MAT1511
ASSIGNMENT 1
2025
, QUESTION 1
a).
P(x) = −x 3 + 7x + 6 ∴ factor x + 2 ⇒ x = −2
P(−2) = −(−2)3 + 7(−2) + 6
P(−2) = 8 − 14 + 6
P(−2) = 0
By the factor theorem: x + 2 is a factor of P(x)
b).
P(x) = −x 3 + 7x + 6
P(x) = −x 3 + 0x 2 + 7x + 6
By sythetic division
−2 −1 0 7 6
↓ 2 −4 −6
−1 2 3 0
P(x) = (x + 2)(−x 2 + 2x + 3)
P(x) = −(x + 2)(x 2 − 2x − 3)
P(x) = −(x + 2)(x + 1)(x − 3)
Solution: x = −2, x = −1 and x = 4
ASSIGNMENT 1
2025
, QUESTION 1
a).
P(x) = −x 3 + 7x + 6 ∴ factor x + 2 ⇒ x = −2
P(−2) = −(−2)3 + 7(−2) + 6
P(−2) = 8 − 14 + 6
P(−2) = 0
By the factor theorem: x + 2 is a factor of P(x)
b).
P(x) = −x 3 + 7x + 6
P(x) = −x 3 + 0x 2 + 7x + 6
By sythetic division
−2 −1 0 7 6
↓ 2 −4 −6
−1 2 3 0
P(x) = (x + 2)(−x 2 + 2x + 3)
P(x) = −(x + 2)(x 2 − 2x − 3)
P(x) = −(x + 2)(x + 1)(x − 3)
Solution: x = −2, x = −1 and x = 4
