Question

If (x - 2) and (x+3) are the factors of p(x) = ax^3+3x^2-bx-12. Find the value of 'a' and b​

Answer 1

Answer:

a=1,b=4

Step-by-step explanation:

p(x) = ax^3+3x^2-bx-12

x-2=0

x=2

p(2)=0

a(2)^3+3(2)^2-2b-12=0

8a+12-12-2b=0

8a-2b=0

4a-b=0

b=4a.....(1)

x+3=0

x=-3

p(-3)=0

a(-3)^3+3(-3)^2-b(-3)-12=0

-27a+27+3b-12=0

-27a+3b=-15

-9a+b=-5

Put b

-9a+4a=-5

-5a=-5

a=1

So,b=4

Answer 2

Given,

(x-2) and (x+3) factors of p(x) = $ax^{3}+3x^{2}-bx-12$.

To Find,

the value of 'a' and 'b'.

Solution,

p(x) = $ax^{3}+3x^{2}-bx-12$                              (1)

one of the factor is (x-2)

x - 2 =0

x=2

putting x = 2 in equation (1)

p(2)=

$a(2)^{3}+3(2)^{2}-2b-12 = 0\\ \\ 8a+12-12-2b=0\\\\8a-2b=0\\\\4a-b=0\\\\b=4a$                           (2)

now, the other factor is (x+3)

x+3=0

x = -3

p(-3)=

$a(-3)^{3}+3(-3)^{2}-(-3)b-12 = 0\\ \\ -27a+27-12+3b=0\\\\-27a+3b=-15\\\\-9a+b=-5$

now using (2), we get

-9a+4a = -5

-5a = -5

a=1

so, b= 4a

     b= 4(1)

     b=4

Hence the value of 'a=1' and 'b=4'.