Question

Show that -1 and 3 are zero of x3-3x2-x+3

Answer 1

Step-by-step explanation:

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Answer 2

$\sf \blue{Answer★}$

If the polynomial p(x)=x^3−3x^2+2^x−6

If the question is, "show that p(x) = x3 – 3x2 + 2x – 6 has only one real zero".

Then the solution is as follows :

Given polynomial is:

$p(x) = {x}^{3} – {3x}^{2} + 2x – 6$

Put x = 3, we get

$p(3) = {3}^{3} - 3( {3})^{2} + 2 \times 3 - 6 \: = 0$

⇒ x = 3 is a zero of p(x).

∴ (x – 3) must be a factor of p(x).

So, we can write,

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

To find another zeros, we must have

${x}^{2} + 2 = 0 \: \: \: \: \: \: ..........(1)$

From (1), we don't get any real value of x.

Hence, the only real zero of p(x) is x = 3.