Question

if x+1 and x-2 are the factors of ax3+x2-2x+b find the values of a and b​

Answer 1

Answer:

Step-by-step explanation:

let p(x)=ax³+x²-2x+b

(x+1) and (x-2) are the factors of p(x)

so, 2 and -1 are the zeroes

p(-1)=a(-1)³+(-1)²-2*-1+b

     =-a+1+2+b

     =-a+3+b

p(2)=a(2)³+(2)²-2*2+b

     =8a+4-4+b

     =8a+b

p(-1)=0

⇒-a+3+b=0

⇒b=a-3

p(2)=0

⇒8a+b=0

⇒8a+(a-3)=0                        (as b=a-3)

⇒8a+a-3=0

⇒9a=3

⇒a=1/3

now

b=a-3

 =1/3-3

 =(1-9)/3

 =-8/3

Answer 2

if,x=-1

then,p(x)=ax^3+x^2-2x+b

p(-1)=a(-1)^3+(-1)2-2(-1)+b

=-a+1+2+b=0

-a+b=-3

b=-3+a.......(1)

if x=2,

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

=8a+4-4+b=0

8a+b=0......(2)

the 1st value of b is substitute in 2nd

then,8a+(-3+a)=0

8a-3+a=0

9a=3

a=3/9
a=1/3

then,b=-3+(1/3)

b=-8/3