Question

Find the value of the polynomial 2x^3-3x^2+4x-1 at :
(1) x=0
(2) x=1
(3) x=2​

Answer 1

Answer:

1. x = 0

${2x}^{3} - {3x}^{2} + 4x + 1$

$2 {(0)}^{3} - 3 {(0)}^{2} + 4(0) + 1$

$0 - 0 + 0 + 1$

$1$

Ans. 1

2. x = 1

${2x}^{3} - {3x}^{2} + 4x + 1$

$2 {(1)}^{3} - 3 {(1)}^{2} + 4(1) + 1$

$1 - 3 + 4 + 1$

$- 2 + 4 + 1$

$2 + 1$

$3$

Ans. 3

3. x = 2

${2x}^{3} - {3x}^{2} + 4x + 1$

$2 {(2)}^{3} - 3 {(2)}^{2} + 4(2) + 1$

$16 - 12 + 8 + 1$

$4 + 8 + 1$

$12 + 1$

$13$

Ans. 13

Hope it helps.

Answer 2

Step-by-step explanation:

2x^3-3x^2+4x-1

(1) x=0

0-0+-1= -1

(2) x=1

2x^3=8

3x^2=6

4x-1=3

2x^3-3x^2+4x-1=8-9+3=11-9=2

(3) x=2

2x^3=64

3x^2=36

4x-1=7

2x^3-3x^2+4x-1=64-36+7=71-36=35

hope you got it

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