Question

3. Find p(o), p(2) and p(3) for the polynomial :
p(x)=2x3 - 13x2 +17x + 12.
4. Verify that 1, 2 and 3 are zeroes of the
polynomial y3 - 6y2 +11y -6.​

Answer 1

Step-by-step explanation:

$p(x) = 2 {x}^{3} - 13 {x}^{2} + 17x + 12 \\ p(0) = 0 - 0 - 0 + 12 = 12 \\ p(2) = 2 \times 8 - 13 \times 4 + 17 \times 2 + 12 \ \\ = 16 - 52 + 34 + 12 \\ = 62 - 52 = 10 \\ p(3) = 2 \times 27 - 13 \times 9 + 17 \times 3 + 12 \\ = 54 - 117 + 51 + 12 \\ = 117 - 117 = 0$

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${y}^{3} - 6 {y}^{2} + 11y - 6 = p(x) \\ p(1) = 1 - 6 + 11 - 6 \\ = 0 \\ p(2) = 8 - 24 + 22 - 6 \\ = 0 \\ p(3) = 27 - 54 + 33 - 6 \\ = 60 - 60 = 0 \\ \\ \: thank \: u \: this \: is \: corrcet \: answer \: .... \: \: i \: hope \: ths \: help \: u \: plz \: thank \: and \: \: follow \: me \: \: thank \: u \: my \: dear \: friend$