For each of the following polynomial, find p(1), p(0) and p(- 2). i.

1.	For each of the following polynomial, find p(1), p(0) and p(- 2). i. p(x) = x3ii. p(y) = y2 – 2y + 5 iii. p(y) = x4 – 2x2 + x
Answer» i. p(x) = x³ ∴ p(1) = 1³ = 1 p(x) = x³ ∴ p(0) = 0³ = 0 p(x) = x³ ∴ p(- 2) = (- 2)³ = - 8 ii. p(y) = y² – 2y + 5 ∴ p(1) = 1² – 2(1) + 5 = 1 – 2 + 5 ∴ P(1) = 4 p(y) = y² – 2y + 5 ∴ p(0) = 0² – 2(0) + 5 = 0 – 0 + 5 ∴ p(0) = 5 p(y) = y² – 2y + 5 ∴ p(- 2) = (- 2)² – 2(- 2) + 5 = 4 + 4 + 5 ∴ p(- 2) = 13 iii. p(x) = x⁴ – 2x² – x ∴ p(1) = (1)⁴ – 2(1)² – 1 = 1 – 2 – 1 ∴ p(1) = - 2 ∴ p(x) = x⁴ – 2x² – x ∴ p(0) = (0)⁴ – 2(0)² – 0 = 0 – 0 – 0 ∴ p(0) = 0 p(x) = x⁴ – 2x² – x ∴ p(-2) = (-2)⁴– 2(-2)² – (-2) = 16 – 2(4) + 2 = 16 – 8 + 2 ∴ p(-2) = 10

For each of the following polynomial, find p(1), p(0) and p(- 2). i. p(x) = x3ii. p(y) = y2 – 2y + 5 iii. p(y) = x4 – 2x2 + x

Answer»

i. p(x) = x³

∴ p(1) = 1³ = 1

p(x) = x³

∴ p(0) = 0³ = 0

p(x) = x³

∴ p(- 2) = (- 2)³ = - 8

ii. p(y) = y² – 2y + 5

∴ p(1) = 1² – 2(1) + 5

= 1 – 2 + 5

∴ P(1) = 4

p(y) = y² – 2y + 5

∴ p(0) = 0² – 2(0) + 5

= 0 – 0 + 5

∴ p(0) = 5

p(y) = y² – 2y + 5

∴ p(- 2) = (- 2)² – 2(- 2) + 5

= 4 + 4 + 5

∴ p(- 2) = 13

iii. p(x) = x⁴ – 2x² – x

∴ p(1) = (1)⁴ – 2(1)² – 1

= 1 – 2 – 1

∴ p(1) = - 2

∴ p(x) = x⁴ – 2x² – x

∴ p(0) = (0)⁴ – 2(0)² – 0

= 0 – 0 – 0

∴ p(0) = 0

p(x) = x⁴ – 2x² – x

∴ p(-2) = (-2)⁴– 2(-2)² – (-2)

= 16 – 2(4) + 2

= 16 – 8 + 2

∴ p(-2) = 10