Question

if p(x) = x^3-4x^2+5x-6 find, p(0), p(-1), p(2)​

Answer 1

Answer:

The value of x³ - 4x² + 5x - 6 at x = 0 is - 6

The value of x³ - 4x² + 5x - 6 at x = - 1 is - 16

The value of x³ - 4x² + 5x - 6 at x = 2 is - 4

Step-by-step explanation:

The value of x³ - 4x² + 5x - 6 at x = 0 is - 6 :

$p(x) = {x}^{3} - 4 {x}^{2} + 5x - 6$

$p(0) = ( {0})^{3} - 4( {0})^{2} + 5(0) - 6$

$p(0) = 0 - 4(0) + 0 - 6 \\ = 0 - 0 + 0 - 6$

$p(0) = - 6$

The value of x³ - 4x² + 5x - 6 at x = - 1 is - 16:

$p(x) = {x}^{3} - 4 {x}^{2} + 5x - 6$

$p( - 1) = ( { - 1})^{3} - 4( { - 1})^{2} + 5( - 1) - 6$

$p( - 1) = - 1 - 4(1) - 5 - 6 \\ = - 1 - 4 - 5 - 6$

$p( - 1) = - 16$

The value of x³ - 4x² + 5x - 6 at x = 0 is - 4:

$p(x) = {x}^{3} - 4 {x}^{2} + 5x - 6$

$p(2) = ( {2})^{3} - 4( {2})^{2} + 5(2) - 6$

$p(2) = 8 - 4(4) + 10 - 6 \\ = 8 - 16 + 10 - 6$

$p(2) = - 4$

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Answer 2

Step-by-step explanation:

it was good

r u from Bihar..