation, we get the values as

${1}^{3} + {1}^{2} + 1 - 3 \\ = 3 - 3 \\ = 0$

On substituting the value of x =1, we get the answer as 0.

Therefore, x - 1 is the factor of the given polynomial f(x).

let us find some other factors of this polynomial.

Divide the polynomial by the factor obtained.

$x - 1) {x}^{3} + {x}^{2} + x - 3( {x}^{2} + 2x + 3 \\ \: - {x}^{3} + {x}^{2} \\ - - - - - - - \\ 2 {x}^{2} + x - 3 \\ - 2 {x}^{2} + 2x \\ - - - - - - - - \\ 3x - 3 \\ - 3x + 3 \\ - - - - - - - - \\ 0$

$p(x) = (x - 1)( {x}^{2} + 2x + 3)$

The above are the factors of the given polynomial x.

The new factor

${x}^{2} + 2x + 3 doesnot \: give \: the \: integral \: zeroes$

so, integral zeroes are only one.

Hence, therefore the factors for the given polynomial was found

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