Question

3.Show that p(x)=x^3-3x^2+2x-6 has only one real zcro. pls explain will mark as brainliest​

Answer 1

Step-by-step explanation:

p(x)=x^3-3x^2+2x-6

By using trial and error method:

Let us put x = 1, 2, 3, 4, 5 and so on...

We find that:

$p(1) = {(1)}^{3} - 3 {(1)}^{2} + 2 \times 1 - 6 \\ = 1 - 3 + 2 - 6 = - 6 \: which \: is \\ \: not \: equal \: to \: zero. \: hence \: 1 \: is \: no \: a \: zero \: of \: the \: given \: polynomial \\ now \: put \: x = 2 \\ we \: find \\ p(2) = {(2)}^{3} - 3 {(2)}^{2} + 2 \times 2 - 6 \\ = 8 - 12 + 4 - 6 = - 6 \: which \: is \: \\ again \: not \: equal \: to \: zero \\hence \: 2 \: is \: also \: not \: a \: zero \: of \\ \: the \: polynomial. \\ next \: we \: put \: x = 3 \\ p(3) = {(3)}^{3} - 3 {(3)}^{2} + 2 \times 3 - 6 \\ = 27 - 27 + 6 - 6 = 0 \\ p(3) = 0 \\ hence \: 3 \: is \: a \: zero \: of \: given \\ \: polynmial \\ \: next \: if \: you \: check \: by \: putting \: \\ any \: other \: value \: for \: x \: it \: will \: not \\ \: give \: zero. \: \\ thus \: 3 \: is \: the \: only \: zero \: of \: given \: \\ polynomial \\ therefore \: given \: polynomial \: has \\ only \: one \: zero.$