Question

(X+2)&(x-3)are factors of xcube +ax+be. Find the values of a and b.Also find the remaining factors

Answer 1

Hello Mate!

Polynomial we have is p(x) = x³ + ax + b

Since two of the factors are ( x + 2 ) and ( x - 3 ) so respective value of x = - 2 and 3.

Keeping values if x we get,

p(x) = -2³ + a(-2) + b

0 = - 8 - 2a + b

8 = - 2a + b or - 8 = 2a - b __(1)

p(x) = 3³ + a(3) + b

- 27 = 3a + b __(2)

On adding (1) and (2)

- 35 = 5a

- 7 = a

Keeping value of a in equation (1) we get

- 8 = 2(-7) -  b

- 8 + 14 = - b

- 6 = b

So, p(x) = x³ + ax + b

p(x) = x³ + 0x² - 7x - 6

Please refer to attatchment above!

Hence other factor is ( x + 1 ).

So zeros if polynomial are x = - 2, + 3 and - 1.

Have marvelous future ahead!

Answer 2

here is your answer OK ☺☺☺☺



using factor theorem:
x+1 = 0 or x = -1
x+2 = 0 or x = -2
so, p(-1) = 0
or (-1)3 + 3(-1)2 - 3a (-1) + b = 0
or -1 + 3 + 3a + b = 0
or 3a + b = -2 -----(1)

also p(-2) = 0
or (-2)3 + 3(-2)2 - 3a(-2) + b = 0
-8 + 12 + 6a + b = 0
or 6a + b = -4 --------(2)
(1) * 2 gives 6a + 2b = -4 ---------(3)
(3) - (1) gives
6a + 2b - (6a+b) = -4 - (-4)
6a + 2b -6a -b = -4 + 4
b = 0
putting b = 0 in (2) gives
6a + 0 = -4
a = -4 / 6
a = - 2/3