✔ 最佳答案
a)Let r be the remainder , then
f(x) = (x + 2) (2x² - x - 10) + r is divisible by (x+3) , so
f(- 3) = (-3 + 2) (2(-3)² - (-3) - 10) + r
0 = - 11 + r
r = 11 b)Not agree.
By part a) , f(x) = (x + 2) (2x² - x - 10) + 11.
Solve (x + 2) (2x² - x - 10) + 11 = 0 :
(x + 3) (2x² - x - 10) - (2x² - x - 10) + 11 = 0
(x + 3) (2x² - x - 10) - (2x² - x - 21) = 0
(x + 3) (2x² - x - 10) - (x + 3) (2x - 7) = 0
(x + 3) (2x² - 3x - 3) = 0
But the roots of 2x² - 3x - 3 are not rational since Δ = 3² - 4(2)(-3) = 33 is not a perfect square. Therefore f(x) = 0 have one rational root = - 3 and two irrational roots.